作业帮求值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,
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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,
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,
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,
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)= 1/4 =
[cos(pai/4)*cosx + sin(pai/4)sinx)][cos(pai/4)cosx - sin(pai/4)sinx]
=[cos(pai/4)cosx]^2 - [sin(pai/4)sinx]^2
=[(cosx)^2 - (sinx)^2]/2
=cos2x/2
所以 cos2x = 1/2
sin2x = ±√[1-(cos2x)^2 = ±√3/2
(sin2x)^2 = 3/4
因此
（sinx)^4 + (cosx)^4
= 1 - (sin2x)^2/2
= 1 - (3/4)/2
= 1 -3/8
= 5/8
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