Rút gọn bt: a) sin10 độ.cos20 độ.cos40 độ b)cos^3sinx-sin^3cosx.Mọi ng giúp e vs ạ e cần gấp cẻm ơn mn

Đáp án:

a.$\dfrac18$

b.$\dfrac14\sin4x$

Giải thích các bước giải:

a.Ta có:

$\sin10^o\cos20^o\cos40^o$

$=\dfrac{2\cos10^o\sin10^o\cos20^o\cos40^o}{2\cos10^o}$

$=\dfrac{\sin20^o\cos20^o\cos40^o}{2\cos10^o}$

$=\dfrac{2\sin20^o\cos20^o\cos40^o}{4\cos10^o}$

$=\dfrac{\sin40^o\cos40^o}{4\cos10^o}$

$=\dfrac{2\sin40^o\cos40^o}{8\cos10^o}$

$=\dfrac{\sin80^o}{8\cos10^o}$

$=\dfrac{\cos(90^o-80^o)}{8\cos10^o}$

$=\dfrac{\cos10^o}{8\cos10^o}$

$=\dfrac18$

b.Ta có:

$\cos^3x\sin x-\sin^3x\cos x$

$=\sin x\cos x(\cos^2x-\sin^2x)$

$=\dfrac12\cdot 2\sin x\cos x(\cos^2x-\sin^2x)$

$=\dfrac12\sin2x\cos2x$

$=\dfrac14\cdot 2\sin2x\cos2x$

$=\dfrac14\sin4x$