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ǰλãҳ >> ΪʲôCos^{2}2x=(1+Cos4x)/2 >>

ΪʲôCos^{2}2x=(1+Cos4x)/2

Ǹ2ǹʽ cos2x=cos(x+x)=cosxcosx-sinxsinx=cos²x-sin²x=2cos²x-1 cos²x=cos2x+1/2 x2x ôcos²2x=1+cos4x/2

cos2x=cos(x+x) cosx^2-sinx^2=cos2x +1 cosx^2+1-sinx^2=cos2x 2cosx^2=1+cos2x cosx^2=(1+cos2x)/2

cos4x=cos2x-sin2x=1-2sin2xcos2x+sin2x=1 sin2x=(1-cos4x)/2 𣿲

sin⁴x+cos⁴x =sin⁴x+2sin²xcos²x+cos⁴x-2sin²xcos²x =(sin²x+cos²x)²-4sin²xcos²x/2 =1-(2sinxcosx)²/2 =1-(sin2x)²/2

cos²x dx = (1 + cos2x)/2 dx = x/2 + (1/4)sin4x + C cos³x dx = (1 - sin²x) dsinx = sinx - (1/3)sin³x + C cos⁴x dx = (cos²x)² dx = [(1 + cos2x)/2]² dx = (1/4) (1 + 2cos2x + ...

Ͳʽ sinsin=-[cos(+)-cos(-)]/2 coscos= [cos(+)+cos(-)]/2 sincos= [sin(+)+sin(-)]/2 cossin= [sin(+)-sin(-)]/2 sin2x*sin4x=-(cos6x-cos2x)/2 sin2x*cos4x= (sin6x-sin2x)/2 cos2x*sin4x= (sin6x+sin2x)/2

dzýۣͼ

sin2x=2sinxcosx,sinĶǹʽ sin2xcos2xΪʲô1/2sin4x"ֻǽx滻Ϊ2x sin(2*2x)=2sin2xcos2x sin2xcos2x=1/2sin4x

֪ô =cos2x[(sin²x+cos²x)²-2sin²xcos²x]+1/4sin2x*2sin2xcos2x =cos2x[1-1/2*(2sinxcosx)²]+1/2sin²2xcos2x =cos2x-1/2*sin²2xcos2x+1/2sin²2xcos2x =cos2x

cos2x=1-2(sinx) sinx=1/2 (sin2x)+cos2x=1 (sin2x)=3/4 cos4x=1-2(sin2x)=-1/2 sinx-cos4x=1/2+1/2=1

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