phmg.net
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

ͼ

dzýۣͼ

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

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

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,

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 )

վҳ | վͼ
All rights reserved Powered by www.phmg.net
copyright ©right 2010-2021
磬ַϵͷzhit325@qq.com