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Question Number 134446 by physicstutes last updated on 03/Mar/21

prove that ((cos^4 x + sin^4 x)/(cos^4 x−sin^4 x)) = ((cos2A + sec 2A)/2)

$$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\:\frac{\mathrm{cos}^{\mathrm{4}} {x}\:+\:\mathrm{sin}^{\mathrm{4}} {x}}{\mathrm{cos}^{\mathrm{4}} {x}−\mathrm{sin}^{\mathrm{4}} {x}}\:=\:\frac{\mathrm{cos2}{A}\:+\:\mathrm{sec}\:\mathrm{2}{A}}{\mathrm{2}} \\ $$

Answered by Ar Brandon last updated on 03/Mar/21

cos^4 x=(((e^(ix) +e^(−ix) )/2))^4 =(1/(16))(e^(4ix) +4e^(2ix) +6+4e^(−2ix) +e^(−4ix) ) =(1/(16))(2cos4x+8cos2x+6) sin^4 x=(((e^(ix) −e^(−ix) )/(2i)))^4 =(1/(16))(e^(4ix) −4e^(2ix) +6−4e^(−2ix) +e^(−4ix) ) =(1/(16))(2cos4x−8cos2x+6) ((cos^4 x+sin^4 x)/(cos^4 x−sin^4 x))=((4cos4x+12)/(16cos2x))=((8cos^2 2x+8)/(16cos2x)) =((cos2x+sec2x)/2)

$$\mathrm{cos}^{\mathrm{4}} \mathrm{x}=\left(\frac{\mathrm{e}^{\mathrm{ix}} +\mathrm{e}^{−\mathrm{ix}} }{\mathrm{2}}\right)^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{e}^{\mathrm{4ix}} +\mathrm{4e}^{\mathrm{2ix}} +\mathrm{6}+\mathrm{4e}^{−\mathrm{2ix}} +\mathrm{e}^{−\mathrm{4ix}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{2cos4x}+\mathrm{8cos2x}+\mathrm{6}\right) \\ $$$$\mathrm{sin}^{\mathrm{4}} \mathrm{x}=\left(\frac{\mathrm{e}^{\mathrm{ix}} −\mathrm{e}^{−\mathrm{ix}} }{\mathrm{2i}}\right)^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{e}^{\mathrm{4ix}} −\mathrm{4e}^{\mathrm{2ix}} +\mathrm{6}−\mathrm{4e}^{−\mathrm{2ix}} +\mathrm{e}^{−\mathrm{4ix}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{2cos4x}−\mathrm{8cos2x}+\mathrm{6}\right) \\ $$$$\frac{\mathrm{cos}^{\mathrm{4}} \mathrm{x}+\mathrm{sin}^{\mathrm{4}} \mathrm{x}}{\mathrm{cos}^{\mathrm{4}} \mathrm{x}−\mathrm{sin}^{\mathrm{4}} \mathrm{x}}=\frac{\mathrm{4cos4x}+\mathrm{12}}{\mathrm{16cos2x}}=\frac{\mathrm{8cos}^{\mathrm{2}} \mathrm{2x}+\mathrm{8}}{\mathrm{16cos2x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{cos2x}+\mathrm{sec2x}}{\mathrm{2}} \\ $$

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