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Question Number 129019 by bramlexs22 last updated on 12/Jan/21

Find the value of 3(sin x−cos x)^4 +6(sin x+cos x)^2 + 4(sin^6 x+cos^6 x).

$$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\mathrm{3}\left(\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{4}} +\mathrm{6}\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} + \\ $$$$\:\mathrm{4}\left(\mathrm{sin}\:^{\mathrm{6}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}\right). \\ $$

Answered by MJS_new last updated on 12/Jan/21

3(s−(√(1−s^2 )))^4 +6(s+(√(1−s^2 )))^2 +4(s^6 +(1−s^2 )^3 )= =−12s(√(1−s^2 ))−12s^4 +12s^2 +3+ +12s(√(1−s^2 )) +6+ +12s^4 −12s^2 +4=13

$$\mathrm{3}\left({s}−\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }\right)^{\mathrm{4}} +\mathrm{6}\left({s}+\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }\right)^{\mathrm{2}} +\mathrm{4}\left({s}^{\mathrm{6}} +\left(\mathrm{1}−{s}^{\mathrm{2}} \right)^{\mathrm{3}} \right)= \\ $$$$=−\mathrm{12}{s}\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }−\mathrm{12}{s}^{\mathrm{4}} +\mathrm{12}{s}^{\mathrm{2}} +\mathrm{3}+ \\ $$$$\:\:\:\:+\mathrm{12}{s}\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{6}+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{12}{s}^{\mathrm{4}} −\mathrm{12}{s}^{\mathrm{2}} +\mathrm{4}=\mathrm{13} \\ $$

Commented by bramlexs22 last updated on 12/Jan/21

amazing prof

$$\mathrm{amazing}\:\mathrm{prof} \\ $$

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