Question

化简cos^2(π/2-α)-sin(α-2π)sin(π+α)-2sin^2(-α)

化简(1) cos^2(π/2-α)-sin(α-2π)sin(π+α)-2sin^2(-α)
(2) p sinπ+q cosπ/2+k cos 2π
(3) a^2 cos2π-b^2 sin3π/2+ab cosπ-ab sinπ/2

Answer 1

1)原式=sin^2 α+sin^2 α-2sin^2α=0
2)原式=p*0+q*0+k*1=1
3）原式=a^2+b^2-ab-ab=(a+b)^2

追问

有过程么，谢谢

追答

利用诱导公式，和特殊角的三角函数值呀？已经很详细了

Answer 2

(1)
cos²(π/2-α)-sin(α-2π)sin(π+α)-2sin²(-α)
=(sinα)²-sinα*(-sinα)-2sin²α
=0

(2)
p*sinπ+q*cosπ/2+k*cos 2π
=p*0+q*0+k*1
=k

(3)
a²cos2π-b²sin3π/2+ab*cosπ-ab*sinπ/2
=a²*1-b²*(-1)+ab*(-1)-ab*1
=a²+b²-2ab
=(a-b)²