Trigonometric Ratios of Multiple Angles

First - Part
Formula of Multiple Angles

\begin{align} (1) & \sin 2A = 2\sin A \cos A \\ (2) & \sin 2A = \frac{2\tan A }{1+\tan^2 A } \\ (3) & \sin 2A = \frac{2\cot A }{1+\cot^2 A } \\ (4) & \cos 2 A = \cos^2 A -\sin^2 A \\ (5) & \cos 2 A = 2\cos^2 A - 1 \\ (6) & \cos 2 A =1 - 2\sin^2 A \\ (7) & \cos 2 A = \frac{1-\tan^2 A}{1+\tan^2 A} \\ (8) & \cos 2 A =\frac{\cos^2 A -1}{\cot^2 A + 1} \\ (9) & \tan 2A = \frac{2\tan A }{1-\tan^2 A}\\ (10) & \cot 2 A = \frac{\cot^2 A -1}{2\cot A } \\ (11) & \sin 3A = 3\sin A - 4\sin^3 A\\ (12) & \cos 3A = 4\cos^3 A - 3\cos A \\ (13) & \tan 3A = \frac{3\tan A - \tan^3 A}{1- 3\tan^2 A} \\ (14) & \cot 3A = \frac{3\cot A - \cot^3 A }{1-3\cot^2 A} \end{align}

1. (a) Write down the formula of \( \sin 2 A \) in terms of \(\tan A \).

\begin{align} \sin 2 A = \frac{2\tan A }{1-\tan^2 A } \end{align}

1. (b) Write down the formula of \( \sin 2\theta \) in terms of \( \cot \theta \).

\begin{align} \sin 2\theta = \frac{ 2\cot \theta }{1+\cot^2 \theta } \end{align}

1. (c) Write down the formula of \( cos 2 A \) in terms of :

(i) \(\sin A\) (ii) \( \cos A \) (iii) \(\tan A\) (iv) \(\cot A\)

\begin{align} & \text{(i)}\ \cos 2A = 1 - 2\sin^2 A \\ & \text{(ii)} \cos 2A = 2\cos^2 A - 1 \\ & \text{(iii)} \cos 2A = \frac{1-\tan^2 A }{1+\tan^2 A } \\ & \text{(iv)} \cos 2A = \frac{ \cot^2 A -1}{\cot^2 A + 1 } \end{align}

2. (a) Express \( \cos 2\theta \) in terms of \( \cos \theta \).

\begin{align} \cos 2\theta & = 2\cos^2 \theta - 1 \end{align}

2. (b) Express \( \cos 2\theta \) in terms of \( \sin \theta \).

\begin{align} \cos 2\theta = 1 - 2\sin^2 \theta \end{align}

2. (c) Express \( \cos 4\theta \) in terms of \( \sin \theta \).

\begin{align} \cos 4\theta & = 1-2\sin^2 2\theta\\ & = 1 - 2( \sin 2\theta )^2 \\ & = 1 -2(2\sin \theta \cos \theta )^2 \\ & = 1 - 2\times 4 \sin^2 \theta \cos^2 \theta \\ & = 1 - 8 \sin^2 \theta ( 1 -\sin^2 \theta ) \\ & = 1 - 8\sin^2 \theta + 8\sin^4 \theta \end{align}

2. (d) Express \( \cos 4\theta \) in terms of \( \cos \theta \).

\begin{align} \cos 4\theta & = 2\cos^2 2\theta - 1\\ & = 2 ( \cos 2\theta )^2 - 1 \\ & = 2 ( 2\cos^2 \theta - 1 )^2 - 1 \\ & = 2\{(2\cos^2 \theta )^2 - 2\times 2\cos^2 \theta \times 1 + 1^2 \} - 1 \\ & = 2\{4\cos^4 \theta - 4\cos^2 \theta + 1 \} - 1 \\ & = 8 \cos^4 \theta - 8 \cos^2 \theta + 2 - 1\\ & = 8 \cos^4 \theta - 8 \cos^2 \theta + 1 \end{align}

2. (e) Express \( \cos 5\theta \) in terms of \( \cos \theta \).

\begin{align} \cos 5\theta & = \cos(3\theta + 2\theta ) \\ & = \cos 3\theta \cos 2\theta - \sin 3\theta \sin 2\theta \\ & = ( 4\cos^3 \theta - 3\cos \theta ) \cos 2\theta - ( 3\sin \theta - 4\sin^3 \theta ) \sin 2\theta \\ & = 4\cos^3 \theta \cos 2\theta - 3\cos \theta \cos 2\theta - 3\sin \theta \sin 2\theta + 4\sin^3 \theta \sin 2\theta \\ & = 4\cos^3 \theta ( 2\cos^2 \theta - 1 ) - 3\cos \theta ( 2\cos^2 \theta - 1) - 3\sin \theta (2\sin \theta \cos \theta ) + 4\sin^3 \theta ( 2\sin \theta \cos \theta ) \\ & = 8 \cos^5 \theta - 4\cos^3 \theta - 6 \cos^3\theta + 3\cos \theta - 6\cos \theta \sin^2 \theta + 8\cos \theta ( \sin^2 \theta )^2 \\ & = 8 \cos ^5 \theta - 10 \cos^3 \theta +3\cos \theta - 6\cos \theta ( 1-\cos^2 \theta ) + 8 \cos \theta ( 1-\cos^2 \theta )^2 \\ & = 8 \cos ^5 \theta - 10 \cos^3 \theta +3\cos \theta - 6\cos \theta + 6 \cos^3 \theta + 8 \cos \theta ( 1 - 2\cos^2 \theta + \cos^4 \theta ) \\ & = 8 \cos^5 \theta - 4 \cos^3 \theta - 3 \cos \theta + 8 \cos \theta - 16 \cos^3 \theta + 8 \cos^5 \theta \\ & = 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta \end{align}

3. (a) Find the value of \( \sin 2A, \cos 2A, \tan 2A \text{ and } \cot 2A \) when

(i) \( \sin A = \frac{4}{5} \) (ii) \( \cos A = \frac{4}{5} \) (iii) \( \tan A = \frac{3}{4} \) (iv) \(\cot A = \frac{12}{13} \)

\begin{align} \text{(i) Given },\sin A & = \frac{4}{5} \\ \cos A & = \sqrt{1-\sin^2 A}\\ & = \sqrt{1-\left(\frac{4}{5} \right)^2}\\ & = \sqrt{1-\frac{16}{25} }\\ & = \sqrt{\frac{25-16}{25} } \\ & = \sqrt{\frac{9}{25} } \\ & = \frac{3}{5} \\ \text{Now} \\ \sin 2A & = 2\sin A \cos A \\ & = 2\times \frac{4}{5} \times \frac{3}{5} \\ & = \frac{24}{25} \\ \therefore \sin 2A & = \frac{24}{25} \end{align} \begin{align} \cos 2A & = 1 - 2\sin^2 A \\ & = 1 -2 \left( \frac{4}{5} \right)^2 \\ & = 1 - 2\left( \frac{16}{25} \right) \\ & = 1 - \frac{32}{25} \\ & = \frac{25-32}{25} \\ & = \frac{-7}{25} \\ \therefore \cos 2A & = \frac{-7}{25} \end{align}

\begin{align} \tan 2A & = \frac{\sin 2A }{\cos 2A} \\ & = \frac{\dfrac{24}{25}}{\dfrac{-7}{25}}\\ & = \frac{24}{-7}\\ \therefore \tan 2A & = -\frac{24}{7} \end{align}

\begin{align} \cot 2A & = \frac{1}{\tan 2A} \\ & = \frac{1}{-\dfrac{24}{7}} \\ \therefore \cot 2A & = - \frac{7}{24} \end{align}

\begin{align} \text{(ii) Given },\cos A & = \frac{4}{5} \\ \sin A & = \sqrt{1-\cos^2 A}\\ & = \sqrt{1-\left(\frac{4}{5} \right)^2}\\ & = \sqrt{1-\frac{16}{25} }\\ & = \sqrt{\frac{25-16}{25} } \\ & = \sqrt{\frac{9}{25} } \\ & = \frac{3}{5} \end{align}

\begin{align} \text{Now}& \\ \sin 2A & = 2\sin A \cos A \\ & = 2\times \frac{3}{5} \times \frac{4}{5} \\ & = \frac{24}{25} \\ \therefore \sin 2A & = \frac{24}{25} \end{align}

\begin{align} \cos 2A & = 2\cos^2 A - 1 \\ & = 2 \left( \frac{4}{5} \right)^2 - 1 \\ & = 2 \times \frac{16}{25} - 1 \\ & = \frac{32}{25}-1\\ & = \frac{32-25}{25} \\ & = \frac{7}{25}\\ \therefore \cos 2A & = \frac{7}{25} \end{align}

\begin{align} \tan 2A & = \frac{\sin 2A }{\cos 2A} \\ & = \frac{\dfrac{24}{25}}{\dfrac{7}{25}} \\ \therefore \tan 2A &= \frac{24}{7} \end{align}

\begin{align} \cot 2A & = \frac{1}{\tan 2A} \\ & = \frac{1}{\dfrac{24}{7}}\\ \therefore \cot 2A & = \frac{7}{24} \end{align}

\begin{align} \text{(iii) Given },\tan A & = \frac{3}{4} \\ \sin 2A & = \frac{2\tan A }{1+\tan^2 A }\\ & = \frac{2\times \dfrac{3}{4}}{1+\left(\dfrac{3}{4}\right)^2}\\ &= \frac{\dfrac{3}{2}}{1+\dfrac{9}{16}}\\ &=\frac{\dfrac{3}{2}}{\dfrac{16+9}{16}}\\ &=\frac{\dfrac{3}{2}}{\dfrac{25}{16}}\\ & = \frac{3}{2}\times \frac{16}{25}\\ \therefore \sin 2A & = \frac{24}{25} \end{align}

\begin{align} \cos 2A & = \frac{1-\tan^2 A }{1+\tan^2 A }\\ & = \frac{1-\left(\dfrac{3}{4}\right)^2}{1+\left(\dfrac{3}{4}\right)^2}\\ &= \frac{1-\dfrac{9}{16}}{1+\dfrac{9}{16}}\\ &=\frac{\dfrac{16-9}{16}}{\dfrac{16+9}{16}}\\ &=\frac{\dfrac{7}{16}}{\dfrac{25}{16}}\\ & = \frac{7}{16}\times \frac{16}{25}\\ \therefore \cos 2A & = \frac{7}{25} \end{align}

\begin{align} \tan 2A & = \frac{2\tan A }{1-\tan^2 A }\\ & = \frac{2\times \dfrac{3}{4}}{1-\left(\dfrac{3}{4}\right)^2}\\ &= \frac{\dfrac{3}{2}}{1-\dfrac{9}{16}}\\ &=\frac{\dfrac{3}{2}}{\dfrac{16-9}{16}}\\ &=\frac{\dfrac{3}{2}}{\dfrac{7}{16}}\\ & = \frac{3}{2}\times \frac{16}{7}\\ \therefore \tan 2A & = \frac{24}{7} \end{align}

\begin{align} \cot 2A & = \frac{1}{\tan 2A} \\ & = \frac{1}{\dfrac{24}{7}}\\ \therefore \cot 2A & = \frac{7}{24} \end{align}

\begin{align} \text{(iv) Given },\cot A & = \frac{12}{13} \\ \text{Then, } \tan A & = \frac{13}{12}\\ \sin 2A & = \frac{2\tan A }{1+\tan^2 A }\\ & = \frac{2\times \dfrac{13}{12}}{1+\left(\dfrac{13}{12}\right)^2}\\ &= \frac{\dfrac{13}{6}}{1+\dfrac{169}{144}}\\ &=\frac{\dfrac{13}{6}}{\dfrac{144+169}{144}}\\ &=\frac{\dfrac{13}{6}}{\dfrac{313}{144}}\\ & = \frac{13}{6}\times \frac{144}{313}\\ \therefore \sin 2A & = \frac{312}{313} \end{align}

\begin{align} \cos 2A & = \frac{1-\tan^2 A }{1+\tan^2 A }\\ & = \frac{1-\left(\dfrac{13}{12}\right)^2}{1+\left(\dfrac{13}{12}\right)^2}\\ &= \frac{1-\dfrac{169}{144}}{1+\dfrac{169}{144}}\\ &=\frac{\dfrac{144-169}{144}}{\dfrac{144+169}{144}}\\ & = \frac{-25}{313}\\ \therefore \cos 2A & = -\frac{25}{313} \end{align}

\begin{align} \tan 2A & = \frac{2\tan A }{1-\tan^2 A }\\ & = \frac{2\times \dfrac{13}{12}}{1-\left(\dfrac{13}{12}\right)^2}\\ &= \frac{\dfrac{13}{6}}{1-\dfrac{169}{144}}\\ &=\frac{\dfrac{13}{6}}{\dfrac{144-169}{144}}\\ &=\frac{\dfrac{13}{6}}{\dfrac{-25}{144}}\\ & = \frac{13}{6}\times \frac{144}{-25}\\ \therefore \tan 2A & = -\frac{312}{25} \end{align}

\begin{align} \cot 2A & = \frac{1}{\tan 2A} \\ & = \frac{1}{-\dfrac{312}{25}}\\ \therefore \cot 2A & = -\frac{25}{312} \end{align}

