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Question Number 156248 by alcohol last updated on 09/Oct/21

Answered by physicstutes last updated on 09/Oct/21

$$e^{i\theta }=\cos \theta +i\sin \theta .......\left(i\right)$$$$e^{-i\theta }=\cos \theta -i\sin \theta .....\left(ii\right)$$$$\mathrm{equation}\left(i\right)\mathrm{and}\left(ii\right)\mathrm{are}\mathrm{the}\mathrm{polar}\mathrm{and}\mathrm{index}\left(\mathrm{euler}\right)\mathrm{form}$$$$\mathrm{of}\mathrm{a}\mathrm{general}\mathrm{complex}\mathrm{number}\mathrm{with}\mathrm{modulus}1.$$$$\left(i\right)+\left(ii\right)\Rightarrow e^{i\theta }+e^{-i\theta }=2\cos \theta$$$$\mathrm{similarly}{e}^{2i\theta }+e^{-2i\theta }=2\cos 2\theta$$$$\Rightarrow \overline{)\begin{array}{c}\cos 2\theta =\frac{1}{2}(e^{2i\theta }+e^{-2i\theta })\end{array}}$$$$\mathrm{also}\left(i\right)-\left(ii\right)\Rightarrow e^{i\theta }-e^{-i\theta }=2i\sin \theta$$$$\mathrm{similarly}{e}^{2i\theta }-e^{-2i\theta }=2i\sin 2\theta$$$$\Rightarrow \overline{)\begin{array}{c}\sin 2\theta =\frac{1}{2i}(e^{2i\theta }-e^{-2i\theta })\end{array}}$$$$\mathbf{Hence}$$$$\mathrm{Now}{\sin }^{2}2\theta ={\left[\frac{1}{2i}(e^{2i\theta }-e^{-2i\theta })\right]}^2=-\frac{1}{4}(e^{4i\theta }-2+e^{-4i\theta })$$$${\cos }^{3}2\theta ={\cos }^{2}2\theta \cos \theta ={\left[\frac{1}{2}(e^{2i\theta }+e^{-2i\theta })\right]}^2\left[\frac{1}{2}(e^{2i\theta }+e^{-2i\theta })\right]=\frac{1}{8}(e^{4i\theta }+2+e^{-4i\theta })(e^{2i\theta }+e^{-2i\theta })$$$${\cos }^{3}2\theta =\frac{1}{8}(e^{6i\theta }+3e^{2i\theta }+3e^{-2i\theta }+e^{-6i\theta })$$$$\Rightarrow {\sin }^{2}2\theta {\cos }^{3}2\theta =-\frac{1}{16}(e^{4i\theta }-2+e^{-4i\theta })(e^{6i\theta }+3e^{2i\theta }+3e^{-2i\theta }+e^{-6i\theta })$$$$=-\frac{1}{16}(e^{10i\theta }+3e^{6i\theta }+3e^{2i\theta }+e^{-2i\theta }-2e^{6i\theta }-6e^{2i\theta }-6e^{-2i\theta }-2e^{-6i\theta }+e^{2i\theta }+3e^{2i\theta }+3e^{-6i\theta }+e^{-10i\theta })$$$$=-\frac{1}{16}(e^{10i\theta }+{10}^{-10i\theta }+e^{6i\theta }+e^{-6i\theta }+e^{2i\theta }+e^{-2i\theta })$$$$\Rightarrow 16{\sin }^{2}2\theta {\cos }^{3}2\theta =2\cos 2\theta -\cos 6\theta -\cos 10\theta \mathbf{as}\mathbf{required}.$$$$$$$$\underset{0}{\stackrel{\frac{\pi }{3}}{\int }}16{\sin }^{2}2\theta {\cos }^{3}2\theta d\theta =\underset{0}{\stackrel{\frac{\pi }{3}}{\int }}(2\cos 2\theta -\cos 6\theta -\cos 10\theta )d\theta ={[\sin 2\theta -\frac{\sin 6\theta }{6}-\frac{\sin 10\theta }{6}]}_0^{\frac{\pi }{3}}$$$$=\frac{\sqrt{3}}{2}-0-0\Rightarrow \overline{)\begin{array}{c}\underset{0}{\stackrel{\frac{\pi }{3}}{\int }}16{\sin }^{2}2\theta {\cos }^{3}2\theta d\theta =\frac{\sqrt{3}}{2}\end{array}}$$$$$$

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