cos^6(π/8)-sin^6(π/8)=求值,[cos^2(π/8)-sin^2(π/8)][cos^4(π/8)+sin^4(π/8)+cos^2(π/8)sin^2(π/8)]=cos(π/4)[1-cos^2(π/8)sin^2(π/8)]=cos(π/4)[1-[sin^2(π/4)]/4]==7√2/16
问题描述：
cos^6(π/8)-sin^6(π/8)=求值,
[cos^2(π/8)-sin^2(π/8)][cos^4(π/8)+sin^4(π/8)+cos^2(π/8)sin^2(π/8)]
=cos(π/4)[1-cos^2(π/8)sin^2(π/8)]
=cos(π/4)[1-[sin^2(π/4)]/4]
==7√2/16
答
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
答
题目有问题，三角函数从来就没有带乘方的。
答
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...