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Question Number 86461 by Rio Michael last updated on 28/Mar/20

$$\mathrm{Use}\mathrm{exponential}\mathrm{representation}\mathrm{of}\sin \theta \mathrm{and}\cos \theta \mathrm{to}\mathrm{show}\mathrm{that}$$$$\mathrm{a}){\sin }^2\theta +{\cos }^2\theta =1\mathrm{b}){\cos }^2\theta -{\sin }^2\theta =\cos 2\theta$$$$\mathrm{c})2\sin \theta \cos \theta =2\sin 2\theta .$$

Answered by TANMAY PANACEA. last updated on 28/Mar/20

$$c)2sin\theta cos\theta$$$$=\frac{2}{i}\left(isin\theta \right)\left(cos\theta \right)$$$$=\frac{2}{i}\times \left(\frac{e^{i\theta }-e^{-i\theta }}{2}\right)\left(\frac{e^{i\theta }+e^{-i\theta }}{2}\right)$$$$=\frac{2}{i}\times \frac{1}{4}\times (e^{i2\theta }-e^{-i2\theta })$$$$=\frac{1}{2i}\times \left(2isin2\theta \right)=sin2\theta$$

Answered by TANMAY PANACEA. last updated on 28/Mar/20

$$e^{i\theta }=cos\theta +isin\theta e^{-i\theta }=cos\theta -isin\theta$$$$cos\theta =\frac{e^{i\theta }+e^{-i\theta }}{2}isin\theta =\frac{e^{i\theta }-e^{-i\theta }}{2}$$$$a){cos}^2\theta +{sin}^2\theta$$$$={cos}^2\theta -{\left(isin\theta \right)}^2$$$$={\left(\frac{e^{i\theta }+e^{-i\theta }}{2}\right)}^2-{\left(\frac{e^{i\theta }-e^{-i\theta }}{2}\right)}^2$$$$=\frac{4\times e^{i\theta }\times e^{-i\theta }}{4}=1$$

Commented by Rio Michael last updated on 28/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by peter frank last updated on 29/Mar/20

$$thankyousir$$

Answered by TANMAY PANACEA. last updated on 28/Mar/20

$$b){cos}^2\theta -{sin}^2\theta$$$$={cos}^2\theta +{\left(isin\theta \right)}^2$$$$={\left(\frac{e^{i\theta }+e^{-i\theta }}{2}\right)}^2+{\left(\frac{e^{i\theta }-e^{-i\theta }}{2}\right)}^2$$$$=\frac{2(e^{i2\theta }+e^{-i2\theta })}{4}=\frac{2cos2\theta }{2}=cos2\theta$$

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