求值cos(pai/4-x)cos(pai/4+x)=1/4 求sin^4x+cos^4x要详细过成,我没有分了,

cos(pai/4-x)cos(pai/4+x)=1/4
1/2〔cos(pai/4-x+pai/4+x)+cos(pai/4-x-pai/4-x)〕=1/4
2〔cos(pai/2)+cos(-2x)〕=1
2〔1+cos(-2x)〕=1
cos(2x)=-1/2,
sin(2x)=√(1-1/4)=√3/2
sin^4x+cos^4x
=(sin^2x+cos^2x)^2-2sin^2xcos^2x
=1-1/2(sin2x)^2
=1-1/2*3/4
=5/8

sin^4x+cos^4x=(sin^2x + cos^2x)^2 - 2*sin^2x*cos^2x=1 - 2*sin^2x*cos^2x为避免混淆,还是把乘方写在外面吧= 1 - 2*(sinx)^2*(cosx)^2= 1 - 2*(sinx*cosx)^2= 1 - (sin2x)^2/2而另一方面cos(pai/4-x)cos(pai/4+x)=...