Giải thích các bước giải:

$(\sin(\dfrac{\pi}2-x)+\sin(10\pi+x))^2+(\cos(\dfrac32\pi-x)+\cos(8\pi-x))^2$ 

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

$=\cos^2x+\sin^2x+2\sin x\cos x+\sin^2x-2\sin x\cos x+\cos^2x$

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

$=2\cdot 1$

$=2$