3. (b) Find the value of \( \sin 3A , \cos 3A \text{ and } \tan 3A \)

when (i) \( \sin A = \frac{3}{5} \) (ii) \( \cos A = \frac{1}{\sqrt{2}} \) (iii) \( \tan A = \sqrt{3} \)

\begin{align} \text{(i) Given } \sin A & = \frac{3}{5}\\ \text{Now} \sin 3A & = 3\sin A - 4\sin^3A \\ & = 3\left(\frac{3}{5} \right) - 4 \left(\frac{3}{5}\right)^3 \\ & = \frac{9}{5} - 4 \times \frac{27}{125}\\ & = \frac{9}{5} - \frac{108}{125}\\ & = \frac{9\times 25 - 108}{125}\\ & = \frac{125-108}{125}\\ & = \frac{117}{125}\\ \therefore \sin 3A & = \frac{117}{125} \end{align}

\begin{align} \text{ Given },\sin A & = \frac{3}{5} \\ \cos A & = \sqrt{1-\sin^2 A}\\ & = \sqrt{1-\left(\frac{3}{5} \right)^2}\\ & = \sqrt{1-\frac{9}{25} }\\ & = \sqrt{\frac{25-9}{25} } \\ & = \sqrt{\frac{16}{25} } \\ & = \frac{4}{5} \\ \therefore \cos A & = \frac{4}{5} \\ \text{Now},&\\ \cos 3A & = 4\cos^3 A - 3\cos A \\ & = 4 \left(\frac{4}{5}\right)^3 -3\left(\frac{4}{5} \right) \\ & = 4 \left(\frac{64}{125}\right) -\frac{12}{5} \\ & = \frac{156}{125} -\frac{12}{5} \\ & = \frac{156 - 12 \times 25}{125} \\ & = \frac{156 - 300 } {125} \\ & = \frac{-144}{125}\\ \therefore \cos 3A & = -\frac{144}{125}\\ \end{align}

\begin{align} \tan 3A & = \frac{\sin 3A}{\cos 3A}\\ & = \frac{\dfrac{117}{125}}{-\dfrac{144}{125}}\\ & =-\frac{117}{144}\\ \therefore \tan 3A & = -\frac{117}{144} \end{align}

\begin{align} \text{(ii) Given, } \cos A & = \frac{1}{\sqrt{2}}\\ \text{or, } \cos A & = \cos 45^\circ \\ \therefore A & = 45^\circ\\ \text{Now}& \\ \sin 3A & = \sin 3(45^\circ ) \\ & = \sin 135^\circ \\ & = \frac{1}{\sqrt{2}}\\ \therefore \sin 3A & = \frac{1}{\sqrt{2}}\\ \cos 3A & = \cos 3(45^\circ) \\ & = \cos 135^\circ \\ & = -\frac{1}{\sqrt{2}}\\ \therefore \cos 3A & = -\frac{1}{\sqrt{2}}\\ \tan 3A & = \tan 3(45^\circ) \\ & =\tan 135^\circ \\ & = -1 \\ \therefore \tan 3A & =-1 \end{align}

\begin{align} \text{(iii) Given, } \tan A & = \sqrt {3} \\ \text{or, } \tan A & = \tan 60^\circ \\ \therefore A & = 60^\circ\\ \text{Now}& \\ \sin 3A & = \sin 3(60^\circ ) \\ & = \sin 180^\circ \\ & = 0 \\ \therefore \sin 3A & = 0\\ \cos 3A & = \cos 3(60^\circ) \\ & = \cos 180^\circ \\ & = -1 \\ \therefore \cos 3A & = -1 \\ \tan 3A & = \tan 3(60^\circ) \\ & =\tan 180^\circ \\ & = 0 \\ \therefore \tan 3A & =0 \end{align}

4. (a) If \( \cos 2A = \frac{7}{8} \), then find the value of \( \sin A \).

\begin{align} \sin^2 A & =\frac{1-\cos 2A}{2}\\ \ & = \frac{1-\dfrac{7}{8}}{2}\\ \ & = \frac{\dfrac{8-7}{8}}{2}\\ \ & = \frac{\dfrac{1}{8}}{2}\\ \ & = \frac{1}{8}\times \frac{1}{2}\\ \ & = \frac{1}{16} \\ \text{or, } \sin^2 A & = \frac{1}{16}\\ \text{or, } \sin A & = \sqrt{\frac{1}{16}}\\ \therefore \sin A & = \frac{1}{4} \end{align}

4. (b) If \( \cos 2\theta = \frac{11}{25} \), then show that \( \cos \theta = \frac{6}{5\sqrt{2}} \).

\begin{align} \cos^2 A &=\frac{1+\cos 2A}{2}\\ \ & = \frac{1+ \dfrac{11}{25} }{2}\\ \ & = \frac{\dfrac{25+11}{25}}{2}\\ \ & = \frac{\dfrac{36}{25}}{2}\\ \ & = \frac{36}{25} \times \frac{1}{2}\\ \text{or, } \cos^2 A & = \frac{36}{50}\\ \text{or, } \cos A & = \sqrt{\frac{36}{50}}\\ \therefore \cos A & = \frac{6}{5\sqrt{2}}\\ \end{align}

4. (c) If \( \tan 2\theta = \frac{24}{7} \), then find the value of \( \tan \theta \).

\begin{align} & \tan 2\theta = \frac{24}{7}\\ \text{or, }& \frac{2\tan \theta }{1-\tan^2 \theta } = \frac{24}{7}\\ \text{or, } & 14 \tan \theta = 24 - 24 \tan^2 \theta \\ \text{or, } & 24 \tan^2 \theta +14 \tan \theta -24 = 0 \\ \text{or, } & 2(12 \tan^2 \theta +7 \tan \theta -12 )= 0 \\ \text{or, } & 12 \tan^2 \theta +7 \tan \theta -12 = \frac{0}{2} \\ \text{or, } & 12 \tan^2 \theta +(16-9) \tan \theta -12 = \frac{0}{2} \\ \text{or, } & 12 \tan^2 \theta +16 \tan \theta- 9 \tan \theta -12 = 0 \\ \text{or, } & 4\tan \theta (3 \tan \theta + 4) - 3( 3\tan \theta + 4 )=0 \\ \text{or, } & (3\tan \theta +4 ) (4\tan \theta - 3) =0 \end{align} \begin{align}& \text{Either} & \text{Or, } \\ & 3\tan \theta +4 = 0 & 4\tan \theta - 3 = 0 \\ & \text{or, } \ 3\tan \theta = -4 & \text{or, }\ 4\tan \theta = 3 \\ &\text{or, } \tan \theta = \frac{-4}{3} & \text{or, } \tan \theta = \frac{3}{4} \\ \end{align} \[ \therefore \tan \theta =\frac{3}{4}, \frac{-4}{3} \]

5. (a) If \( \cos\theta=\frac{1}{2}\left(a+\frac{1}{a}\right)\), then prove that:\(\cos2\theta=\frac{1}{2}\left(a^2+\frac{1}{a^2}\right).\)

\begin{align} \text{LHS } &=\cos2\theta\\ \ &=2\cos^2\theta-1\\ \ &=2\left(\cos\theta\right)^2-1\\ \ &=2\left\{\frac{1}{2}\left(a+\frac{1}{a}\right)\right\}^2-1\\ \ &=2\times\frac{1}{4}\left(a+\frac{1}{a}\right)^2-1\\ \ &=\frac{1}{2}\left(a+\frac{1}{a}\right)^2-1\\ \ &=\frac{1}{2}\left(a^2+2\times a\times\frac{1}{a}+\frac{1}{a^2}\right)-1\\ \ &=\frac{1}{2}\left(a^2+\frac{1}{a^2}+2\right)-1\\ \ &=\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)+\frac{1}{2}\times2-1\\ \ &=\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)+1-1\\ \ &=\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)\\ \ &= \text{ RHS }\\ \end{align}

5. (b) If \( \displaystyle\sin\theta=\frac{1}{2}\left(b+\frac{1}{b}\right)\) Prove that:\(\displaystyle \cos2\theta=-\frac{1}{2}\left(b^2+\frac{1}{b^2}\right)\)

\begin{align} \text{LHS } \ & \displaystyle =\cos2\theta\\ \ & \displaystyle=1-2\sin^2\theta\\ \ & \displaystyle =1-2\left(\sin\theta\right)^{^2}\\ \ & \displaystyle =1-2\left\{\frac{1}{2}\left(b+\frac{1}{b}\right)\right\}^2\\ \ & \displaystyle =1-2\times\frac{1}{4}\left(b+\frac{1}{b}\right)^2\\ \ & \displaystyle =1-\frac{1}{2}\left(b^2+2\times b\times\frac{1}{b}+\frac{1}{b^2}\right)\\ \ & \displaystyle =1-\frac{1}{2}\left(b^2+2+\frac{1}{b^2}\right)\\ \ & \displaystyle =1-\frac{1}{2}\left(b^2+\frac{1}{b^2}+2\right)\\ \ & \displaystyle =1-\frac{1}{2}\left(b^2+\frac{1}{b^2}\right)-\frac{1}{2}\times2\\ \ & \displaystyle =1-\frac{1}{2}\left(b^2+\frac{1}{b^2}\right)-1\\ \ & \displaystyle =-\frac{1}{2}\left(b^2+\frac{1}{b^2}\right)\\ \ & \displaystyle =\text{ RHS } \end{align}

5. (c) If \( \cos\theta=\frac{1}{2}\left(p+\frac{1}{p}\right)\), then prove that: \(\cos3\theta=\frac{1}{2}\left(p^3+\frac{1}{p^3}\right).\)

\begin{align} \text{LHS} &=\cos3\theta\\ &=4\cos^3\theta-3\cos\theta\\ &=4\left(\cos\theta\right)^3-3\cos\theta\\ &=4\left\{\frac{1}{2}\left(a+\frac{1}{a}\right)^3\right\}-3\times\frac{1}{2}\left(a+\frac{1}{a}\right)^2\\ &=4\times\frac{1}{8}\left(a+\frac{1}{a}\right)^3-\frac{3}{2}\left(a+\frac{1}{a}\right)\\ &=\frac{1}{2}\left(a+\frac{1}{a}\right)^3-\frac{3}{2}\left(a+\frac{1}{a}\right)\\ &=\frac{1}{2}\left\{a^3+\frac{1}{a^3}+ 3\times a \times \frac{1}{a}\left(a+\frac{1}{a} \right) \right\}-\frac{3}{2}\left(a+\frac{1}{a}\right)\\ &=\frac{1}{2}\left\{a^3+\frac{1}{a^3}+ 3\left(a+\frac{1}{a} \right) \right\}-\frac{3}{2}\left(a+\frac{1}{a}\right)\\ & = \frac{1}{2} \left( a^3 + \frac{1}{a^3} \right) +\frac{3}{2} \left( a+ \frac{1}{a} \right) - \frac{3}{2} \left( a + \frac{1}{a} \right)\\ & = \frac{1}{2} \left( a^3 + \frac{1}{a^3} \right)\\ &=\text{RHS} \end{align}

Alternative

\begin{align} \text{LHS} &=\cos3\theta\\ &=4\cos^3\theta-3\cos\theta\\ &=4\left(\cos\theta\right)^3-3\cos\theta\\ &=4\left\{\frac{1}{2}\left(a+\frac{1}{a}\right)^3\right\}-3\times\frac{1}{2}\left(a+\frac{1}{a}\right)^2\\ &=4\times\frac{1}{8}\left(a+\frac{1}{a}\right)^3-\frac{3}{2}\left(a+\frac{1}{a}\right)\\ &=\frac{1}{2}\left(a+\frac{1}{a}\right)^3-\frac{3}{2}\left(a+\frac{1}{a}\right)\\ &=\frac{1}{2}\left\{\left(a+\frac{1}{a}\right)^3-3\left(a+\frac{1}{a}\right)\right\}\\ &=\frac{1}{2}\left\{\left(a+\frac{1}{a}\right)^3-3\times a\times\frac{1}{a}\left(a+\frac{1}{a}\right)\right\}\\ &=\frac{1}{2}\left(a^3+\frac{1}{a^3}\right)\\ &\left[\because x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\right]\\ &=\text{RHS} \end{align}

5. (d) If \(\sin\theta=\frac{1}{2}\left(x+\frac{1}{x}\right)\) , show that \( \sin3\theta=-\frac{1}{2}\left(x^3+\frac{1}{x^3}\right)\)

\begin{align} \text{LHS } \ & =\sin3\theta\\ \ & =3\sin\theta-4\sin^3\theta\\ \ & =3\times\frac{1}{2}\left(x+\frac{1}{x}\right)-4\times\left\{\frac{1}{2}\left(x+\frac{1}{x}\right)\right\}^3\\ \ & =\frac{3}{2}\left(x+\frac{1}{x}\right)-4\times\frac{1}{8}\left(x+\frac{1}{x}\right)^3\\ \ & =\frac{3}{2} \left( x + \frac{1}{x} \right) - \frac{1}{2} \left(x+\frac{1}{x}\right)^3\\ \ & =\frac{3}{2} \left( x + \frac{1}{x} \right) - \frac{1}{2} \left\{x^3+ \frac{1}{x^3}+3\times x\times \frac{1}{x} \left( x+ \frac{1}{x} \right) \right\}\\ \ & =\frac{3}{2} \left( x + \frac{1}{x} \right) - \frac{1}{2} \left\{x^3+ \frac{1}{x^3}+3 \left( x + \frac{1}{x} \right) \right\}\\ \ & =\frac{3}{2} \left( x + \frac{1}{x} \right) - \frac{1}{2} \left(x^3+ \frac{1}{x^3}\right) - \frac{3}{2} \left( x+ \frac{1}{x} \right) \\ \ & = - \frac{1}{2} \left(x^3+ \frac{1}{x^3}\right) \\ \ & =\text{ RHS} \end{align}

Alternative

\begin{align} \text{LHS } \ & =\sin3\theta\\ \ & =3\sin\theta-4\sin^3\theta\\ \ & =3\times\frac{1}{2}\left(x+\frac{1}{x}\right)-4\times\left\{\frac{1}{2}\left(x+\frac{1}{x}\right)\right\}^3\\ \ & =\frac{3}{2}\left(x+\frac{1}{x}\right)-4\times\frac{1}{8}\left(x+\frac{1}{x}\right)^3\\ \ & =\frac{1}{2}\left\{3\left(x+\frac{1}{x}\right)-\left(x+\frac{1}{x}\right)^3\right\}\\ \ & =-\frac{1}{2}\left\{\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right\}\\ \ & =-\frac{1}{2}\left\{\left(x+\frac{1}{x}\right)^3-3\times x\times\frac{1}{x}\left(a+\frac{1}{x}\right)\right\}\\ \ & =-\frac{1}{2}\left(x^3+\frac{1}{x^3}\right) \left[\because a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\right]\\ \ & =\text{ RHS}\\ \end{align}

6. (a) Prove that: \( \cos A = \pm \sqrt{\frac{1+\cos 2A}{2}} \)

First Method

\begin{align} \text{We know, }\\\cos 2A &= 2\cos^2 A -1\\ \cos 2A +1&= 2\cos^2 A \\ or, 2 \cos^2 A&=1+\cos 2A\\ or, \cos^2 A & = \frac{1+\cos 2A}{2}\\ \therefore \ \cos A&= \pm\sqrt{\frac{1+\cos 2A}{2}}\\ \end{align}

Second Method

\begin{align} \text{ RHS } & = \pm\sqrt{\frac{1+\cos 2A}{2}}\\ \ & = \pm\sqrt{\frac{1+(2\cos^2 A-1)}{2}}\\ \ & = \pm\sqrt{\frac{1+2\cos^2 A-1}{2}}\\ \ & = \pm\sqrt{\frac{2\cos^2 A}{2}}\\ \ & = \pm\sqrt{\cos^2 A}\\ \ & = \pm \cos A\\ \ & = \cos A\\ \ & = \text{ LHS } \\ \end{align}

6. (b) Prove that: \( \sin A = \pm \sqrt{\frac{1-\cos 2A}{2}} \)

First Method

\begin{align} \cos 2A &= 1-2\sin^2 A\\ or, 2 \sin^2 A&=1-\cos 2A\\ or, \sin^2 A & = \frac{1-\cos 2A}{2}\\ \therefore \ \sin A&= \pm\sqrt{\frac{1 \cos 2A}{2}} \end{align}

