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Question Number 83036 by otchereabdullai@gmail.com last updated on 27/Feb/20

$$\mathrm{Prove}\mathrm{that}{\cos }^4\theta -{\sin }^4\theta ={\cos }^2\theta -{\sin }^2\theta$$

Commented by Tony Lin last updated on 27/Feb/20

$${cos}^4\theta -{sin}^4\theta$$$$={({cos}^2\theta -{sin}^2\theta )}^2+2{sin}^2\theta {cos}^2\theta$$$$=\frac{cos4\theta +sin4\theta +1}{2}\ne cos2\theta$$$$$$

Commented by MJS last updated on 27/Feb/20

$$\cos \theta =c\wedge \sin \theta =s$$$$c^4-s^4=c^2-s^2$$$$(c^2-s^2)(c^2+s^2)=c^2-s^2$$$$\mathrm{but}{c}^2+s^2=1$$$$\Rightarrow \mathrm{true}$$

Commented by otchereabdullai@gmail.com last updated on 27/Feb/20

$$\mathrm{thanks}\mathrm{prof}\mathrm{mjs}$$

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