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

Use exponential representation of sin θ and cos θ to show that a) sin^2 θ + cos^2 θ = 1 b) cos^2 θ − sin^2 θ = cos2θ c) 2 sinθ cosθ = 2sin2θ.

$$\mathrm{Use}\:\mathrm{exponential}\:\mathrm{representation}\:\mathrm{of}\:\mathrm{sin}\:\theta\:\mathrm{and}\:\mathrm{cos}\:\theta\:\mathrm{to}\:\mathrm{show}\:\mathrm{that} \\ $$$$\left.\mathrm{a}\left.\right)\:\mathrm{sin}^{\mathrm{2}} \:\theta\:+\:\mathrm{cos}^{\mathrm{2}} \:\theta\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\right)\:\mathrm{cos}^{\mathrm{2}} \theta\:−\:\mathrm{sin}^{\mathrm{2}} \theta\:=\:\mathrm{cos2}\theta \\ $$$$\left.\mathrm{c}\right)\:\mathrm{2}\:\mathrm{sin}\theta\:\mathrm{cos}\theta\:=\:\mathrm{2sin2}\theta. \\ $$

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

c)2sinθcosθ =(2/i)(isinθ)(cosθ) =(2/i)×(((e^(iθ) −e^(−iθ) )/2))(((e^(iθ) +e^(−iθ) )/2)) =(2/i)×(1/4)×(e^(i2θ) −e^(−i2θ) ) =(1/(2i))×(2isin2θ)=sin2θ

$$\left.{c}\right)\mathrm{2}{sin}\theta{cos}\theta \\ $$$$=\frac{\mathrm{2}}{{i}}\left({isin}\theta\right)\left({cos}\theta\right) \\ $$$$=\frac{\mathrm{2}}{{i}}×\left(\frac{{e}^{{i}\theta} −{e}^{−{i}\theta} }{\mathrm{2}}\right)\left(\frac{{e}^{{i}\theta} +{e}^{−{i}\theta} }{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{2}}{{i}}×\frac{\mathrm{1}}{\mathrm{4}}×\left({e}^{{i}\mathrm{2}\theta} −{e}^{−{i}\mathrm{2}\theta} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}×\left(\mathrm{2}{isin}\mathrm{2}\theta\right)={sin}\mathrm{2}\theta \\ $$

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

e^(iθ) =cosθ+isinθ e^(−iθ) =cosθ−isinθ cosθ=((e^(iθ) +e^(−iθ) )/2) isinθ=((e^(iθ) −e^(−iθ) )/2) a)cos^2 θ+sin^2 θ =cos^2 θ−(isinθ)^2 =(((e^(iθ) +e^(−iθ) )/2))^2 −(((e^(iθ) −e^(−iθ) )/2))^2 =((4×e^(iθ) ×e^(−iθ) )/4)=1

$${e}^{{i}\theta} ={cos}\theta+{isin}\theta\:\:\:\:\:{e}^{−{i}\theta} ={cos}\theta−{isin}\theta \\ $$$${cos}\theta=\frac{{e}^{{i}\theta} +{e}^{−{i}\theta} }{\mathrm{2}}\:\:\:\:\:\:{isin}\theta=\frac{{e}^{{i}\theta} −{e}^{−{i}\theta} }{\mathrm{2}} \\ $$$$\left.{a}\right){cos}^{\mathrm{2}} \theta+{sin}^{\mathrm{2}} \theta \\ $$$$={cos}^{\mathrm{2}} \theta−\left({isin}\theta\right)^{\mathrm{2}} \\ $$$$=\left(\frac{{e}^{{i}\theta} +{e}^{−{i}\theta} }{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{{e}^{{i}\theta} −{e}^{−{i}\theta} }{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{4}×{e}^{{i}\theta} ×{e}^{−{i}\theta} }{\mathrm{4}}=\mathrm{1} \\ $$

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

thank you sir

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

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

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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

b)cos^2 θ−sin^2 θ =cos^2 θ+(isinθ)^2 =(((e^(iθ) +e^(−iθ) )/2))^2 +(((e^(iθ) −e^(−iθ) )/2))^2 =((2(e^(i2θ) +e^(−i2θ) ))/4)=((2cos2θ)/2)=cos2θ

$$\left.{b}\right){cos}^{\mathrm{2}} \theta−{sin}^{\mathrm{2}} \theta \\ $$$$={cos}^{\mathrm{2}} \theta+\left({isin}\theta\right)^{\mathrm{2}} \\ $$$$=\left(\frac{{e}^{{i}\theta} +{e}^{−{i}\theta} }{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{e}^{{i}\theta} −{e}^{−{i}\theta} }{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}\left({e}^{{i}\mathrm{2}\theta} +{e}^{−{i}\mathrm{2}\theta} \right)}{\mathrm{4}}=\frac{\mathrm{2}{cos}\mathrm{2}\theta}{\mathrm{2}}={cos}\mathrm{2}\theta \\ $$

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