Second Method

\begin{align} \text{ RHS } & = \pm\sqrt{\frac{1-\cos 2A}{2}}\\ \ & = \pm\sqrt{\frac{1-(1-2\sin^2 A)}{2}}\\ \ & = \pm\sqrt{\frac{1-1+2\sin^2 A}{2}}\\ \ & = \pm\sqrt{\frac{2\sin^2 A}{2}}\\ \ & = \pm\sqrt{\sin^2 A}\\ \ & = \pm \sin A\\ \ & = \sin A\\ \ & = \text{ LHS } \end{align}

6. (c) Prove that: \( \tan A = \pm \sqrt{\frac{1-\cos 2A}{1+\cos 2A }} \)

\begin{align} \text{ RHS } & = \pm\sqrt{\frac{1-\cos 2A}{1+\cos 2A}}\\ \ & = \pm\sqrt{\frac{1-(1-2\sin^2 A)}{1+(2cos^2A+1)}}\\ \ & = \pm\sqrt{\frac{1-1+2\sin^2 A}{1+2\cos^2A-1}}\\ \ & = \pm\sqrt{\frac{2\sin^2 A}{2\cos^2 A}}\\ \ & = \pm\sqrt{\tan^2 A}\\ \ & = \pm \tan A\\ \ & = \tan A\\ \ & = \text{ LHS } \\ \end{align}

6. (d) Prove that: \( \cot A = \pm \sqrt{\frac{1+\cos 2A}{-\cos 2A }} \)

\begin{align} \text{ RHS } & = \pm\sqrt{\frac{1+\cos 2A}{1-\cos 2A}}\\ \ & = \pm\sqrt{\frac{1+(2\cos^2 A - 1 )}{1-(1- 2\sin^2 A)}}\\ \ & = \pm\sqrt{\frac{1+2\cos^2 A - 1}{1-1+2\sin^2 A }}\\ \ & = \pm\sqrt{\frac{2\cos^2 A}{2\sin^2 A}}\\ \ & = \pm\sqrt{\cot^2 A}\\ \ & = \pm \cot A\\ \ & = \cot A\\ \ & = \text{ LHS } \\ \end{align}

7. Prove that:

7. (a) \( \frac{1+\cos 2A }{2} = \cos^2 A \)

\begin{align} \text{LHS } & =\frac{1+\cos 2A }{2} \\ & = \frac{1+2\cos^2 A - 1 }{2}\\ & = \frac{2\cos^2 A }{2}\\ & = \cos^2 A \\ & = \text{ RHS} \end{align}

7. (b) \( \frac{1-\cos 2A }{2} = \sin^2 A \)

\begin{align} \text{LHS } & =\frac{1-\cos 2A }{2} \\ & = \frac{1-( 1 - 2\sin^2 A ) }{2}\\ & = \frac{1- 1 + 2\sin^2 A ) }{2}\\ & = \frac{2\sin^2 A }{2}\\ & = \sin^2 A \\ & = \text{ RHS} \end{align}

7. (c) \( \frac{1+\cos 2A }{\sin 2A} = \cot A \)

\begin{align} \text{LHS } & =\frac{1+\cos 2A }{\sin 2A} \\ & = \frac{1+2\cos^2 A - 1 }{2\sin A \cos A}\\ & = \frac{2\cos^2 A }{2\sin A \cos A }\\ & = \frac{\cos A }{\sin A }\\ & = \cot A \\ & = \text{ RHS} \end{align}

7. (d) \( \frac{1-\cos 2A }{\sin 2A} = \tan A \)

\begin{align} \text{LHS } & =\frac{1-\cos 2A }{\sin 2A} \\ & = \frac{1-( 1 - 2\sin^2 A ) }{2\sin A \cos A }\\ & = \frac{1- 1 + 2\sin^2 A }{2\sin A \cos A}\\ & = \frac{2\sin^2 A }{2\sin A \cos A }\\ & = \frac{\sin A }{\cos A}\\ & = \tan A \\ & = \text{ RHS} \end{align}

7. (e) \( \frac{1-\cos 2A }{1+\cos 2A} = \tan^2 A \)

\begin{align} \text{LHS } & =\frac{1-\cos 2A }{1+\cos 2A} \\ & = \frac{1-( 1 - 2\sin^2 A ) }{1+2\cos^2 A - 1 }\\ & = \frac{1- 1 + 2\sin^2 A }{2\cos^2 A }\\ & = \frac{2\sin^2 A }{2\cos^2 A }\\ & = \frac{\sin^2 A }{\cos^2 A}\\ & = \tan^2 A \\ & = \text{ RHS} \end{align}

7. (f) \( \frac{1+\cos 2A }{1-\cos 2A} = \cot^2 A \)

\begin{align} \text{LHS } & =\frac{1+\cos 2A }{1-\cos 2A} \\ & = \frac{1+2\cos^2 A - 1 }{1-( 1 - 2\sin^2 A ) }\\ & = \frac{2\cos^2 A }{1-1+2\sin^2 A }\\ & = \frac{2\cos^2 A }{2\sin^2 A }\\ & = \frac{\cos^2 A }{\sin^2 A}\\ & = \cot^2 A \\ & = \text{ RHS} \end{align}

7. (g) \( \frac{\sin 2\theta }{1-\cos 2\theta } = \cot \theta \)

\begin{align} \text{LHS } & =\frac{\sin 2\theta }{1-\cos 2\theta} \\ & = \frac{2\sin \theta \cos \theta }{1-( 1 - 2\sin^2 \theta ) }\\ & = \frac{2\sin \theta \cos \theta }{1-1+2\sin^2 \theta }\\ & = \frac{2\sin \theta \cos \theta }{2\sin^2 \theta }\\ & = \frac{\cos \theta }{\sin\theta}\\ & = \cot \theta \\ & = \text{ RHS} \end{align}

7. (h) \( \frac{\sin 2\theta }{1+\cos 2\theta } = \tan \theta \)

\begin{align} \text{LHS } & =\frac{\sin 2\theta }{1+\cos 2\theta} \\ & = \frac{2\sin \theta \cos \theta }{1+2\cos^2 \theta -1 }\\ & = \frac{2\sin \theta \cos \theta }{2\cos^2 \theta }\\ & = \frac{\sin \theta }{\cos \theta}\\ & = \tan \theta \\ & = \text{ RHS} \end{align}

7. (i) \( \frac{1+\sin 2\theta}{\cos 2\theta} = \frac{\cot \theta + 1}{\cot \theta - 1 } \)

\begin{align} \text{LHS } & = \frac{1+\sin 2\theta}{\cos 2\theta} \\ & = \frac{\sin^2 \theta + \cos^2 \theta +2\sin \theta\cos \theta }{\cos ^2\theta-\sin^2 \theta } \\ & =\frac{(\cos\theta + \sin \theta )^2 }{(\cos \theta - \sin \theta ) (\cos \theta + \sin \theta ) } \\ & = \frac{\cos \theta + \sin \theta }{ \cos \theta - \sin \theta } \\ & = \frac{\dfrac{\cos \theta}{\sin \theta} + \dfrac{\sin \theta}{\sin \theta} }{ \dfrac{\cos \theta}{\sin \theta} - \dfrac{\sin \theta}{\sin \theta } } \\ & =\frac{\cot \theta + 1}{\cot \theta - 1 }\\ & = \text{ RHS} \end{align}

7. (j) \( \frac{1-\sin 2A}{\cos 2A} = \frac{1- \tan A }{1+\tan A } \)

\begin{align} \text{LHS } & = \frac{1-\sin 2A}{\cos 2A} \\ & = \frac{\sin^2 A + \cos^2 A -2\sin A \cos A }{\cos^2 A -\sin^2 A } \\ & = \frac{(\cos A - \sin A )^2 }{ (\cos A - \sin A ) ( \cos A + \sin A ) }\\ & = \frac{\cos A - \sin A }{ \cos A + \sin A }\\ & = \frac{\dfrac{\cos A}{\cos A} - \dfrac{\sin A}{\cos A } }{ \dfrac{\cos A}{\cos A} +\dfrac{ \sin A}{\cos A} }\\ & = \frac{1- \tan A }{1+\tan A } \\ & = \text{ RHS} \end{align}

7. (k) \( \tan \theta + \cot \theta = 2\text{cosec}2\theta \)

\begin{align} \text{LHS } & = \tan \theta + \cot \theta \\ & = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta } \\ & = \frac{\sin^2 \theta + \cos^2 \theta }{\sin \theta \cos \theta } \\ & = \frac{1}{\sin \theta \cos \theta }\times \frac{2}{2} \\ & = \frac{2}{2\sin \theta \cos \theta } \\ & = \frac{2}{\sin 2\theta }\\ & = 2\text{cosec}2\theta \\ & = \text{ RHS} \end{align}

7. (l) \( \frac{1-\sin 2A}{\cos 2A} = \tan (45^\circ - A ) \)

7. (m) \( 2\cos^2 (45^{\circ} -A ) = 1+ \sin 2A \)

\begin{align} \text{ LHS } & = 2 \cos^2 (45^{\circ} -A)\\ \ & = 2 \times \frac{1+\cos 2(45^{\circ}-A)}{2}\\ \ & \left[ \because \cos^2 A = \frac{ 1+\cos 2A}{2} \right]\\ \ & = 1+\cos(90^{\circ} -2A)\\ \ & = 1+\sin 2A\\ \ & = \text{ RHS} \end{align}

7. (n) \( \cos^2 ( 45^{\circ} -A) -\sin^2 (45^{\circ} -A) = \sin 2A \)

\begin{align}\text{LHS } & = \cos^2 ( 45^{\circ} -A) -\sin^2 (45^{\circ} -A)\\ \ & = \cos 2(45^{\circ} - A) \\ \ & = \cos (90^{\circ} - 2A) \\ \ & = \sin 2A \\ \ & = \text { RHS }\end{align}

7. (o) \(\cot \theta - \tan \theta = 2\cot 2\theta \)

\begin{align} \text{LHS } & = \cot \theta - \tan \theta\\ & = \frac{\cos \theta }{ \sin \theta } - \frac{\sin \theta }{\cos \theta }\\ & = \frac{\cos^2 \theta - \sin^2 \theta }{\sin \theta \cos \theta }\\ & = \frac{\cos 2\theta}{\sin \theta \cos \theta }\\ & = \frac{2\cos 2 \theta }{2\sin \theta \cos \theta}\\ & = \frac{2\cos 2\theta}{\sin 2\theta }\\ & = 2\cot 2\theta \\ & = \text { RHS }\end{align}

7. (p) \(1+\tan \theta . \tan 2\theta = \sec 2\theta \)

\begin{align} \text{LHS } & = 1+\tan \theta . \tan 2\theta \\ & = 1+ \frac{\sin \theta } {\cos \theta } . \frac{\sin 2\theta}{\cos 2\theta } \\ & = \frac{\cos2 \theta \cos \theta + \sin 2\theta \sin \theta}{\cos \theta \cos 2\theta }\\ & = \frac{\cos (2\theta - \theta )}{\cos \theta \cos 2\theta }\\ & = \frac{\cos \theta }{\cos \theta \cos 2\theta }\\ & = \frac{1}{\cos 2\theta }\\ & = \sec 2\theta \\ & = \text { RHS }\end{align}

7. (q) \(\frac{\sin^3 A + \cos^3 A }{\sin A + \cos A } = 1 - \frac{1}{2} \sin 2A \)

\begin{align} \text{LHS } & =\frac{\sin^3 A + \cos^3 A }{\sin A + \cos A } \\ & =\frac{(\sin A + \cos A )(\sin^2 A - \sin A \cos A + \cos^2 A)}{\sin A + \cos A } \\ & =\sin^2 A - \sin A \cos A + \cos^2 A \\ & = \sin^2 A + \cos^2 A - \frac{1}{2} (2\sin A \cos A ) \\ & = 1 - \frac{1}{2} \sin 2A \\ & = \text { RHS }\end{align}

7. (r) \(\cos^4 \theta - \sin^4 \theta = \cos 2\theta \)

\begin{align} \text{LHS } & =\cos^4 \theta - \sin^4 \theta \\ & = (\cos^2 \theta )^2 - (\sin^2 \theta )^2 \\ & = (\cos^2 \theta - \sin^2 \theta ) (\cos^2 \theta + \sin^2 \theta ) \\ & = (\cos 2\theta)(1)\\ & = \cos 2\theta \\ & = \text { RHS }\end{align}

7. (s) \(\frac{\sin 3A}{\sin A} - \frac{\cos 3A}{\cos A}= 2 \)

\begin{align} \text{LHS } & =\frac{\sin 3A}{\sin A} - \frac{\cos 3A}{\cos A} \\ & = \frac{\sin 3A \cos A - \cos 3A \sin A }{\sin A \cos A } = 2 \\ & = \frac{\sin (3A - A ) } { \sin A \cos A } \\ & = \frac{\sin 2A }{\sin A \cos A }\\ & = \frac{2\sin A \cos A }{\sin A \cos A}\\ & = 2 \\ & = \text { RHS }\end{align}

7. (t) \(\frac{\sin 5\theta }{\sin \theta } - \frac{\cos 5\theta}{\cos \cos \theta }= 4\cos 2\theta \)

\begin{align} \text{LHS } & =\frac{\sin 5\theta }{\sin \theta } - \frac{\cos 5\theta}{\cos \cos \theta } \\ & = \frac{\sin 5\theta \cos \theta - \cos 5\theta \sin \theta }{\sin \theta \cos \theta } \\ & = \frac{\sin (5\theta - \theta ) } { \sin \theta \cos \theta} \\ & = \frac{\sin 4\theta }{\sin \theta \cos \theta }\\ & = \frac{2\sin 2\theta \cos 2\theta }{\sin \theta \cos \theta}\\ & = \frac{2(2\sin \theta \cos \theta ) \cos 2\theta }{\sin \theta \cos \theta } \\ & = 4\cos 2\theta \\ & = \text { RHS }\end{align}

8. Prove that:

8. (a) \( \frac{1+\sin 2\theta - \cos 2\theta }{1+\sin 2\theta + \cos 2\theta } = \tan \theta \)

\begin{align} \text{LHS } & = \frac{1+\sin 2\theta - \cos 2\theta }{1+\sin 2\theta + \cos 2\theta } \\ & = \frac{1+2\sin \theta \cos \theta - (1-2\sin^2 \theta) }{1+2\sin \theta \cos \theta + (2\cos^2\theta -1)} \\ & = \frac{2\sin \theta \cos \theta +2\sin^2 \theta }{2\sin \theta \cos \theta + 2\cos^2\theta } \\ & = \frac{2\sin \theta ( \cos \theta + \sin \theta ) }{2\cos \theta ( \sin \theta + \cos \theta )}\\ & = \frac{\sin \theta ( \sin \theta + \cos \theta ) }{\cos \theta ( \sin \theta + \cos \theta )}\\ & = \frac{\sin \theta }{\cos \theta} \\ & = \tan \theta \\ & = \text { RHS }\end{align}

