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Question Number 75655 by mr W last updated on 15/Dec/19

if z ∈ C is sin^2 z+cos^2 z=1 also valid?

$${if}\:{z}\:\in\:\mathbb{C} \\ $$$${is}\:\mathrm{sin}^{\mathrm{2}} \:{z}+\mathrm{cos}^{\mathrm{2}} \:{z}=\mathrm{1}\:{also}\:{valid}? \\ $$

Answered by MJS last updated on 15/Dec/19

yes. sin^2 z =(((e^(iz) −e^(−iz) )/(2i)))^2 =((e^(2iz) −2+e^(−2iz) )/(−4)) cos^2 z =(((e^(iz) +e^(−iz) )/2))^2 =((e^(2iz) +2+e^(−2iz) )/4) ((e^(2iz) −2+e^(−2iz) )/(−4))+((e^(2iz) +2+e^(−2iz) )/4)=1

$$\mathrm{yes}. \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{z}\:=\left(\frac{\mathrm{e}^{\mathrm{i}{z}} −\mathrm{e}^{−\mathrm{i}{z}} }{\mathrm{2i}}\right)^{\mathrm{2}} =\frac{\mathrm{e}^{\mathrm{2i}{z}} −\mathrm{2}+\mathrm{e}^{−\mathrm{2i}{z}} }{−\mathrm{4}} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:{z}\:=\left(\frac{\mathrm{e}^{\mathrm{i}{z}} +\mathrm{e}^{−\mathrm{i}{z}} }{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{e}^{\mathrm{2i}{z}} +\mathrm{2}+\mathrm{e}^{−\mathrm{2i}{z}} }{\mathrm{4}} \\ $$$$\frac{\mathrm{e}^{\mathrm{2i}{z}} −\mathrm{2}+\mathrm{e}^{−\mathrm{2i}{z}} }{−\mathrm{4}}+\frac{\mathrm{e}^{\mathrm{2i}{z}} +\mathrm{2}+\mathrm{e}^{−\mathrm{2i}{z}} }{\mathrm{4}}=\mathrm{1} \\ $$

Commented by mr W last updated on 15/Dec/19

thank you for comfirming, sir.

$${thank}\:{you}\:{for}\:{comfirming},\:{sir}. \\ $$

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