y=sin^2 x 如何化成y=(1-cos2x)/2

sin(x+\u03a0/4\uff09sin(x-\u03a0/4\uff09\u600e\u4e48\u5316\u6210cos2x/2

cos((x+\u03c0/4)+(x-\u03c0/4))=cos(x+\u03c0/4)cos(x-\u03c0/4)-sin(x+\u03c0/4)sin(x-\u03c0/4)
cos((x+\u03c0/4)-(x-\u03c0/4))=cos(x+\u03c0/4)cos(x-\u03c0/4)+sin(x+\u03c0/4)sin(x-\u03c0/4)
\u6240\u4ee5\uff0csin(x+\u03c0/4)sin(x-\u03c0/4)=1/2(cos((x+\u03c0/4)-(x-\u03c0/4))-cos((x+\u03c0/4)+(x-\u03c0/4)))=1/2(cos(\u03c0/2)-cos2x)=-1/2cos2x

\u5e0c\u671b\u5bf9\u4f60\u6709\u6240\u5e2e\u52a9\uff01
\u6ee1\u610f\u8bf7\u522b\u5fd8\u4e86\u91c7\u7eb3\u54e6\uff01

Y=3sin2x+(3\u221a3)cos2x+1
=6\uff081\uff0f2\u00d7sin2X+\u221a3\uff0f2cos2X\uff09+1 [\u63d0\u51fa\u6765\u7684\u6570=\uff08sin\u524d\u7684\u7cfb\u6570²\uff0bcos\u524d\u7684\u7cfb\u6570²\ufe5a½]
=6\uff08cos3\uff0f\u03c0\u00d7sin2X+sin3\uff0f\u03c0cos2X\uff09+1
\uff1d6sin\uff082X+3\uff0f\u03c0\uff09+1

y=sin^2 x
cos(2x)=(cosx)²-(sinx)²=(1-sinx)²-(sinx)²=1-2(sinx)²

所以 sin²x=(1-cos2x)/2

cos2x=2cos^2x-1=1-2sin^2x
带进去就行了

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