设sin(π/4+θ)=1/3,则sin2θ=sin(π/4+θ)=1/3cos(π/2+2θ)=1-2sin²(π/4+θ)=7/9而cos(π/2+2θ)=-sin2θ所以sin2θ=-7/9 为什么cos(π/2+2θ)=-sin2θ

设sin(π/4+θ)=1/3,则sin2θ= 设5sinα+3sinβ=4,5cosα-3sinβ=1,则sin(α-β)=手贱打错了设5sinα+3COSβ=4,5cosα-3sinβ=1,则sin(α-β)=原来的sin改成COS。 设sinθ+cosθ=√2/3,π/2sinθ^3是sinθ的3次 提问sin(π/4-θ)=1/3,则sin(π/4+θ)= 设θ是第三象限的角,sin(θ/2+3π/2)>0,则[√(1-sinθ)]/[cos(θ/2)-sin(θ/2)]的值为? 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 设f(x)=(1+cos2x)/2sin(π/2-x)+sinx+a^2sin(x+π/4)的最大值为√2+3,则常数a= 设α,β,γ∈(0,π/2)且(sinα)^2+(sinβ)^2+(sinγ)^2=1求函数y=(sinα)^3/sinβ+(sinβ)^3/sinγ+(sinγ)^3/sinα 的最小值. 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sin(π/4+θ)=1/2,求sin2θ. 设sin(π/4+α)+cos(π/4+α)=-1/5,则cos2α= 设cosθ+cos^2θ=1,则sin^2θ+sin^6θ+sin^8θ的值为 设θ为第二象限角,若tan(θ+π/4)=1/2,则sinθ+cosθ=? 设向量a=(sinθ,1)与b=(1,2sinθ)平行,则cos2θ= 设sin(x)-sin(y)=1/3,T=cos(y)^2+2sin(x),则T的取值范围 设sinα+sinβ=1/3,求sinα-cos²β的取值范围 设sinα-sinβ=1/3,cosα+cosβ=1/2,则cos(α+β)=? 傅里叶级数作图f（x）=2sin[x] - sin[2x] + 2/3sin[3x] - 1/2sin[4x]我用mathematica输入程序Plot[{2sin[x],-2sin[x],2sin[x] - sin[2x],-2sin[x] + sin[2x],2sin[x] - sin[2x] + 2/3sin[3x],-2sin[x] + sin[2x] - 2/3sin[3x],2sin[x] - sin[2x] + 2/3si