Đáp án:
$\sin^22x+\sin623x+\sin^24x+\sin^25x-2=-2\cos x.\cos2x.\cos7x$
Giải thích các bước giải:
$\sin^22x+\sin623x+\sin^24x+\sin^25x-2$
$=\dfrac{1-\cos4x}{2}+\dfrac{1-\cos6x}{2}+\dfrac{1-\cos8x}{2}+\dfrac{1-\cos10x}{2}-2$
$=\dfrac{4-\cos4x-\cos6x-\cos8x-\cos10x}{2}-2$
$=2-\dfrac{\cos4x+\cos6x+\cos8x+\cos10x}{2}-2$
$=-\dfrac{\cos4x+\cos6x+\cos8x+\cos10x}{2}$
$=-\dfrac{(\cos4x+\cos10x)+(\cos6x+\cos8x)}{2}$
$=-\dfrac{2\cos7x.\cos3x+2\cos7x.\cos x}{2}$
$=-\cos7x(\cos3x+\cos x)$
$=-\cos7x.2.\cos2x.\cos x$
$=-2\cos x.\cos2x.\cos7x$.
