化简：sin(x+π/3)-√3cos(2π/3-x)+2sin(x-π/3)=

化简sin(x)(cos(x)+cos(3x)+cos(5x)+cos(7x)) 化简（1)√3sin x+cos x (2）√2(sin x-cos x) （3）√2cos x-√6sin x √3cos^2x-sin^2x化简 化简√3sin (x/2)+cos(x/2) 化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x 化简：sin(x+π/3)-√3cos(2π/3-x)+2sin(x-π/3)= sin(x+π/3)+2sin(x-π/3)-√3cos(2π/3 - x)化简、、、 已知sin[a-b]cos a-cos[b-a]sin a=3/5,b是第三象限角,求sin[b+5π/4]的值第一题1/2cos x-√3/2sin x第二题√3sin x+cos x第三题√2[sin x-cos x]第四题√2cos x-√6sin x √3 cos²x-sin²x-√3sin²x化简 化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π) 化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π) 化简 1/2cos x-根号3/2sin x 根3sin x+cos x 化简2sin^2[(π/4)+x]+根号3（sin^x-cos^x）-1 化简y=sin^2(x)+2sin(x)cos(x)+3cos^2(x) 化简sin(x+7π/4)+cos(x-3π/4）步骤我已经找到撒sin(x+7π/4)+cos(x-3π/4)=sin(x+2π-π/4)+cos(x-π+π/4)我问下这一步是在干嘛?=sin(x-π/4)+cos[-(π-x-π/4)]=sin(x-π/4)+cos(π-x-π/4)=sin(x-π/4)-cos(x+π/4)=sin(x-π/4)-cos(x+π/2- 化简3sin(2x+π/3)-√3cos(2x+π/3) 若方程12x²+πx-12x=0的两的根分别是α,β,则cosαcosβ-√3sinαcosβ-√cosαsinβ-sinαsinβ= 化简：1、sin(x-π/3)-cos(x+π/6)+√3cosx=?2、已知,sinα+sinβ=√2/2,求cosα+cosβ的取值范围