8. (b) \( \frac{1+\sin 2\theta + \cos 2\theta }{1+\sin 2\theta - \cos 2\theta } = \tan \theta \)

\begin{align} \text{LHS } & = \frac{1+\sin 2\theta + \cos 2\theta }{1+\sin 2\theta - \cos 2\theta } \\ & = \frac{1+2\sin \theta \cos \theta +(2\cos^2 \theta -1 ) }{1+2\sin \theta \cos \theta - (1 - 2\sin^2 \theta)} \\ & = \frac{2\sin \theta \cos \theta +2\cos^2 \theta }{2\sin \theta \cos \theta + 2\sin^2\theta } \\ & = \frac{2\cos \theta ( \sin \theta + \cos \theta ) }{2\sin \theta ( \cos \theta + \sin \theta )}\\ & = \frac{\cos \theta ( \sin \theta + \cos \theta ) }{\sin \theta ( \sin \theta + \cos \theta )}\\ & = \frac{\cos \theta }{\sin \theta} \\ & = \cot \theta \\ & = \text { RHS }\end{align}

8. (c) \( \frac{\sin \alpha + \sin 2\alpha }{1+\cos \alpha + \cos 2\alpha }= \tan \alpha \)

\begin{align} \text{LHS } & =\frac{\sin \alpha + \sin 2\alpha }{1+\cos \alpha + \cos 2\alpha } \\ & = \frac{\sin \alpha + 2\sin \alpha \cos \alpha }{1+\cos \alpha + 2\cos^2\alpha -1 } \\ & = \frac{\sin \alpha( 1+ 2 \cos \alpha) }{\cos \alpha + 2\cos^2\alpha } \\ & = \frac{\sin \alpha( 1+ 2 \cos \alpha) }{\cos \alpha (1+ 2\cos \alpha) } \\ & = \frac{\sin \alpha}{\cos \alpha} \\ & = \tan \alpha \\ & = \text { RHS }\end{align}

8. (d) \( (1+\sin 2A + \cos 2A )^2 = 4\cos^2 A (1+\sin 2A) \)

\begin{align} \text{LHS } & =(1+\sin 2A + \cos 2A )^2 \\ & = (1+2\sin A\cos A + 2\cos ^2A- 1 )^2\\ & = (2\sin A\cos A + 2\cos ^2A )^2\\ & = \left\{ 2\cos A ( \cos A + \sin A ) \right\}^2\\ & = 4\cos^2 A ( \cos A + \sin A)^2 \\ & = 4\cos^2 A ( \cos^2 A + 2\cos A \sin A + \sin^2 A )\\ & = 4\cos^2 A ( \sin^2 A + \cos^2 A + 2\sin A \cos A) \\ & = 4\cos^2 A (1+ \sin 2A) \\ & = \text { RHS }\end{align}

9. (a) \( \sin 8 \theta = 8 \sin \theta \cos \theta \cos 2\theta \cos 4\theta \)

\begin{align} \text{LHS } & = \sin 8 \theta \\ & = 2\sin 4\theta \cos 4\theta \\ & = 2 \times 2\sin 2\theta \cos 2\theta \cos 4\theta \\ & = 2\times 2\times 2\sin \theta \cos \theta \cos 2\theta \cos 4\theta \\ & = 8\sin \theta \cos \theta \cos 2\theta \cos 4\theta \\ & = \text { RHS }\end{align}

9. (b) \( \sin 4 \theta = 4 \sin \theta \cos \theta \cos 2\theta \)

\begin{align} \text{LHS } & = \sin 4 \theta \\ & = 2\sin 2\theta \cos 2\theta \\ & = 2\times 2\sin \theta \cos \theta \cos 2\theta \\ & = 4\sin \theta \cos \theta \cos 2\theta \\ & = \text {RHS }\end{align}

9. (c) \( \cos 4 \theta = 1-8\sin^2 \theta + 8 \sin^4 \theta \)

\begin{align} \text{LHS } & = \cos 4 \theta \\ & = 1-2\sin^2 2\theta \\ & = 1 - 2 ( \sin 2\theta )^2 \\ & = 1 -2 ( 2\sin \theta \cos \theta )^2 \\ & = 1 -2 ( 4\sin^2 \theta \cos^2 \theta ) \\ & = 1 - 8 \sin^2 \theta \cos^2 \theta \\ & = 1 -8\sin^2 \theta (1- \sin^2 \theta ) \\ & = 1 - 8 \sin^2 \theta + 8 \sin^4 \theta \\ & = \text {RHS }\end{align}

9. (d) \( \cos 4 \theta = 8\cos^4 \theta - 8 \cos^2 \theta + 1 \)

\begin{align} \text{LHS } & = \cos 4 \theta \\ & = 2\cos^2 2\theta - 1 \\ & = 2(\cos 2\theta )^2 - 1 \\ & = 2(2\cos^2 \theta - 1)^2 - 1 \\ & = 2(4\cos^4 \theta - 4\cos^2 \theta +1 ) - 1 \\ & = 8\cos^4 \theta - 8 \cos^2 \theta + 2 - 1 \\ & = 8\cos^4 \theta - 8 \cos^2 \theta + 1 \\ & = \text {RHS }\end{align}

9. (e) \( \frac{\sin 8 \theta}{8\sin \theta} = \cos \theta \cos 2\theta \cos 4\theta \)

\begin{align} \text{LHS } & = \frac{\sin 8 \theta}{8\sin \theta} \\ & = \frac{2\sin 4\theta \cos 4\theta}{8\sin \theta} \\ & = \frac{2 \times 2\sin 2\theta \cos 2\theta \cos 4\theta }{8\sin \theta}\\ & = \frac{2\times 2\times 2\sin \theta \cos \theta \cos 2\theta \cos 4\theta}{8\sin \theta} \\ & = \frac{8\sin \theta \cos \theta \cos 2\theta \cos 4\theta}{8\sin \theta} \\ & = \cos \theta \cos 2\theta \cos 4\theta \\ & = \text { RHS }\end{align}

9. (f) \( \sin A \cos 2A = \frac{1}{4} \sin 4A. \sec A \)

\begin{align} \text{RHS } & = \frac{1}{4} \sin 4A. \sec A \\ & = \frac{1}{4} \times 2 \sin 2A. \cos 2A \frac{1}{\cos A} \\ & = \frac{1}{4} \times 2 \times 2\sin A. \cos A \cos 2A \frac{1}{\cos A} \\ & = \sin A \cos 2A \\ & = \text { LHS }\end{align}

9. (g) \( 2\cos 2\theta + 1 = ( 2\cos \theta + 1)(2\cos \theta - 1) \)

\begin{align} \text{LHS } & = 2\cos 2\theta + 1 \\ & = 2 (2\cos^2 \theta - 1 ) + 1 \\ & = 4\cos^2 \theta - 2 + 1 \\ & = 4\cos^2 \theta - 1 \\ & = (2\cos \theta )^2 - 1^2 \\ & = ( 2\cos \theta + 1)(2\cos \theta - 1 ) \\ & = \text { RHS }\end{align}

9. (h) \( \sqrt{2+\sqrt{2+2\cos 4\theta }} = 2\cos \theta \)

\begin{align} \text{LHS } & = \sqrt{2+\sqrt{2+2\cos 4\theta }} \\ & = \sqrt{2+\sqrt{2+2(2\cos^2 2\theta - 1) }} \\ & = \sqrt{2+\sqrt{2+4\cos^2 2\theta - 2}} \\ & = \sqrt{2+\sqrt{4\cos^2 2\theta }} \\ & = \sqrt{2+ 2\cos 2\theta } \\ & = \sqrt{2+ 2(2\cos^2 \theta - 1)}\\ & = \sqrt{2+4\cos^2 \theta -2 }\\ & = \sqrt{4\cos^2 \theta }\\ & = 2\cos \theta \\ & = \text {RHS }\end{align}

9. (i) \( \frac{1}{2}\sqrt{\frac{1-\cos 4\theta}{2}} = \sin \theta \cos \theta \)

\begin{align} \text{LHS } & = \frac{1}{2}\sqrt{\frac{1-\cos 4\theta}{2}} \\ & = \frac{1}{2}\sqrt{\frac{1-(1-2\sin^2 2\theta)}{2}} \\ & = \frac{1}{2}\sqrt{\frac{1-1+2\sin^2 2\theta)}{2}}\\ & = \frac{1}{2}\sqrt{\frac{2\sin^2 2\theta)}{2}}\\ & = \frac{1}{2} \sqrt{\sin^2 2\theta} \\ & = \frac{1}{2} \times \sin 2\theta \\ & = \frac{1}{2} \times 2\sin \theta \cos \theta \\ & =\sin \theta \cos \theta \\ & = \text {RHS }\end{align}

9. (j) \( \tan 2\alpha + \sin 2\alpha = \frac{4\tan \alpha}{1-\tan^4 \alpha} \)

\begin{align} \text{LHS } & = \tan 2\alpha + \sin 2\alpha \\ & = \frac{2\tan \alpha}{1-\tan^2 \alpha} + \frac{2\tan \alpha}{1+\tan^2 \alpha} \\ & = \frac{2\tan \alpha( 1+ \tan^2 \alpha ) + 2\tan \alpha (1-\tan^2 \alpha)}{(1-\tan^2 \alpha)(1+\tan^2 \alpha)} \\ & = \frac{2\tan \alpha+ \tan^3 \alpha + 2\tan \alpha -\tan^3 \alpha}{1^2-(\tan^2 \alpha)^2} \\ & = \frac{4\tan \alpha}{1-\tan^4 \alpha} \\ & = \text {RHS }\end{align}

10. Prove that:

10. (a) \( 4(\cos^3 10^\circ + \sin^3 20^\circ ) = 3(\cos 10^\circ + \sin 20^\circ) \)

\begin{align} \text{LHS } & = 4(\cos^3 10^\circ + \sin^3 20^\circ ) \\ & = 4\cos^3 10^\circ + 4\sin^3 20^\circ \\ & = \{3\cos 10^\circ + \cos 3(10^\circ)\} + \{3\sin 20^\circ - \sin 3(20^\circ) \}\\ & = 3\cos 10^\circ + \cos 30^\circ + 3\sin20^\circ - \sin 60^\circ \\ & = 3\cos 10^\circ + \frac{\sqrt{3}}{2} + 3\sin20^\circ -\frac{\sqrt{3}}{2} \\ & = 3(\cos 10^\circ + \sin20^\circ) \\ & = \text {RHS }\end{align}

10. (c) \( 4(\cos^3 20^\circ + \sin^3 50^\circ ) = 3(\cos 20^\circ + \sin 50^\circ) \)

\begin{align} \text{LHS } & = 4(\cos^3 20^\circ + \sin^3 50^\circ ) \\ & = 4\cos^3 20^\circ + 4\sin^3 50^\circ \\ & = \{3\cos 20^\circ + \cos 3(20^\circ)\} + \{3\sin 50^\circ - \sin 3(50^\circ) \}\\ & = 3\cos 20^\circ + \cos 60^\circ + 3\sin50^\circ - \sin 150^\circ \\ & = 3\cos 20^\circ + \frac{1}{2} + 3\sin50^\circ -\frac{1}{2} \\ & = 3(\cos 20^\circ + \sin50^\circ) \\ & = \text {RHS }\end{align}

11. (a) \( (2\cos \theta + 1)(2\cos \theta - 1)(2\cos 2\theta -1)= 2\cos 4\theta + 1 \)

\begin{align} \text{LHS} \ & =(2\cos\theta+1)(2\cos\theta-1)\left(2\cos2\theta-1\right)\\ \ & =\left\{\left(2\cos\theta\right)^2-1^2\right\}\left(2\cos2\theta-1\right)\\ \ & =\left(4\cos^2\theta-1\right)\left(2\cos2\theta-1\right)\\ \ & =\left(4\times\frac{1+\cos2\theta}{2}-1\right)\left(2\cos2\theta-1\right)\\ \ & =\left(2+2\cos2\theta-1\right)\left(2\cos2\theta-1\right)\\ \ & =\left(2\cos2\theta+1\right)\left(2\cos2\theta-1\right)\\ \ & =\left(2\cos2\theta\right)^2-1^2\\ \ & =4\cos^22\theta-1\\ \ & =4\times\frac{1+\cos4\theta}{2}-1\\ \ & =2\left(1+\cos4\theta\right)-1\\ \ & =2+2\cos4\theta-1\\ \ & = 2\cos 4\theta + 1 \\ \ & = \text{ RHS } \end{align}

11. (b) Prove that: \( \frac{2\cos8\theta+1}{2\cos\theta+1}=\left(2\cos\theta-1\right)\left(2\cos2\theta-1\right)\left(2\cos4\theta-1\right)\)

\begin{align} \text{LHS } & =\frac{2\cos8\theta+1}{2\cos\theta+1}\\ & =\frac{2\left(2\cos^24\theta-1\right)+1}{2\cos\theta+1}\\ & =\frac{4\cos^24\theta-2+1}{2\cos\theta+1}\\ & =\frac{4\cos^24\theta-1}{2\cos\theta+1}\\ & =\frac{\left(2\cos4\theta\right)^2-1^2}{2\cos\theta+1}\\ & =\frac{\left(2\cos4\theta-1\right)\left(2\cos4\theta+1\right)}{2\cos\theta+1}\\ & =\frac{\left(2\cos4\theta-1\right)\left\{2\left(2\cos^22\theta-1\right)+1\right\}}{2\cos\theta+1}\\ & =\frac{\left(2\cos4\theta-1\right)\left(4\cos^22\theta-2+1\right)}{2\cos\theta+1}\\ & =\frac{\left(2\cos4\theta-1\right)\left(4\cos^22\theta-1\right)}{2\cos\theta+1}\\ & =\frac{\left(2\cos4\theta-1\right)\left\{\left(2\cos2\theta\right)^2-1^2\right\}}{2\cos\theta+1}\\ & =\frac{\left(2\cos4\theta-1\right)\left(2\cos2\theta-1\right)\left(2\cos2\theta+1\right)}{2\cos\theta+1}\\ & =\frac{\left(2\cos4\theta-1\right)\left(2\cos2\theta-1\right)\left\{2\left(2\cos^2\theta-1\right)+1\right\}}{2\cos\theta+1}\\ & =\frac{\left(2\cos4\theta-1\right)\left(2\cos2\theta-1\right)\left(4\cos^2\theta-2+1\right)}{2\cos\theta+1}\\ & =\frac{\left(2\cos4\theta-1\right)\left(2\cos2\theta-1\right)\left(4\cos^2\theta-1\right)}{2\cos\theta+1}\\ & =\frac{\left(2\cos4\theta-1\right)\left(2\cos2\theta-1\right)\left\{\left(2\cos\theta\right)^2-1^2\right\}}{2\cos\theta+1}\\ & =\frac{\left(2\cos4\theta-1\right)\left(2\cos2\theta-1\right)\left(2\cos\theta-1)(2\cos\theta+1)\right)}{2\cos\theta+1}\\ & =\left(2\cos4\theta-1\right)\left(2\cos2\theta-1\right)\left(2\cos\theta-1\right) \\ & =\left(2\cos\theta-1\right)\left(2\cos2\theta-1\right)\left(2\cos4\theta-1\right) \\ & = \text{RHS } \end{align}

11. (c) \( \frac{\sec8A-1}{\sec4A-1}=\frac{\tan8A}{\tan2A} \)

\begin{align} \text{LHS} & =\frac{\sec8A-1}{\sec4A-1}\\ & =\frac{\frac{1}{\cos8A}-1}{\frac{1}{\cos4A}-1}\\ & =\frac{\frac{1-\cos8A}{\cos8A}}{\frac{1-\cos4A}{\cos4A}}\\ & =\frac{1-\cos8A}{\cos8A}\times\frac{\cos4A}{1-\cos4A}\\ & =\frac{1-\left(1-2\sin^24A\right)}{\cos8A}\times\frac{\cos4A}{1-\left(1-2\sin^22A\right)}\\ & =\frac{2\sin^24A}{\cos8A}\times\frac{\cos4A}{2\sin^22A}\\ & =\frac{2\sin4A\cos4A}{\cos8A}\times\frac{\sin4A}{2\sin^22A}\\ & =\frac{\sin8A}{\cos8A}\times\frac{2\sin2A\cos2A}{2\sin^22A}\\ & =\tan8A\times\frac{\cos2A}{\sin2A}\\ & =\tan8A.\cot2A\\ & =\frac{\tan8A}{\tan2A}\\ & =\text{RHS} \end{align}

12.

