询问一道三角函数的题,(sin20)^2+(cos50)^2+sin30*sin70请问这道题如何做?

(sin20)^2+(cos50)^2+sin30*sin70
=(sin20)^2+(sin40)^2+1/2*cos20
=[sin(30-10)]^2+[sin(10+30)]^2+1/2*cos20
=[sin30cos10-sin10cos30]^2+[sin30cos10+sin10cos30]^2+1/2cos20
=[1/2cos10-根号3/2sin10]^2+[1/2cos10+根号3/2sin10]^2+1/2cos20
=(1/2cos^2 10+3/2sin^2 10)+1/2cos20
=(1/2cos^2 10+3/2sin^2 10)+1/2(cos^2 10-sin^2 10)
=cos^2 10+sin^2 10
=1