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Question 1.23: Prove the identities (a) sin³ θ= 3/4 sinθ  - 1/4 sin^3 θ, (b...

Prove the identities (a) \sin ^3 \theta=\frac{3}{4} \sin \theta-\frac{1}{4} \sin 3 \theta , (b) \cos ^4 \theta=\frac{1}{8} \cos 4 \theta+\frac{1}{2} \cos 2 \theta+\frac{3}{8} \text {. }

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\begin{aligned}(a)  \sin ^3 \theta &=\left(\frac{e^{i \theta}-e^{-i \theta}}{2 i}\right)^3=\frac{\left(e^{i \theta}-e^{-i \theta}\right)^3}{8 i^3}=-\frac{1}{8 i}\left\{\left(e^{i \theta}\right)^3-3\left(e^{i \theta}\right)^2\left(e^{-i \theta}\right)+3\left(e^{i \theta}\right)\left(e^{-i \theta}\right)^2-\left(e^{-i \theta}\right)^3\right\} \\ &=-\frac{1}{8 i}\left(e^{3 i \theta}-3 e^{i \theta}+3 e^{-i \theta}-e^{-3 i \theta}\right)=\frac{3}{4}\left(\frac{e^{i \theta}-e^{-i \theta}}{2 i}\right)-\frac{1}{4}\left(\frac{e^{3 i \theta}-e^{-3 i \theta}}{2 i}\right) \\ &=\frac{3}{4} \sin \theta-\frac{1}{4} \sin 3 \theta \end{aligned} \begin{aligned}(b) \cos ^4 \theta &=\left(\frac{e^{i \theta}+e^{-i \theta}}{2}\right)^4=\frac{\left(e^{i \theta}+e^{-i \theta}\right)^4}{16} \\ &=\frac{1}{16}\left\{\left(e^{i \theta}\right)^4+4\left(e^{i \theta}\right)^3\left(e^{-i \theta}\right)+6\left(e^{i \theta}\right)^2\left(e^{-i \theta}\right)^2+4\left(e^{i \theta}\right)\left(e^{-i \theta}\right)^3+\left(e^{-i \theta}\right)^4\right\} \\ &=\frac{1}{16}\left(e^{4 i \theta}+4 e^{2 i \theta}+6+4 e^{-2 i \theta}+e^{-4 i \theta}\right)=\frac{1}{8}\left(\frac{e^{4 i \theta}+e^{-4 i \theta}}{2}\right)+\frac{1}{2}\left(\frac{e^{2 i \theta}+e^{-2 i \theta}}{2}\right)+\frac{3}{8} \\ &=\frac{1}{8} \cos 4 \theta+\frac{1}{2} \cos 2 \theta+\frac{3}{8} \end{aligned}

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