cos^6(π/8)-sin^6(π/8)=求值,

知识储备：a³ +b³ = (a+b)(a²-ab +b²)
a³ -b³= (a-b)(a² +ab +b²)
cos^6(π/8)-sin^6(π/8)=[cos³(π/8)-sin³(π/8)]·[cos³(π/8)+sin³(π/8)]
={[cos(π/8)-sin(π/8)][cos²(π/8) +cos(π/8)·sin(π/8)+sin²(π/8)]}· {[cos(π/8)+sin(π/8)][cos²(π/8) -cos(π/8)·sin(π/8)+sin²(π/8)]}
=[cos²(π/8)-sin²(π/8)]·[1+cos(π/8)·sin(π/8)]·[1-cos(π/8)·sin(π/8)]
=cos(π/4)·[1+(1/2)sin(π/4)]·[1-(1/2)sin(π/4)
=(√2/2)·[1+(√2/4)]·[1-(√2/4)]
=7√2/16

题目有问题，三角函数从来就没有带乘方的。

cos^6(π/8)-sin^6(π/8)
=【cos^3(π/8)】^2-【sin^3(π/8)】^2 (cos^2B-sin^2 B=cos(2B)
=cos^3(π/8+π/8)
=cos^3(π/4)
=(√2/2)^3
=√2/4