cos(2x+π/3)+sinx的平方

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cos(2x+π/3)+sinx的平方cos(2x+π/3)+sinx的平方cos(2x+π/3)+sinx的平方=cos(π/3)cos2x-sin(π/3)sin2x+(1-cos2x)/2=(1

cos(2x+π/3)+sinx的平方
cos(2x+π/3)+sinx的平方

cos(2x+π/3)+sinx的平方
=cos(π/3)cos2x-sin(π/3)sin2x+(1-cos2x)/2
=(1/2)cos2x-(√3/2)sin2x+1/2-(1/2)cos2x
=1/2-(√3/2)sin2x


cos(2x+π/3)+sin²x
= cos2xcos(π/3) - sin2xsin(π/3) + sin²x
= (1-2sin²A)/2 - 2sinxcosx * (√3)/2 + sin²x
= 1/2 - (√3)sinxcosx

cos2xcos排/3-sin2xsin(排/3)+sin^2x
=cos2x/2-根3(sin2x)/2+sin^2x
=sin(2x-排/6)+(1-sin2x)/2
后面会了吧