一道高中数学必修四 简单三角恒等变换的题：

Question

函数f(x)=cos^4 x – sin^4 x – 2sin x cos x可化为（ ）
A.f(x)=根号2 sin2x
B.f(x)=根号2 sin(2x+兀/4)
C.f(x)=根号2 sin(2x–兀/4)
D.f(x)=根号2 sin(兀/4–2x)
最好有详细的解题思路或过程，谢谢！

Answer 1

f(x)=cos^4 x – sin^4 x – 2sin x cos x
=（cos^2 x – sin^2 x）(cos^2 x +sin^2 x)– 2sin x cos x
=cos^2 x – sin^2 x– 2sin x cos x
=cos2x-sin2x
=√2[sin(兀/4)cos2x-cos(兀/4)sin2x]
=√2sin(兀/4–2x)本回答被提问者采纳

Answer 2

f(x)=cos^4(x)-sim^4(x)-2sinxcosx
=[cos^2(x)+sin^2(x)][cos^2(x)-sin^2(x)]-sin2x
=cos2x-sin2x
=根号2 sin(兀/4–2x)