Question

Вычислите:
sin^6 1 - 3sin^4 1 + 3sin^2 1 + cos^6 1 + 1

Answer 1

$\sin^6x - 3 \sin^4x + 3 \sin^2x + \cos^6x + 1$

$\sin^6x+\cos^6x=\left ( \sin^2x+\cos^2x \right )\left ( \sin^4x-\sin^2x\cos^2x+\cos^4x \right )=\\=\sin^4x-\sin^2x\cos^2x+\cos^4x=\sin^4x-\sin^2\left ( 1-\sin^2x \right )+\left ( \cos^2x \right )^2=\\=\sin^4x-\sin^2x+\sin^4x+\left ( 1-\sin^2x \right )^2=\\=2\sin^4x-\sin^2x+1-2\sin^2x+\sin^4x=\\3\sin^4x-3\sin^2x+1$

Подставляем в наше выражение

$3\sin^4x-3\sin^2x+1-3\sin^4x+3\sin^2x+1=2$