Question Number 86461 by Rio Michael last updated on 28/Mar/20
$$\mathrm{Use}\:\mathrm{exponential}\:\mathrm{representation}\:\mathrm{of}\:\mathrm{sin}\:\theta\:\mathrm{and}\:\mathrm{cos}\:\theta\:\mathrm{to}\:\mathrm{show}\:\mathrm{that} \\ $$$$\left.\mathrm{a}\left.\right)\:\mathrm{sin}^{\mathrm{2}} \:\theta\:+\:\mathrm{cos}^{\mathrm{2}} \:\theta\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\right)\:\mathrm{cos}^{\mathrm{2}} \theta\:−\:\mathrm{sin}^{\mathrm{2}} \theta\:=\:\mathrm{cos2}\theta \\ $$$$\left.\mathrm{c}\right)\:\mathrm{2}\:\mathrm{sin}\theta\:\mathrm{cos}\theta\:=\:\mathrm{2sin2}\theta. \\ $$
Answered by TANMAY PANACEA. last updated on 28/Mar/20
$$\left.{c}\right)\mathrm{2}{sin}\theta{cos}\theta \\ $$$$=\frac{\mathrm{2}}{{i}}\left({isin}\theta\right)\left({cos}\theta\right) \\ $$$$=\frac{\mathrm{2}}{{i}}×\left(\frac{{e}^{{i}\theta} −{e}^{−{i}\theta} }{\mathrm{2}}\right)\left(\frac{{e}^{{i}\theta} +{e}^{−{i}\theta} }{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{2}}{{i}}×\frac{\mathrm{1}}{\mathrm{4}}×\left({e}^{{i}\mathrm{2}\theta} −{e}^{−{i}\mathrm{2}\theta} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}×\left(\mathrm{2}{isin}\mathrm{2}\theta\right)={sin}\mathrm{2}\theta \\ $$
Answered by TANMAY PANACEA. last updated on 28/Mar/20
$${e}^{{i}\theta} ={cos}\theta+{isin}\theta\:\:\:\:\:{e}^{−{i}\theta} ={cos}\theta−{isin}\theta \\ $$$${cos}\theta=\frac{{e}^{{i}\theta} +{e}^{−{i}\theta} }{\mathrm{2}}\:\:\:\:\:\:{isin}\theta=\frac{{e}^{{i}\theta} −{e}^{−{i}\theta} }{\mathrm{2}} \\ $$$$\left.{a}\right){cos}^{\mathrm{2}} \theta+{sin}^{\mathrm{2}} \theta \\ $$$$={cos}^{\mathrm{2}} \theta−\left({isin}\theta\right)^{\mathrm{2}} \\ $$$$=\left(\frac{{e}^{{i}\theta} +{e}^{−{i}\theta} }{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{{e}^{{i}\theta} −{e}^{−{i}\theta} }{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{4}×{e}^{{i}\theta} ×{e}^{−{i}\theta} }{\mathrm{4}}=\mathrm{1} \\ $$
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
$${thank}\:{you}\:{sir} \\ $$
Answered by TANMAY PANACEA. last updated on 28/Mar/20
$$\left.{b}\right){cos}^{\mathrm{2}} \theta−{sin}^{\mathrm{2}} \theta \\ $$$$={cos}^{\mathrm{2}} \theta+\left({isin}\theta\right)^{\mathrm{2}} \\ $$$$=\left(\frac{{e}^{{i}\theta} +{e}^{−{i}\theta} }{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{e}^{{i}\theta} −{e}^{−{i}\theta} }{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}\left({e}^{{i}\mathrm{2}\theta} +{e}^{−{i}\mathrm{2}\theta} \right)}{\mathrm{4}}=\frac{\mathrm{2}{cos}\mathrm{2}\theta}{\mathrm{2}}={cos}\mathrm{2}\theta \\ $$