作业帮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
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![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](/uploads/image/z/3740453-53-3.jpg?t=cos%5E6%28%CF%80%2F8%29-sin%5E6%28%CF%80%2F8%29%3D%E6%B1%82%E5%80%BC%2C%5Bcos%5E2%28%CF%80%2F8%29-sin%5E2%28%CF%80%2F8%29%5D%5Bcos%5E4%28%CF%80%2F8%29%2Bsin%5E4%28%CF%80%2F8%29%2Bcos%5E2%28%CF%80%2F8%29sin%5E2%28%CF%80%2F8%29%5D%3Dcos%28%CF%80%2F4%29%5B1-cos%5E2%28%CF%80%2F8%29sin%5E2%28%CF%80%2F8%29%5D%3Dcos%28%CF%80%2F4%29%5B1-%5Bsin%5E2%28%CF%80%2F4%29%5D%2F4%5D%3D%3D7%E2%88%9A2%2F16)
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
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
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
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