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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 6A−cos 2A=−2sin 4Asin 2A ⇔−2sin 4Asin 2A−2cos 4A+2= ⇔ −4sin^2 2A cos 2A−2(1−2sin^2 2A)+2 ⇔−4sin^2 2Acos 2A−2+4sin^2 2A+2 ⇔4sin^2 2A (1−cos 2A) ⇔4(2sin Acos A)^2 (1−(1−2sin^2 A)) ⇔16sin^2 Acos^2 A (2sin^2 A) ⇔32sin^4 Acos^2 A

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

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