Prove that:

12. (a) \( 4\sin \theta \sin (60^\circ - \theta ) \sin (60^\circ + \theta ) = \sin 3\theta \)

\begin{align} \text{LHS} & =4\sin \theta \sin (60^\circ - \theta ) \sin (60^\circ + \theta ) \\ & = 4\sin \theta ( \sin 60^\circ \cos \theta -\cos 60^\circ \sin \theta )( \sin 60^\circ \cos \theta + \cos 60^\circ \sin \theta ) \\ & = 4\sin \theta [ ( \sin 60^\circ \cos \theta)^2 -(\cos 60^\circ \sin \theta )^2 ]\\ & = 4\sin \theta \left[ \left( \frac{\sqrt{3}}{2} \cos \theta\right)^2 -\left( \frac{1}{2} \sin \theta\right)^2 \right]\\ & = 4\sin \theta \left[ \frac{3}{4} \cos^2 \theta - \frac{1}{4} \sin^2 \theta \right]\\ & = 4\sin \theta \times \frac{3\cos^2 \theta - \sin^2 \theta }{4} \\ & = \sin \theta ( 3\cos^2 \theta -\sin^2 \theta ) \\ & = \sin \theta [ 3(1-\sin^2 \theta ) - \sin^2 \theta ]\\ & = \sin \theta [ 3 - 3\sin^2 \theta - \sin^2 \theta ] \\ & = \sin \theta [ 3 - 4\sin^2 \theta ] \\ & = 3\sin \theta - 4\sin^3 \theta \\ & = \sin 3\theta \\ & =\text{RHS} \end{align}

Alternative

\begin{align} \text{LHS} & =4\sin \theta \sin (60^\circ - \theta ) \sin (60^\circ + \theta ) \\ & =2\sin \theta \{2 \sin (60^\circ - \theta ) \sin (60^\circ + \theta ) \} \\ & = 2\sin \theta [ \cos \{60^\circ-\theta - (60^\circ + \theta ) \} - \cos \{60^\circ-\theta + (60^\circ + \theta ) \}] \\ & [ \because 2\sin A \sin B = \cos(A-B) - \cos(A+B) ]\\ & = 2\sin \theta [ \cos(60^\circ -\theta -60^\circ - \theta) - \cos(60^\circ -\theta +60^\circ + \theta) ] \\ & = 2\sin \theta [ \cos (-2\theta) - \cos (120^\circ) ]\\ & = 2\sin \theta [ \cos 2\theta - \cos 120^\circ]\\ & = 2 \cos 2\theta \sin \theta - 2\sin \theta \cos 120^\circ]\\ & = \sin (2\theta + \theta ) - \sin (2\theta - \theta ) - 2\sin \theta \left(\frac{-1}{2}\right) \\ & [\because 2\cos A \sin B = \sin ( A + B) - \sin (A - B )]\\ & = \sin 3\theta - \sin \theta + \sin \theta \\ & = \sin 3\theta \\ & =\text{RHS} \end{align}

12. (b) \( \sin \theta ( \sin (120^\circ - \theta ) \sin (120^\circ + \theta ) = \frac{1}{4} \sin 3\theta \)

\begin{align} \text{LHS} & =\sin \theta ( \sin (120^\circ - \theta ) \sin (120^\circ + \theta ) \\ & = \sin \theta ( \sin 120^\circ \cos \theta -\cos 120^\circ \sin \theta )( \sin 120^\circ \cos \theta + \cos 120^\circ \sin \theta ) \\ & = \sin \theta [ ( \sin 120^\circ \cos \theta)^2 -(\cos 120^\circ \sin \theta )^2 ]\\ & = \sin \theta \left[ \left( \frac{\sqrt{3}}{2} \cos \theta\right)^2 -\left( \frac{-1}{2} \sin \theta\right)^2 \right]\\ & = \sin \theta \left[ \frac{3}{4} \cos^2 \theta - \frac{1}{4} \sin^2 \theta \right]\\ & = \sin \theta \times \frac{3\cos^2 \theta - \sin^2 \theta }{4} \\ & = \frac{\sin \theta ( 3\cos^2 \theta -\sin^2 \theta )}{4} \\ & = \frac{1}{4}\sin \theta [ 3(1-\sin^2 \theta ) - \sin^2 \theta ]\\ & = \frac{1}{4}\sin \theta [ 3 - 3\sin^2 \theta - \sin^2 \theta ] \\ & =\frac{1}{4} \sin \theta [ 3 - 4\sin^2 \theta ] \\ & = \frac{1}{4}[3\sin \theta - 4\sin^3 \theta] \\ & = \frac{1}{4}\sin 3\theta \\ & =\text{RHS} \end{align}

12. (c) \(\cos A \cos ( 60^\circ - A ) \cos (60^\circ + A ) = \frac{1}{4} \cos 3A \)

\begin{align} \text{LHS} & = \cos A \cos ( 60^\circ - A ) \cos (60^\circ + A ) \\ & = \cos A ( \cos 60^\circ \cos A + \sin 60^\circ \sin A )( \cos 60^\circ \cos A - \sin 60^\circ \sin A ) \\ & = \cos A [ ( \cos 60^\circ \cos A)^2 -(\sin 60^\circ \sin A )^2 ]\\ & = \cos A \left[ \left( \frac{1}{2} \cos A \right)^2 -\left( \frac{\sqrt{3}}{2} \sin A \right)^2 \right]\\ & = \cos A \times \frac{ \cos^2 A - 3 \sin^2 A }{4} \\ & = \frac{\cos A ( \cos^2 A -3\sin^2 A)}{4} \\ & = \frac{1}{4} \cos A [ \cos^2 A - 3( 1 - \cos^2 A ) ]\\ & = \frac{1}{4} \cos A [ \cos^2 A - 3 + 3\cos^2 A ]\\ & = \frac{1}{4} \cos A [ 4\cos^2 A - 3]\\ & = \frac{1}{4} [4\cos^3 A - 3\cos A ] \\ & = \frac{1}{4} \cos 3A \\ & =\text{RHS} \end{align}

12. (d) \( 4\sin \theta. \sin \left( \theta + \frac{\pi^c}{3} \right) . \sin \left( \theta + \frac{2\pi^c}{3} \right) = \sin 3\theta \)

\begin{align} \text{LHS} & = 4\sin \theta. \sin \left( \theta + \frac{\pi^c}{3} \right) . \sin \left( \theta + \frac{2\pi^c}{3} \right)\\ & = 4\sin \theta. \sin \left( \theta + 60^\circ \right) . \sin \left( \theta + 120^\circ \right)\\ & = 4\sin \theta (\sin \theta \cos 60^\circ + \cos \theta \sin 60^\circ )( \sin \theta \cos 120^\circ + \cos \theta \sin 120^\circ ) \\ & = 4\sin \theta \left(\sin \theta . \frac{1}{2} + \cos \theta \frac{\sqrt{3}}{2} )( \sin \theta . \frac{-1}{2} + \cos \theta \frac{\sqrt{3}}{2} \right) \\ & = 4\sin \theta \left( \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta \right).\left( \frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta \right)\\ & = 4\sin \theta \left[ \left( \frac{\sqrt{3}}{2} \cos \theta\right)^2 -\left( \frac{1}{2} \sin \theta\right)^2 \right]\\ & = 4\sin \theta \left[ \frac{3}{4} \cos^2 \theta - \frac{1}{4} \sin^2 \theta \right]\\ & = 4\sin \theta \times \frac{3\cos^2 \theta - \sin^2 \theta }{4} \\ & =\sin \theta ( 3\cos^2 \theta -\sin^2 \theta ) \\ & = \sin \theta [ 3(1-\sin^2 \theta ) - \sin^2 \theta ]\\ & = \sin \theta [ 3 - 3\sin^2 \theta - \sin^2 \theta ] \\ & = \sin \theta [ 3 - 4\sin^2 \theta ] \\ & = 3\sin \theta - 4\sin^3 \theta \\ & = \sin 3\theta \\ & =\text{RHS} \end{align}

12. (e) \(\tan A . \tan ( 60^\circ + A) .\tan (120^\circ + A) = - \tan 3A \)

\begin{align} \text{LHS} & =\tan A . \tan ( 60^\circ + A) .\tan (120^\circ + A) \\ & = \tan A \left(\frac{\tan 60^\circ + \tan A }{1-\tan 60^\circ. \tan A } \right) \left(\frac{\tan 120^\circ + \tan A }{1-\tan 120^\circ. \tan A } \right) \\ & = \tan A \left(\frac{\sqrt{3} + \tan A }{1-\sqrt{3}. \tan A } \right) \left(\frac{-\sqrt{3} + \tan A }{1+ \sqrt{3} \tan A } \right) \\ & = \tan A \left(\frac{\tan^2 A - (\sqrt{3})^2}{1^2 - (\sqrt{3} \tan A )^2 }\right) \\ & = \tan A \left( \frac{\tan^2 A -3 }{1 - 3\tan^2 A } \right)\\ & = \frac{\tan^3 A - 3\tan A }{1-3\tan^2 A }\\ & = - \left( \frac{3\tan A -\tan^3 A }{1-3\tan^2 A } \right) \\ & = -\tan 3A \\ & =\text{RHS} \end{align}

12. (f) \( \tan A + \tan (60^\circ + A ) + \tan (120^\circ + A ) = 3\tan 3A \)

\begin{align} \text{LHS} & = \tan A + \tan (60^\circ + A ) + \tan (120^\circ + A ) \\ & = \tan A + \frac{\tan 60^ \circ + \tan A }{1-\tan 60^ \circ \tan A } + \frac{\tan 120^ \circ + \tan A }{1-\tan 120^ \circ \tan A } \\ & = \tan A + \frac{\sqrt{3} + \tan A }{1-\sqrt{3} \tan A } + \frac{-\sqrt{3} + \tan A }{1+\sqrt{3} \tan A } \\ & = \tan A + \frac{ \tan A + \sqrt{3} }{1-\sqrt{3} \tan A } + \frac{ \tan A -\sqrt{3} }{1+\sqrt{3} \tan A } \\ & = \tan A + \frac{(\tan A + \sqrt{3} ) (1+\sqrt{3} \tan A ) + (\tan A - \sqrt{3} ) (1-\sqrt{3} \tan A ) }{1^2-\left(\sqrt{3} \tan A \right)^2 } \\ & = \tan A + \frac{\tan A + \sqrt{3} \tan^2 A + \sqrt{3} + 3\tan A + \tan A - \sqrt{3} \tan^2 A -\sqrt{3} + 3\tan A }{1-3\tan^2 A } \\ & = \tan A + \frac{8\tan A }{1-3\tan^2 A } \\ & = \frac{\tan A ( 1-3\tan^2 A ) +8\tan A }{1-3\tan^2 A }\\ & = \frac{ \tan A - 3\tan^3 A +8 \tan A }{1-3\tan^2 A} \\ & = \frac{ 9\tan A - 3\tan^3 A }{1-3\tan^2 A} \\ & = 3\left( \frac{3\tan A - \tan^3 A }{1-3\tan^2 A } \right) \\ & = 3\tan 3A \\ & =\text{RHS} \end{align}

12. (g) \( \cot \theta \cos (30^\circ - \theta ) - \sin (30^\circ - \theta ) = \frac{\sqrt{3}}{2} \text{ cosec} \theta \)

\begin{align} \text{LHS} & =\cot \theta \cos (30^\circ - \theta ) - \sin (30^\circ - \theta ) \\ & = \frac{\cos \theta } {\sin \theta } \cos (30^\circ - \theta ) - \sin (30^\circ -\theta ) \\ & = \frac{\cos \theta \cos (30^\circ - \theta ) - \sin \theta \sin (30^\circ - \theta)}{\sin \theta } \\ & = \frac{\cos(\theta + 30^\circ - \theta ) }{\sin \theta}\\ & = \frac{\cos 30^\circ }{\sin \theta }\\ & = \cos 30^\circ \text{cosec} \theta \\ & = \frac{\sqrt{3}}{{2}} \text{ cosec} \theta \\ & =\text{RHS} \end{align}

13.(a) \(\sin^2 A + \sin^2 (A + 120^\circ) + \sin^2 ( A - 120^\circ ) =\frac{3}{2} \)

\begin{align} \text{LHS } & =\sin^2A+\sin^2(A+120^{\circ})+\sin^2(A-120^{\circ})\\ &=\frac{1-\cos2A}{2}+\frac{1-\cos2\left(A+120^\circ\right)}{2}+\frac{1-\cos2\left(A-120^\circ\right)}{2}\\ &=\frac{1-\cos2A+1-\cos\left(2A+240^\circ\right)+1-\cos\left(2A-240^\circ\right)}{2}\\ &=\frac{3-\cos2A-\left\{\cos\left(2A+240^\circ\right)+\cos\left(2A-240^\circ\right)\right\}}{2}\\ &=\frac{3-\cos2A-2\cos2A\cos240^\circ}{2}\\ &=\frac{3-\cos2A-2\cos2A\times\left(-\frac{1}{2}\right)}{2}\\&\left[\because\cos240^\circ=\cos\left(180^\circ+60^\circ\right)=-\cos60^\circ=-\frac{1}{2}\right]\\ &=\frac{3-\cos2A+\cos2A}{2}\\ &=\frac{3}{2}\\ &=\text{RHS } \end{align}

