sin(θ-π/6)=1/2..则cosθ=

若sinθ+sin^2θ=1,则cos^2θ+cos^4θ+cos^6θ 已知cosθ+cosθ^2=1,则sinθ^2+sinθ^6+sinθ^8= 已知cosθ+cosθ=1,则sinθ+sin∧6θ+sin∧8θ= 设角a=-35π/6,则2sin(π+α)cos(π-α)-cos(π+α)/ 1+sin^2α+sin(π-α)-cos^2(π+α)sina=sin(-35π/6)=sin(-6π+π/6)=1/2 cosα=根号3/22sin(π＋α)cos(π-α)-cos(π+α)/ [1+sin^2α+sin(π-α)-cos^2(π+α)]=2（-sinα)(-cosα）+cosα/[1+sin² sin(θ-π/6)=1/2..则cosθ= cot(π+θ)=2,则(3sinθ-cosθ)/(sinθ+cosθ)=? sinθ-cosθ=1/2,则sin^3θ-cos^3θ=?. 若sinθ-cosθ=1/5,则cosθ乘以sinθ= 若tanθ=1/2则(cosθ+sinθ/cosθ-sinθ),1-sinθcosθ+2cos^2θ= 由cos(a+b)=cos a cos b-sin a sin b cos(a-b)=cos a cos b+sin a sin b解题设a为锐角,证：1、2分之根3乘cos a + 2分之1乘sin a=cos(6分之π-a)2、cos a-sin a=根号2cos(4分之π+a) 设cosθ+cos^2θ=1,则sin^2θ+sin^6θ+sin^8θ的值为 若sinα+sin^2α=1,则cos^2α+cos^6α+cos^8α=? sin α+sin^2α=1,则cos^2 α+cos^6 α+cos^8α=多少?, 若sinα+sin^2α=1,则cos^2α+cos^4α+cos^6α=? sin(π-θ）+cos(2π-θ）/cos(5π/2-θ)+sin(3π/2+θ)=2,则sinθcosθ=______ 如sinθ+2cosθ=0,则有：tanθ=-2.将（cos2θ-sinθ）/（1+cos²θ ）变为tanθ的函数,（cos2θ-sinθ）/（1+cos²θ ）=（cos²θ-sin²θ-sinθ）/（1+cos²θ ）=（1-tan²θ-sinθ/cos²θ）/[（1/cos² 求证sinθ/(1+cosθ)+(1+cosθ)/sinθ=2/sinθ 求值:已知sinθ+cosθ=1/5,已知θ∈(0,π).求(1)sinθ*cosθ(2)sinθ-cosθ(3)tanθ