Question 8 and 9, Exercise 10.2

Solutions of Question 8 and 9 of Exercise 10.2 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Write ${{\cos }^{4}}\theta$ in term of first power of one or more cosine functions.

$$\begin{align}{{\cos}^{4}}\theta &={{\left( {{\cos }^{2}}\theta \right)}^{2}}\\ &={{\left( \dfrac{1+\cos 2\theta }{2} \right)}^{2}}\\ &=\dfrac{1+2\cos 2\theta +{{\cos }^{2}}2\theta }{4}\\ &=\dfrac{1}{4}\left[ 1+2\cos 2\theta +{{\cos }^{2}}2\theta \right]\\ &=\dfrac{1}{4}\left[ 1+2\cos 2\theta +\dfrac{1+\cos 4\theta }{2} \right]\\ &=\dfrac{1}{8}\left[ 2+4\cos 2\theta +1+\cos 4\theta \right]\\ &=\dfrac{1}{8}\left[ 3+4\cos 2\theta +\cos 4\theta \right] \end{align}$$ $$\implies \bbox[4px,border:2px solid black]{\cos^4 \theta=\dfrac{3}{8}+\dfrac{1}{2}\cos 2\theta +\dfrac{1}{8}\cos 4\theta}$$

Prove the identity $\sin 4\theta =8\sin \theta {{\cos }^{3}}\theta -4\sin \theta \cos \theta$ .

$$\begin{align}L.H.S.&=\sin 4\theta \\ &=\sin 2\left( 2\theta \right)\\ &=2\sin 2\theta \cos 2\theta \\ &=4\sin \theta \cos \theta \left( 2{{\cos }^{2}}\theta -1 \right)\\ &=8\sin \theta {{\cos }^{3}}\theta -4\cos \theta \sin \theta \\ & = R.H.S\end{align}$$

Prove the identity $\cot 4\theta =\dfrac{1-{{\tan }^{2}}2\theta }{2\tan 2\theta }$ .

$$\begin{align}L.H.S.&=\cot 4\theta =\dfrac{\cos 4\theta }{\sin 4\theta }\\ &=\dfrac{\cos 2\left( 2\theta \right)}{\sin 2\left( 2\theta \right)}\\ &=\dfrac{2{{\cos }^{2}}2\theta -1}{2\sin 2\theta \cos 2\theta}\quad\text{(by using double angle identity)}\\ &=\dfrac{2{{\cos }^{2}}2\theta -\left( {{\cos }^{2}}2\theta +{{\sin }^{2}}2\theta \right)}{2\sin 2\theta \cos 2\theta }\\ &=\dfrac{{{\cos }^{2}}2\theta -{{\sin }^{2}}2\theta }{2\sin 2\theta \cos 2\theta }\\ &=\dfrac{{{\cos }^{2}}2\theta \left( 1-{{\tan }^{2}}2\theta \right)}{2\sin 2\theta \cos 2\theta }\\ &=\dfrac{1-{{\tan }^{2}}2\theta }{2\tan 2\theta }=R.H.S.\end{align}$$

Prove the identity $\cot 3\theta =\dfrac{\cot^3\theta-3\cot\theta}{3\cot^2\theta -1}$ .

$$\begin{align}L.H.S.&=\cot 3\theta =\dfrac{1}{\tan3\theta}\\ &=\dfrac{1}{\tan( 2\theta +\theta)}\\ &=\dfrac{1-\tan 2\theta \tan \theta }{\tan 2\theta +\tan \theta }\\ &=\dfrac{1-\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }\tan \theta }{\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }+\tan \theta }\\ &= \dfrac{{\dfrac{{1 - {{\tan }^2}\theta - 2{{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}}}{{\dfrac{{2\tan \theta + \tan \theta - {{\tan }^3}\theta }}{{1 - {{\tan }^2}\theta }}}} \\ &=\dfrac{1-{{\tan }^{2}}\theta -2{{\tan }^{2}}\theta }{2\tan \theta +\tan \theta -{{\tan }^{3}}\theta }\\ &=\dfrac{1-3{{\tan }^{2}}\theta }{3\tan \theta -{{\tan }^{3}}\theta } \\ &= \dfrac{{{{\tan }^3}\theta \left( {\frac{1}{{{{\tan }^3}\theta }} - \frac{3}{{\tan \theta }}} \right)}}{{{{\tan }^3}\theta \left( {\frac{3}{{{{\tan }^2}\theta }} - 1} \right)}}\\ &=\dfrac{\cot^3\theta -3\cot\theta}{3\cot^2\theta -1}=R.H.S.\end{align}$$

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