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Question Number 97680 by bemath last updated on 09/Jun/20

Commented by bobhans last updated on 09/Jun/20

$$\cos 6\mathrm{A}-\cos 2\mathrm{A}=-2\sin 4\mathrm{Asin}2\mathrm{A}$$$$\iff -2\sin 4\mathrm{Asin}2\mathrm{A}-2\cos 4\mathrm{A}+2=$$$$\iff -4\sin {}^{2}2\mathrm{A}\cos 2\mathrm{A}-2(1-2\sin {}^{2}2\mathrm{A})+2$$$$\iff -4\sin {}^{2}2\mathrm{Acos}2\mathrm{A}-2+4\sin {}^{2}2\mathrm{A}+2$$$$\iff 4\sin {}^{2}2\mathrm{A}(1-\cos 2\mathrm{A})$$$$\iff 4{\left(2\sin \mathrm{Acos}\mathrm{A}\right)}^2(1-(1-2\sin {}^2\mathrm{A}))$$$$\iff 16\sin {}^2\mathrm{Acos}{}^2\mathrm{A}\left(2\sin {}^2\mathrm{A}\right)$$$$\iff 32\sin {}^4\mathrm{Acos}{}^2\mathrm{A}$$

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