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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

ʽӵļô-1/2 1/cos^2xй̶ֵ 4cos4x0զ-1/2 ڸв3.14ǻȡ

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

ʽд̫ˣдЩӦòѣ һЩʾɣ [cos²(x)]²cos²(x)cos²(x)[cos²(x)1]sin²(x)cos²(x) ʽӿ塣

ΪxҪֻ-xxĻΪ϶ηҪ

f(x)=cos2xsin2x+1/2cos4x=1/2sin4x+1/2cos4x=2/2sin(4x+/4)

a moment later,

f(x)=4sin(2x)*(1-cos(2x+/2))/2+cos4x-1=2sin(2x)*(1+sin(2x))+cos(4x)-1=2sin(2x)+2sin^2(x)+2cos^2(2x)-2=2sin(2x)Ϊ μסǺƽתʽʱҪƽηתʱ򷴹Ǹ߿ϲ

ǹʽ sin22sincos

consider sin(A+B)= sinAcosB+cosAsinB (1) sin(A-B)= sinAcosB-cosAsinB (2) (1)-(2) cosAsinB = (1/2)( sin(A+B)- sin(A-B) ) A=3x, B=x cos3xsinx = (1/2)( sin(4x)- sinx )

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