13.(b) \(\cos^2A+\cos^2(A+120^{\circ})+\cos^2(A-120^{\circ}) =\frac{3}{2} \)

\begin{align} \text{LHS } & =\cos^2A+\cos^2(A+120^{\circ})+\cos^2(A-120^{\circ})\\ &=\frac{1+\cos2A}{2}+\frac{1+\cos2\left(A+120^\circ\right)}{2}+\frac{1+\cos2\left(A-120^\circ\right)}{2}\\ &=\frac{1+\cos2A+1+\cos\left(2A+240^\circ\right)+1+\cos\left(2A-240^\circ\right)}{2}\\ &=\frac{3+\cos2A+\left\{\cos\left(2A+240^{\circ}\right)+\cos\left(2A-240^{\circ}\right)\right\}}{2}\\ &=\frac{3+\cos2A+2\cos2A\cos240^\circ}{2}\\ &=\frac{3+\cos2A+2\cos2A\left(-\frac{1}{2}\right)}{2}\\&\left[\because\cos240^{\circ}=\cos\left(180^{\circ}+60^{\circ}\right)=-\cos60^{\circ}=-\frac{1}{2}\right]\\ &=\frac{3+\cos2A-\cos2A}{2}\\ &=\frac{3}{2}\\ &=\text{RHS } \end{align}

13.(c) \(-\sin^3\theta+\sin^3(60^{\circ}+\theta)+\sin^3(240^{\circ}-\theta) = \frac{3}{4}\sin3\theta\)

\begin{align} \text{LHS} & = -\sin^3\theta+\sin^3(60^{\circ}+\theta)+\sin^3(240^{\circ}-\theta)\\ &=-\sin^3\theta+\sin^3(60^{\circ}+ \theta)+\sin^3(180^\circ+60^\circ- \theta )\\ &=-\sin^3\theta+\sin^3(60^{\circ}+\theta)-\sin^3(60^{\circ}-\theta)\\ &=\frac{-\left(3\sin\theta-\sin3\theta\right)}{4}+\frac{3\sin\left(60^{\circ}+\theta\right)+\sin3\left(60^{\circ}+\theta\right)}{4}-\frac{3\sin\left(60^{\circ}-A\right)-\sin3\left(60^{\circ}-\theta\right)}{4}\\ &=\frac{-3\sin\theta+\sin3\theta+3\sin\left(60^{\circ}+\theta\right)-\sin\left(180^{\circ}+3\theta\right)-3\sin\left(60^{\circ}-\theta\right)+\sin\left(180^{\circ}-3\theta\right)}{4}\\ &=\frac{-3\sin\theta+\sin3\theta+3\sin\left(60^{\circ}+\theta\right)+\sin3\theta-3\sin\left(60^{\circ}-\theta\right)+\sin3\theta}{4}\\ &=\frac{-3\sin\theta+3\sin3\theta+3\left[\sin\left(60^{\circ}+\theta\right)-\sin\left(60^{\circ}-\theta\right)\right]}{4}\\ &=\frac{-3\sin\theta+3\sin3\theta+3\left(2\cos60^\circ\sin\theta\right)}{4}\\ &=\frac{-3\sin\theta+3\sin3\theta+3\left(2\times\frac{1}{2}\sin\theta\right)}{4}\\ &=\frac{-3\sin\theta+3\sin3\theta+3\sin\theta}{4}\\ &=\frac{3}{4}\sin3\theta\\ &=\text{RHS } \end{align}

14. (a) \(\text{cosec }2A+\cot4A = \cot A-\text{cosec }4A \)

\begin{align} \text{LHS } &=\text{cosec }2A+\cot4A\\ &=\frac{1}{\sin2A}+\frac{\cos4A}{\sin4A}\\ &=\frac{1}{\sin2A}+\frac{2\cos^22A-1}{2\sin2A\cos2A}\\ &=\frac{2\cos2A+2\cos^22A-1}{2\sin2A\cos2A}\\ &=\frac{2\cos2A\left(1+\cos2A\right)-1}{\sin4A}\\ &=\frac{2\cos2A\left(1+\cos2A\right)}{2\sin2A\cos2A}-\frac{1}{2\sin2A\cos2A}\\ &=\frac{1+\cos2A}{\sin2A}-\frac{1}{\sin4A}\\ &=\frac{1+2\cos^2A-1}{2\sin A\cos A}-\text{cosec }4A\\ &=\frac{2\cos^2A}{2\sin A\cos A}-\text{cosec }4A\\ &=\frac{\cos A}{\sin A}-\text{cosec }4A\\ &=\cot A-\text{cosec }4A\\ &=\text{RHS } \end{align}

14. (b) \(\text{cosec } 4A+\cot 8A = \cot 2A-\text{cosec } 8A \)

\begin{align} \text{LHS } &=\text{cosec } 4A+\cot 8A \\ &=\frac{1}{\sin4A}+\frac{\cos8A}{\sin8A}\\ &=\frac{1}{\sin4A}+\frac{2\cos^24A-1}{2\sin4A\cos4A}\\ &=\frac{2\cos4A+2\cos^24A-1}{2\sin4A\cos4A}\\ &=\frac{2\cos4A\left(1+\cos4A\right)-1}{\sin8A}\\ &=\frac{2\cos4A\left(1+\cos4A\right)}{2\sin4A\cos4A}-\frac{1}{2\sin4A\cos4A}\\ &=\frac{1+\cos4A}{\sin4A}-\frac{1}{\sin8A}\\ &=\frac{1+2\cos^2 2A-1}{2\sin 2A\cos 2A}-\text{cosec }8A\\ &=\frac{2\cos^2 2A}{2\sin 2A\cos 2A}-\text{cosec } 8A\\ &=\frac{\cos 2A}{\sin 2A}-\text{cosec } 8A\\ &=\cot 2A-\text{cosec } 8A\\ &=\text{RHS } \end{align}

15. (a) \( \frac{\cos^3\theta-\cos3\theta}{\cos\theta}+\frac{\sin^3\theta+\sin3\theta}{\sin\theta}=3 \)

\begin{align} \text{LHS} & =\frac{\cos^3\theta-\cos3\theta}{\cos\theta}+\frac{\sin^3\theta+\sin3\theta}{\sin\theta}\\ &=\frac{\cos^3\theta-\left(4\cos^3\theta-3\cos\theta\right)}{\cos\theta}+\frac{\sin^3\theta+\left(3\sin\theta-4\sin^3\theta\right)}{\sin\theta}\\ &=\frac{\cos^3\theta-4\cos^3\theta+3\cos\theta}{\cos\theta}+\frac{\sin^3\theta+3\sin\theta-4\sin^3\theta}{\cos\theta}\\ &=\frac{3\cos\theta-3\cos^3\theta}{\cos\theta}+\frac{3\sin\theta-3\sin^3\theta}{\sin\theta}\\ &=\frac{3\cos\theta}{\cos\theta}-\frac{3\cos^3\theta}{\cos\theta}+\frac{3\sin\theta}{\sin\theta}-\frac{3\sin^3\theta}{\sin\theta}\\ &=3-3\cos^2\theta+3-3\sin^2\theta\\ &=6-3\left(\sin^2\theta+\cos^2\theta\right)\\ &=6-3\left(1\right)\\ &=6-3\\ &=3\\ &=\text{RHS} \end{align}

15. (b) \(\cos^3\alpha.\cos3\alpha+\sin^3\alpha.\sin3\alpha = \cos^3\alpha \)

\begin{align} \text{LHS} & =\cos^3\alpha.\cos3\alpha+\sin^3\alpha.\sin3\alpha\\ &=\frac{3\cos\alpha+\cos3\alpha}{4}\times\cos3\alpha+\frac{3\sin\alpha-\sin3\alpha}{4}\times\sin3\alpha\\ &=\frac{3\cos3\alpha\cos\alpha+\cos^23\alpha}{4}+\frac{3\sin3\alpha\sin\alpha-\sin^23\alpha}{4}\\ &=\frac{3\cos3\alpha\cos\alpha+\cos^23\alpha+3\sin3\alpha\sin\alpha-\sin^23\alpha}{4}\\ &=\frac{3\left(\cos3\alpha\cos\alpha+\sin3\alpha\sin\alpha\right)+\left(\cos^23\alpha-\sin^23\alpha\right)}{4}\\ &=\frac{3\cos\left(3\alpha-\alpha\right)+\cos2\left(3\alpha\right)}{4}\\ &=\frac{3\cos\left(3\alpha-\alpha\right)+\cos6\alpha}{4}\\ &=\frac{3\cos2\alpha+\cos3\left(2\alpha\right)}{4}\\ &=\frac{3\cos2\alpha+4\cos^3 2\alpha-3\cos 2\alpha}{4}\\ &=\frac{4\cos^3 2\alpha}{4}\\ &=\cos^3 2\alpha\\ &=\text{RHS} \end{align}

15. (c) \( 4\sin^3\alpha\cos3\alpha+4\cos^3\alpha\sin3\alpha=3\sin4\alpha \)

\begin{align} \text{LHS} &=4\sin^3\alpha\cos3\alpha+4\cos^3\alpha\sin3\alpha\\ &=4\left(\frac{3\sin\alpha-\sin3\alpha}{4}\right)\times\cos3\alpha+4\left( \frac{3\cos\alpha+\cos3\alpha}{4} \right) \times\sin3\alpha\\ &= 3\sin\alpha\cos3\alpha -\sin3\alpha\cos3\alpha + 3\cos\alpha\sin3\alpha +\cos3\alpha\sin3\alpha \\ &= 3\sin\alpha \cos3\alpha -\sin3\alpha\cos3\alpha+3\cos\alpha\sin3\alpha+\sin3\alpha\cos3\alpha \\ &= 3\left(\sin\alpha\cos3\alpha+\cos\alpha\sin3\alpha\right)\\ &=3\sin\left(\alpha+3\alpha\right) \\ &=3\sin4\alpha\\ &=\text{RHS} \end{align}

16. (a) \( \cos^2\frac{\pi^c}{8}+\cos^2\frac{3\pi^c}{8}+\cos^2\frac{5\pi^c}{8}+\cos^2\frac{7\pi^c}{8} = 2 \)

\begin{align} \text{LHS} &=\cos^2\frac{\pi^c}{8}+\cos^2\frac{3\pi^c}{8}+\cos^2\frac{5\pi^c}{8}+\cos^2\frac{7\pi^c}{8}\\ &=\cos^2\frac{\pi^c}{8}+\cos^2\frac{3\pi^c}{8}+\cos^2\left(\frac{8\pi^c-3\pi^c}{8}\right)+\cos^2\left(\frac{8\pi^c-\pi^c}{8}\right)\\ &=\cos^2\frac{\pi^c}{8}+\cos^2\frac{3\pi^c}{8}+\cos^2\left(\frac{8\pi^c}{8}-\frac{3\pi^c}{8}\right)+\cos^2\left(\frac{8\pi^c}{8}-\frac{\pi^c}{8}\right)\\ &=\cos^2\frac{\pi^c}{8}+\cos^2\frac{3\pi^c}{8}+\cos^2\left(\pi^c-\frac{3\pi^c}{8}\right)+\cos^2\left(\pi^c-\frac{\pi^c}{8}\right)\\ &=\cos^2\frac{\pi^c}{8}+\cos^2\frac{3\pi^c}{8}+\cos^2\frac{3\pi^c}{8}+\cos^2\frac{\pi^c}{8}\\ &=2\cos^2\frac{\pi^c}{8}+2\cos^2\frac{3\pi^c}{8}\\ &=1+\cos2\left(\frac{\pi^c}{8}\right)+1+\cos2\left(\frac{3\pi^c}{8}\right)\\ &=1+\cos\frac{\pi^c}{4}+1+\cos\frac{3\pi^c}{4}\\ &=1+\frac{1}{\sqrt{2}}+1-\frac{1}{\sqrt{2}}\\&\left[\because\cos45^\circ=\frac{1}{\sqrt{2}}\text{ and }\cos135^\circ=-\frac{1}{\sqrt{3}}\right]\\ &=2\\ &=\text{RHS} \end{align}

16. (b) \( \sin^4\frac{\pi^c}{8}+\sin^4\frac{3\pi^c}{8}+\sin^4\frac{5\pi^c}{8}+\sin^4\frac{7\pi^c}{8} = \frac{3}{2} \)

\begin{align} \text{LHS} &=\sin^4\frac{\pi^c}{8}+\sin^4\frac{3\pi^c}{8}+\sin^4\frac{5\pi^c}{8}+\sin^4\frac{7\pi^c}{8}\\ &=\sin^4\frac{\pi^c}{8}+\sin^4\frac{3\pi^c}{8}+\sin^4\left(\frac{8\pi^c}{8}-\frac{3\pi^c}{8}\right)+\sin^4\left(\frac{8\pi^c}{8}-\frac{\pi^c}{8}\right)\\ &=\sin^4\frac{\pi^c}{8}+\sin^4\frac{3\pi^c}{8}+\sin^4\frac{3\pi^c}{8}+\sin^4\frac{\pi^c}{8}\\ &=2\sin^4\frac{\pi^c}{8}+2\sin^4\frac{3\pi^c}{8}\\ &=2\left(\sin^2\frac{\pi^c}{8}\right)^2+2\left(\sin^2\frac{3\pi^c}{8}\right)^2\\ &=2\left(\frac{1-\cos2\left(\frac{\pi^c}{8}\right)}{2}\right)^2+2\left(\frac{1-\cos2\left(\frac{3\pi^c}{8}\right)}{2}\right)^2\\ &=2\left(\frac{1-\cos45^\circ}{2}\right)^2+2\left(\frac{1-\cos135^\circ}{2}\right)^2\\ &=2\times\frac{1}{4}\left(1-\cos45^\circ\right)^2+2\times\frac{1}{4}\left(1-\cos135^\circ\right)^2\\ &=\frac{1}{2}\left(1-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}\left(1+\frac{1}{\sqrt{2}}\right)^2\\ &=\frac{1}{2}\left\{1^2-2\times1\times\frac{1}{\sqrt{2}}+\left(\frac{1}{\sqrt{2}}\right)^2\right\}+\frac{1}{2}\left\{1^2+2\times1\times\frac{1}{\sqrt{2}}+\left(\frac{1}{\sqrt{2}}\right)^2\right\}\\ &=\frac{1}{2}\left(1-\frac{2}{\sqrt{2}}+\frac{1}{2}+1+\frac{2}{\sqrt{2}}+\frac{1}{2}\right)\\ &=\frac{1}{2}\left(1+\frac{1}{2}+1+\frac{1}{2}\right)\\ &=\frac{1}{2}\left(3\right)\\ &=\frac{3}{2}\\ &=\text{RHS} \end{align}

