设cos(π/4+α)=-√3/2(π/2

设cos(π/4+α)=-√3/2(π/2 设α属于(0,π/2),若sinα=3/5,则√2cos(α+π/4)等于 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cosα=-√5/5 π 设cos(α-β)=-4/5,cos(α+β)=12/13,α-β∈(π/2,π),α+β∈(3π/2,2π),求cos 设f(x)=(sin^2(6π+x)+cosx-2cos^3(3π+x)-3)/2+cos^2(x-4π)-cos(-x)设f（x）=（sin^2(6π+x）+cosx-2cos^3(3π+x）-3）/2+cos^2（x-4π）-cos（-x）求f（π/3）的值 设α∈(0,π/3),β(π/6,π/2),且α,β满足5√3sinα+5cosα=8√2sinβ+√6cosβ =2 求cos(α+β )设α∈(0,π/3),β(π/6,π/2),且α,β满足53sinα+5cosα=8 /2sinβ+/6cosβ =2 求cos(α+β ) 设sinθ+cosθ=√2/3,π/2 设sinθ+cosθ=√2/3,π/2 设α∈(0,π/2),若sinα=4/5,则cos(α-π/3)等于? 设cos(a+π/4)=2/3,求sin2a的值~ 设cos(a+π/4)=2/3,求sin2a的值 设f(x)=2cos三次方α+sin²(2π-α)+cos(2π-α)-3/2+2cos²(π+α)+cos(-α),求f(π/3)详细过程 设tan(π+α)=2,则sin(α-π)+cos(π-α)/sin(π+α)-cos(π+α)= 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α为锐角 若cos (α+π/3)=1/5则cos(α-π/6)= 设α+β=π/3,tanα+tanβ=3,则cosαcosβ= 设α∈（0,π/3) β∈（π/6,π/2) 且5/3sinα+5cosα=8 ,/2sinβ+/6cosβ =2 求cos（α+β ）设α∈（0,π/3) β∈（π/6,π/2) 且5/3sinα+5cosα=8 ,/2sinβ+/6cosβ =2 求cos（α+β ）的值且5√3sinα+5cosα=8 ,√2sinβ+√6cosβ =2看