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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

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

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dzýۣͼ

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

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 + ...

f(x)=[1+cos(2x)](sinx)^2 =[1+cos(2x)][1-cos(2x)]/2 ={1-[cos(2x)]^2}/2 =[sin(2x)]^2/2 =(1/4)[1-cos(4x)] =(-1/4)cos(4x) +1/4 ΪR,ԭԳ. f(-x)=(-1/4)cos(-4x)+1/4=(-1/4)cos(4x)+1/4=f(x),ż СT=2/...

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

ʵsin^4x-cos^4x-1+2cos^2x=0 (sin^4x-cos^4x)+(cos^2x-sin^2x)=(cos^2x+sin^2x)(cos^2x-sin^2x)=0 ö-

a moment later,

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