16. (c) \(\cos^4\frac{\pi^c}{8}+\cos^4\frac{3\pi^c}{8}+\cos^4\frac{5\pi^c}{8}+\cos^4\frac{7\pi^c}{8} = \frac{3}{2} \)

\begin{align} \text{LHS} &=\cos^4\frac{\pi^c}{8}+\cos^4\frac{3\pi^c}{8}+\cos^4\frac{5\pi^c}{8}+\cos^4\frac{7\pi^c}{8}\\ &=\cos^4\frac{\pi^c}{8}+\cos^4\frac{3\pi^c}{8}+\cos^4\left(\frac{8\pi^c}{8}-\frac{3\pi^c}{8}\right)+\cos^4\left(\frac{8\pi^c}{8}-\frac{\pi^c}{8}\right)\\ &=\cos^4\frac{\pi^c}{8}+\cos^4\frac{3\pi^c}{8}+\cos^4\frac{3\pi^c}{8}+\cos^4\frac{\pi^c}{8}\\ &=2\cos^4\frac{\pi^c}{8}+2\cos^4\frac{3\pi^c}{8}\\ &=2\left(\cos^2\frac{\pi^c}{8}\right)^2+2\left(\cos^2\frac{3\pi^c}{8}\right)^2\\ &=2\left(\frac{1+\cos2\left(\frac{\pi^c}{8}\right)}{2}\right)^2+2\left(\frac{1+\cos2\left(\frac{3\pi^c}{8}\right)}{2}\right)^2\\ &=2\left(\frac{1+\cos45^{\circ}}{2}\right)^2+2\left(\frac{1+\cos135^{\circ}}{2}\right)^2\\ &=2\times\frac{1}{4}\left(1+\cos45^{\circ}\right)^2+2\times\frac{1}{4}\left(1+\cos135^{\circ}\right)^2\\ &=\frac{1}{2}\left(1+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}\left(1-\frac{1}{\sqrt{2}}\right)^2\\ &=\frac{1}{2}\left\{1^2+2\times1\times\frac{1}{\sqrt{2}}+\left(\frac{1}{\sqrt{2}}\right)^2\right\}+\frac{1}{2}\left\{1^2-2\times1\times\frac{1}{\sqrt{2}}+\left(\frac{1}{\sqrt{2}}\right)^2\right\}\\ &=\frac{1}{2}\left(1+\frac{2}{\sqrt{2}}+\frac{1}{2}+1-\frac{2}{\sqrt{2}}+\frac{1}{2}\right)\\ &=\frac{1}{2}\left(1+\frac{1}{2}+1+\frac{1}{2}\right)\\ &=\frac{1}{2}\left(3\right)\\ &=\frac{3}{2}\\ &=\text{RHS} \end{align}

16. (d) \( \left(1+\cos\frac{\pi^c}{8}\right)\left(1+\cos\frac{3\pi^c}{8}\right) \left(1+\cos\frac{5\pi^c}{8}\right)\left(1 + \cos\frac{7\pi^c}{8}\right) = \frac{1}{8} \)

\begin{align} \text{LHS} &= \left(1+\cos\frac{\pi^c}{8}\right)\left(1+\cos\frac{3\pi^c}{8}\right) \left(1+\cos\frac{5\pi^c}{8}\right)\left(1 + \cos\frac{7\pi^c}{8}\right)\\ &=\left(1+\cos\frac{\pi^c}{8}\right)\left(1+\cos\frac{3\pi^c}{8}\right) \left\{1+\cos\left(\pi^c-\frac{3\pi^c}{8}\right)\right\}\left\{1 + \cos\left(\pi^c-\frac{\pi^c}{8}\right)\right\}\\ &=\left(1+\cos\frac{\pi^c}{8}\right)\left(1+\cos\frac{3\pi^c}{8}\right) \left(1-\cos\frac{3\pi^c}{8}\right)\left(1-\cos\frac{\pi^c}{8}\right)\\ &=\left(1^2-\cos^2\frac{\pi^c}{8}\right)\left(1^2-\cos^2\frac{3\pi^c}{8}\right)\\ &= \left(1-\cos^2\frac{\pi^c}{8}\right)\left(1-\cos^2\frac{3\pi^c}{8}\right)\\ &= \sin^2\frac{\pi^c}{8}\times\sin^2\frac{3\pi^c}{8}\\ &= \frac{1-\cos2\left(\frac{\pi^c}{8}\right)}{2}\times\frac{1-\cos2\left(\frac{3\pi^c}{8}\right)}{2}\\ &=\frac{1}{4}\left(1-\cos\frac{\pi^c}{4}\right)\left(1 - \cos\frac{3\pi^c}{4}\right)\\ &=\frac{1}{4}\left(1- \cos45^\circ\right)\left(1 - \cos135^\circ\right)\\ &=\frac{1}{4}\left(1-\frac{1}{\sqrt{2}}\right)\left(1+\frac{1}{\sqrt{2}}\right)\\ &=\frac{1}{4}\left\{1^2-\left(\frac{1}{\sqrt{2}}\right)^2\right\}\\ &=\frac{1}{4}\left(1-\frac{1}{2}\right)\\ &=\frac{1}{4}\left(\frac{1}{2}\right)\\ &=\frac{1}{8}\\ &=\text{RHS} \end{align}

17. (a) \( 2\cos\theta = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+2\cos16\theta}}}} \)

\begin{align} \text{RHS} &=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+2\cos16\theta}}}}\\ &=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+2\left(2\cos^28\theta-1\right)}}}}\\ &=\sqrt{2+\sqrt{2+\sqrt{2+}\sqrt{2+4\cos^28\theta-2}}}\\ &=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{4\cos^28\theta}}}}\\ &=\sqrt{2+\sqrt{2+\sqrt{2+2\cos8\theta}}}\\ &=\sqrt{2+\sqrt{2+\sqrt{2+2\left(2\cos^24\theta-1\right)}}}\\ &=\sqrt{2+\sqrt{2+\sqrt{2+4\cos^24\theta-2}}}\\ &=\sqrt{2+\sqrt{2+\sqrt{4\cos^24\theta}}}\\ &=\sqrt{2+\sqrt{2+2\cos4\theta}}\\ &=\sqrt{2+\sqrt{2+2\left(2\cos^22\theta-1\right)}}\\ &=\sqrt{2+\sqrt{2+4\cos^22\theta-2}}\\ &=\sqrt{2+\sqrt{4\cos^22\theta}}\\ &=\sqrt{2+2\cos2\theta}\\ &=\sqrt{2+2\left(2\cos^2\theta-1\right)}\\ &=\sqrt{2+4\cos^2\theta-2}\\ &=\sqrt{4\cos^2\theta}\\ &=2\cos\theta\\ &=\text{LHS} \end{align}

17. (b) \( \sqrt{2+\sqrt{2+\sqrt{2+2\cos8 A}}} = 2\cos A \)

\begin{align} \text{LHS} &=\sqrt{2+\sqrt{2+\sqrt{2+2\cos8 A}}}\\ &=\sqrt{2+\sqrt{2+\sqrt{2+2\left(2\cos^24 A-1\right)}}}\\ &=\sqrt{2+\sqrt{2+\sqrt{2+4\cos^24 A-2}}}\\ &=\sqrt{2+\sqrt{2+\sqrt{4\cos^24 A}}}\\ &=\sqrt{2+\sqrt{2+2\cos4 A}}\\ &=\sqrt{2+\sqrt{2+2\left(2\cos^22 A-1\right)}}\\ &=\sqrt{2+\sqrt{2+4\cos^22 A-2}}\\ &=\sqrt{2+\sqrt{4\cos^22 A}}\\ &=\sqrt{2+2\cos2 A}\\ &=\sqrt{2+2\left(2\cos^2 A-1\right)}\\ &=\sqrt{2+4\cos^2 A-2}\\ &=\sqrt{4\cos^2 A}\\ &=2\cos A\\ &=\text{RHS} \end{align}

18.

\( \cos^2\theta+\sin^2\theta.\cos2\beta=\cos^2\beta+\sin^2\beta.\cos2\theta \)

\begin{align} \text{ LHS} \ & =\cos^2\theta+\sin^2\theta.\cos2\beta\\ \ & =\cos^2\theta+\sin^2\theta\left(1-2\sin^2\beta\right)\\ \ & =\cos^2\theta+\sin^2\theta-2\sin^2\theta\sin^2\beta\\ \ & =1-2\sin^2\theta\sin^2\beta\\ \ & =\cos^2\beta+\sin^2\beta-2\sin^2\theta\sin^2\beta\\ \ & =\cos^2\beta+\sin^2\beta\left(1-2\sin^2\theta\right)\\ \ & =\cos^2\beta+\sin^2\beta\cos2\theta\\ \ & = \text{ RHS}\\ \end{align}

19. (a) Prove that:

\( \tan\theta+2\tan2\theta+4\tan4\theta+8\cot8\theta = \cot\theta \)

\begin{align} \text{ LHS} &=\tan\theta+2\tan2\theta+4\tan4\theta+8\cot8\theta\\ &=\tan\theta+2\tan2\theta+4\tan4\theta+\frac{8}{\tan8\theta}\\ &=\tan\theta+2\tan2\theta+4\tan4\theta+\frac{8}{\frac{2\tan4\theta}{1-\tan^24\theta}}\\ &=\tan\theta+2\tan2\theta+4\tan4\theta+\frac{8\left(1-\tan^24\theta\right)}{2\tan4\theta}\\ &=\tan\theta+2\tan2\theta+4\tan4\theta+\frac{4\left(1-\tan^24\theta\right)}{\tan4\theta}\\ &=\tan\theta+2\tan2\theta+\frac{4\tan^24\theta+4-4\tan^24\theta}{\tan4\theta}\\ &=\tan\theta+2\tan2\theta+\frac{4}{\tan4\theta}\\ &=\tan\theta+2\tan2\theta+\frac{4}{\frac{2\tan2\theta}{1-\tan^22\theta}}\\ &=\tan\theta+2\tan2\theta+\frac{4\left(1-\tan^22\theta\right)}{2\tan2\theta}\\ &=\tan\theta+2\tan2\theta+\frac{2\left(1-\tan^22\theta\right)}{\tan2\theta}\\ &=\tan\theta+2\tan2\theta+\frac{2-2\tan^22\theta}{\tan2\theta}\\ &=\tan\theta+\frac{2\tan^22\theta+2-2\tan^22\theta}{\tan2\theta}\\ &=\tan\theta+\frac{2}{\tan2\theta}\\ &=\tan\theta+\frac{2}{\frac{2\tan\theta}{1-\tan^2\theta}}\\ &=\tan\theta+\frac{2\left(1-\tan^2\theta\right)}{2\tan\theta}\\ &=\tan\theta+\frac{1-\tan^2\theta}{\tan\theta}\\ &=\frac{\tan^2\theta+1-\tan^2\theta}{\tan\theta}\\ &=\frac{1}{\tan\theta}\\ &=\cot\theta\\ & = \text{RHS} \end{align}

19. (b) Prove that:

\( \frac{2\sin A}{\cos3 A}+\frac{2\sin3 A}{\cos9 A}+\frac{2\sin9 A}{\cos27 A} = \tan27 A-\tan A \)

\begin{align} \text{ LHS} & =\frac{2\sin A}{\cos3 A}+\frac{2\sin3 A}{\cos9 A}+\frac{2\sin9 A}{\cos27 A}\\ & =\frac{2\sin A\cos A}{\cos3 A\cos A}+\frac{2\sin3 A\cos3 A}{\cos9 A\cos3 A}+\frac{2\sin9 A\cos9 A}{\cos27 A\cos9 A}\\ & =\frac{\sin2 A}{\cos3 A\cos A}+\frac{\sin6 A}{\cos9 A\cos3 A}+\frac{\sin18 A}{\cos27 A\cos9 A}\\ & =\frac{\sin\left(3 A- A\right)}{\cos3 A\cos A}+\frac{\sin\left(9 A-3 A\right)}{\cos9 A\cos3 A}+\frac{\sin\left(27 A-9 A\right)}{\cos27 A\cos9 A}\\ & =\frac{\sin3 A\cos A-\cos3 A\sin A}{\cos3 A\cos A} +\frac{\sin9 A\cos3 A-\cos9 A\sin3 A}{\cos9 A\cos3 A} +\frac{\sin27 A\cos9 A-\cos27 A\sin9 A}{\cos27 A\cos9 A}\\ & =\frac{\sin3 A\cos A}{\cos3 A\cos A}-\frac{\cos3 A\sin A}{\cos3 A\cos A}+\frac{\sin9 A\cos3 A}{\cos9 A\cos3 A}-\frac{\cos9 A\sin3 A}{\cos9 A\cos3 A}+\frac{\sin27 A\cos9 A}{\cos27 A\cos9 A}-\frac{\cos27 A\sin9 A}{\cos27 A\cos9 A}\\ & =\frac{\sin3 A}{\cos3 A}-\frac{\sin A}{\cos A}+\frac{\sin9 A}{\cos9 A}-\frac{\sin3 A}{\cos3 A}+\frac{\sin27 A}{\cos27 A}-\frac{\sin9 A}{\cos9 A}\\ & =\frac{\sin27 A}{\cos27 A}-\frac{\sin A}{\cos A}\\ & =\tan27 A-\tan A\\ & = \text{RHS} \end{align}

19. (c) Prove that:

\( \frac{2\sin 2\theta}{\cos 6\theta}+\frac{2\sin 6\theta}{\cos 18\theta}+\frac{2\sin 18\theta}{\cos 54\theta} = \tan54\theta-\tan2\theta \)

\begin{align} \text{ LHS} & =\frac{2\sin 2\theta}{\cos 6\theta}+\frac{2\sin 6\theta}{\cos 18\theta}+\frac{2\sin 18\theta}{\cos 54\theta}\\ & =\frac{2\sin2\theta\cos2\theta}{\cos6\theta\cos2\theta}+\frac{2\sin6\theta\cos6\theta}{\cos18\theta\cos6\theta}+\frac{2\sin18\theta\cos18\theta}{\cos54\theta\cos18\theta}\\ & =\frac{\sin4\theta}{\cos6\theta\cos2\theta}+\frac{\sin12\theta}{\cos18\theta\cos6\theta}+\frac{\sin36\theta}{\cos54\theta\cos18\theta}\\ & =\frac{\sin\left(6\theta-2\theta\right)}{\cos6\theta\cos2\theta}+\frac{\sin\left(18\theta-6\theta\right)}{\cos18\theta\cos6\theta}+\frac{\sin\left(54\theta-18\theta\right)}{\cos54\theta\cos18\theta}\\ & =\frac{\sin6\theta\cos2\theta-\cos6\theta\sin2\theta}{\cos6\theta\cos2\theta} +\frac{\sin18\theta\cos6\theta-\cos18\theta\sin6\theta}{\cos18\theta\cos6\theta}+\frac{\sin54\theta\cos18\theta-\cos54\theta\sin18\theta}{\cos54\theta\cos18\theta}\\ & =\frac{\sin6\theta\cos2\theta}{\cos6\theta\cos2\theta}-\frac{\cos6\theta\sin2\theta}{\cos6\theta\cos3\theta}+\frac{\sin18\theta\cos6\theta}{\cos18\theta\cos6\theta}-\frac{\cos18\theta\sin6\theta}{\cos18\theta\cos6\theta}+\frac{\sin54\theta\cos18\theta}{\cos54\theta\cos18\theta}-\frac{\cos54\theta\sin18\theta}{\cos54\theta\cos18\theta}\\ & =\frac{\sin6\theta}{\cos6\theta}-\frac{\sin2\theta}{\cos2\theta}+\frac{\sin18\theta}{\cos18\theta}-\frac{\sin6\theta}{\cos6\theta}+\frac{\sin54\theta}{\cos54\theta}-\frac{\sin18\theta}{\cos18\theta}\\ & =\frac{\sin54\theta}{\cos54\theta}-\frac{\sin2\theta}{\cos2\theta}\\ & =\tan54\theta-\tan2\theta\\ & = \text{RHS} \end{align}

Prove that:

20. (a) \(\cos^6\theta-\sin^6\theta = \frac{1}{3}\left(\cos^32\theta+3\cos2\theta\right) \)

\begin{align} \text{ LHS} &=\cos^6\theta-\sin^6\theta\\ &=\left(\cos^2\right)^3-\left(\sin^2\right)^3\\ &=\left(\cos^2\theta-\sin^2\theta\right)\left\{\left(\cos^2\theta\right)^2+\cos^2\theta\sin^2\theta+\left(\sin^2\right)^2\right\}\\ &=\cos2\theta\left\{\left(\cos^2\theta\right)^2+2\cos^2\theta\sin^2\theta+\left(\sin^2\right)^2-\cos^2\theta\sin^2\theta\right\}\\ &=\cos2\theta\left\{\left(\cos^2\theta+\sin^2\theta\right)^2-\frac{1}{4}\left(2\sin\theta\cos\theta\right)^2\right\}\\ &=\cos2\theta\left\{\left(1\right)^2-\frac{1}{4}\left(\sin2\theta\right)^2\right\}\\ &=\cos2\theta\left\{1-\frac{1}{4}\sin^22\theta\right\}\\ &=\cos2\theta\left\{\frac{4-\sin^22\theta}{4}\right\}\\ &=\cos2\theta\left\{\frac{4-\left(1-\cos^22\theta\right)}{4}\right\}\\ &=\cos2\theta\left(\frac{4-1+\cos^22\theta}{4}\right)\\ &=\cos2\theta\left(\frac{3+\cos^22\theta}{4}\right)\\ &=\frac{3\cos2\theta+\cos^3\theta}{4}\\ &=\frac{1}{4}\left(\cos^32\theta+3\cos2\theta\right)\\ & = \text{RHS} \end{align}

20. (b) \( \sin^6 \theta -\cos^6 \theta =-\frac{1}{4} ( \cos^3 2\theta + 3\cos 2\theta ) \)

\begin{align} \text{ LHS} & = \sin^6 \theta -\cos^6 \theta \\ & = (\sin^2 \theta )^3 - (\cos^2 \theta )^3\\ & = ( \sin^2 \theta - \cos^2 \theta )\{ ( \sin^2 \theta )^2 + \sin^2 \theta \cos^2 \theta + (\cos^2 \theta)^2 \}\\ & = -( \cos^2 \theta - \sin^2 \theta ) \left\{ ( \sin^2 \theta )^2 +2 \sin^2 \theta \cos^2 \theta + (\cos^2 \theta)^2 - \sin^2 \theta \cos^2 \theta \right\}\\ & = - \cos 2\theta \left\{(\sin^2 \theta + \cos^2 \theta )^2 - \frac{1}{4} ( 2\sin \theta \cos \theta )^2 \right\} \\ & = - \cos 2\theta \left\{(1 )^2 - \frac{1}{4} ( \sin 2\theta)^2 \right\} \\ & = - \cos 2\theta \left\{1 - \frac{1}{4} \sin^2 2\theta \right\} \\ & = - \cos 2\theta \left\{ \frac{4- \sin^2 2\theta }{4} \right\} \\ & = - \cos 2\theta \left\{ \frac{4- (1-\cos^2 2\theta) }{4} \right\} \\ & = - \cos 2\theta \left\{ \frac{4- 1 + \cos^2 2\theta }{4} \right\} \\ & = - \cos 2\theta \left\{ \frac{ \cos^2 2\theta + 3 }{4} \right\} \\ & = -\frac{1}{4} ( \cos^3 2\theta + 3\cos 2\theta )\\ & = \text{RHS} \end{align}

20. (c) \( \sin^6 \theta +\cos^6 \theta = \frac{1}{4} ( 3\cos^2 2\theta + 1 ) \)

\begin{align} \text{ LHS} & = \sin^6 \theta +\cos^6 \theta \\ & = (\sin^2 \theta )^2 + ( \cos^2 \theta )^2 \\ & = (\sin^2 \theta + \cos^2 \theta )\{(\sin^2 \theta)^2 - \sin^2 \theta \cos^2 \theta +(\cos^2 \theta )^2 \}\\ & = (1) \{ (\sin^2 \theta)^2 +2\sin^2 \theta \cos^2 \theta +(\cos^2 \theta )^2 -3 \sin^2 \theta \cos^2 \theta \}\\ & = (\sin^2 \theta + \cos^2 \theta)^2 - \frac{3}{4} (2\sin \theta \cos \theta )^2 \\ & = (1)^2 - \frac{3}{4} (\sin 2\theta )^2 \\ & = 1 -\frac{3}{4} \sin^2 2\theta \\ & = 1 -\frac{3}{4} (1-\cos^2 2\theta ) \\ & = \frac{4-3+3\cos^2 2\theta }{4}\\ & = \frac{1}{4} ( 3\cos^2 2\theta + 1 )\\ & = \text{RHS} \end{align}

20. (d) \( \sin^6 \theta +\cos^6 \theta = \frac{1}{4} ( 4 - 3\sin^2 2\theta ) \)

\begin{align} \text{ LHS} & = \sin^6 \theta +\cos^6 \theta \\ & = (\sin^2 \theta )^2 + ( \cos^2 \theta )^2 \\ & = (\sin^2 \theta + \cos^2 \theta )\{(\sin^2 \theta)^2 - \sin^2 \theta \cos^2 \theta +(\cos^2 \theta )^2 \}\\ & = (1) \{ (\sin^2 \theta)^2 +2\sin^2 \theta \cos^2 \theta +(\cos^2 \theta )^2 -3 \sin^2 \theta \cos^2 \theta \}\\ & = (\sin^2 \theta + \cos^2 \theta)^2 - \frac{3}{4} (2\sin \theta \cos \theta )^2 \\ & = (1)^2 - \frac{3}{4} (\sin 2\theta )^2 \\ & = 1 -\frac{3}{4} \sin^2 2\theta \\ & =\frac{4-3\sin^2 2\theta }{4}\\ & = \frac{1}{4} ( 4 - 3\sin^2 2\theta )\\ & = \text{RHS} \end{align}

20. (e) \( \cos^8\theta+\sin^8\theta = 1-\sin^22\theta+\frac{1}{8}\sin^42\theta \)

\begin{align} \text{ LHS} & =\cos^8\theta+\sin^8\theta\\ & =\left(\cos^4\theta\right)^2+\left(\sin^4\theta\right)^2\\ & =\left(\cos^4\theta-\sin^4\theta\right)^2+2\cos^4\theta\sin^4\theta\\ & =\left\{\left(\cos^2\theta\right)^2-\left(\sin^2\theta\right)^2\right\}^2+\frac{2^4}{2^3}\cos^4\theta\sin^4\theta\\ & =\left\{\left(\cos^2\theta+\sin^2\theta\right)\left(\cos^2\theta-\sin^2\theta\right)\right\}^2\\& +\frac{1}{8}\left(2\sin\theta\cos\theta\right)^4\\ & =\left\{1\times\cos2\theta\right\}^2+\frac{1}{8}\left(\sin2\theta\right)^4\\ & =\cos^22\theta+\frac{1}{8}\sin^42\theta\\ & =1-\sin^22\theta+\frac{1}{8}\sin^42\theta\\ & = \text{RHS} \end{align}

Prove that:

21. (a) \( \cos\frac{\pi^c}{7}.\cos\frac{2\pi^c}{7}.\cos\frac{3\pi^c}{7} = \frac{1}{8} \)

\begin{align} \text{ LHS} & =\cos\frac{\pi^c}{7}.\cos\frac{2\pi^c}{7}.\cos\frac{3\pi^c}{7}\\ & =\frac{\left(2\sin\frac{\pi^c}{7}\cos\frac{\pi^c}{7}\right)\cos\frac{2\pi^c}{7}\cos\frac{3\pi^c}{7}}{2\sin\frac{\pi^c}{7}}\\ & =\frac{\sin\frac{2\pi^c}{7}\cos\frac{2\pi^c}{7}\cos\frac{3\pi^c}{7}}{2\sin\frac{\pi^c}{7}}\\ & =\frac{\left(2\sin\frac{2\pi^c}{7}\cos\frac{2\pi^c}{7}\right)\cos\frac{3\pi^c}{7}}{2\times2\sin\frac{\pi^c}{7}}\\ & =\frac{\sin\frac{4\pi^c}{7}\cos\frac{3\pi^c}{7}}{4\sin\frac{\pi^c}{7}}\\ & =\frac{2\sin\frac{4\pi^c}{7}\cos\left(\pi^c-\frac{4\pi^c}{7}\right)}{2\times4\sin\frac{\pi^c}{7}}\\ & =\frac{-2\sin\frac{4\pi^c}{7}\cos\frac{4\pi^c}{7}}{8\sin\frac{\pi^c}{7}}\\ & =\frac{-\sin\frac{8\pi^c}{7}}{8\sin\frac{\pi^c}{7}}\\ & =\frac{-\sin\left(\pi^c+\frac{\pi^c}{7}\right)}{8\sin\frac{\pi^c}{7}}\\ & =\frac{\sin\frac{\pi^c}{7}}{8\sin\frac{\pi^c}{7}}\\ & = \frac{1}{8}\\ & = \text{RHS} \end{align}

21. (b) \(\sin\frac{\pi^c}{18}.\sin\frac{5\pi^c}{18}.\sin\frac{7\pi^c}{18}= \frac{1}{8} \)

\begin{align} \text{ LHS} &=\sin\frac{\pi^c}{18}.\sin\frac{5\pi^c}{18}.\sin\frac{7\pi^c}{18}\\ &=\sin\frac{180^\circ}{18}.\sin\frac{5\times180^{\circ}}{18}.\sin\frac{7\times180^{\circ}}{18}\\ &=\sin10^\circ.\sin50^\circ.\sin70^\circ\\ &=\sin\left(90^\circ-80^\circ\right)\sin\left(90^\circ-40^\circ\right)\sin\left(90^\circ-20^\circ\right)\\ &=\cos80^\circ\cos40^\circ\cos20^\circ\\ &=\frac{2\sin20^\circ\cos20^\circ\cos40^\circ\cos80^\circ}{2\sin20^\circ}\\ &=\frac{\sin40^\circ\cos40^\circ\cos80^\circ}{2\sin20^\circ}\\ &=\frac{2\sin40^\circ\cos40^\circ\cos80^\circ}{2\times2\sin20^\circ}\\ &=\frac{\sin80^\circ\cos80^\circ}{4\sin20^\circ}\\ &=\frac{2\sin80^{\circ}\cos80^{\circ}}{2\times4\sin20^{\circ}}\\ &=\frac{\sin160^{\circ}}{8\sin20^{\circ}}\\ &=\frac{\sin\left(180^{\circ}-60^{\circ}\right)}{8\sin20^{\circ}}\\ &=\frac{\sin20^{\circ}}{8\sin20^{\circ}}\\ & = \frac{1}{8} \\ & = \text{RHS} \end{align}

22. Without using calculator or table find the value of \( \cos 18^\circ \).

\begin{align} \text{ Let, } & \theta = 18^{\circ} \\ \text{or, } & \ 5\theta = 5\times 18^{\circ} \\ \text{or, } & \ 2\theta + 3\theta = 90^{\circ} \\ \text{or, } & \ 3\theta = 90^{\circ} - 2\theta \\ \text{or, } & \ \cos 3\theta = \cos (90^{\circ} - 2\theta )\\ \text{or, } & \ 4\cos^3\theta -3\cos \theta = \sin 2\theta \\ \text{or, } & \ \cos \theta [4\cos^2 \theta - 3 ] = 2\sin \theta \cos \theta \\ \text{or, } & \ 4(1-\sin^2 \theta ) -3 = 2\sin \theta \\ \text{or, } & \ 4 -4\sin^2 \theta -3 -2\sin \theta =0 \\ \text{or, } & \ -4\sin^2 \theta -2\sin \theta +1 = 0 \\ \text{or, } & \ -(4\sin^2 \theta +2\sin \theta -1) =0 \\ \text{or, } & \ 4\sin^2 \theta +2\sin \theta -1=0 \end{align} Now, comparing with \( ax^2+bx+c=0 \), we get \( a=4, b= 2, c= -1, x = \sin \theta \)\\ Now, \begin{align} \sin \theta & = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a} \\ \ & =\frac{-2\pm \sqrt{(2)^2 -4 \times 4 \times (-1)}}{2\times 4 } \\ \ & = \frac{-2\pm \sqrt{4+16}}{8} \\ \ & = \frac{-2\pm\sqrt{20}}{8} \\ \ & = \frac{-2\pm 2\sqrt{5}}{8} \\ \ & = \frac{2(-1\pm \sqrt{5})}{8}\\ \ & =\frac{-1\pm\sqrt{5}}{4} \end{align} Since \( \sin 18^{\circ} \) is positive, \( \sin 18^{\circ} = \frac{-1+\sqrt{5}}{4}=\frac{\sqrt{5}-1}{4} \)\\ Now, \begin{align} \cos 18^{\circ} &=\sqrt{1-\sin^2 18^{\circ}}\\ \ & = \sqrt{1-\left(\frac{\sqrt{5}-1}{4}\right)^2}\\ \ & = \sqrt{1-\frac{\left\{(\sqrt{5})^2-2\times \sqrt{5}\times 1+1^2 \right\}^2}{16}}\\ \ & = \sqrt{\frac{16-(5-2\sqrt{5}+1)}{16}}\\ \ & = \sqrt{\frac{16-5+2\sqrt{5}-1}{16}}\\ \ & = \sqrt{\frac{10+2\sqrt{5}}{16}}\\ \ & = \frac{\sqrt{10+2\sqrt{5}}}{4}\\ \therefore \cos 18^{\circ}& = \frac{\sqrt{10+2\sqrt{5}}}{4} \end{align}

Trigonometric Ratios of Multiple Angles