Question

高中数学，三角函数，求详细过程

Answer 1

1.

tanθ=tan(θ+β-β)

=[tan(θ+β)-tanβ]/[1+tan(θ+β)tanβ]

=(⅔-¼)/(1+⅔·¼)

=5/14

2.

0<x<π/2，则sinx>0

cosx=3/5，sinx=√(1-cos²x)=√[1-(3/5)²]=4/5

0<x<π/2，π/2<y<π，则π/2<x+y<3π/2

又sin(x+y)=5/13>0，因此π/2<x+y<π

cos(x+y)<0

cos(x+y)=-√[1-sin²(x+y)]=-√[1-(5/13)²]=-12/13

cosy=cos(x+y-x)

=cos(x+y)cosx+sin(x+y)sinx

=(-12/13)(3/5)+(5/13)(4/5)

=-16/65

3.

tan(α+β)=tan[(α-π/6)+(π/6+β)]

=[tan((α-π/6)+tan(π/6+β)]/[1-tan(α-π/6)tan(π/6+β)]

=(3/7+ 2/5)/[1-(3/7)(2/5)]

=1

4.

sin(2x)=-cos(π/2+ 2x)

=-cos[2(x+π/4)]

=-[1-2sin²(x+π/4)]

=2sin²(x+π/4)-1

=2·(-¾)²-1

=⅛

Answer 2

第一个看图

追答

Answer 3

图

追答

第二题cos(x+y)我写错了，应该是-12/13，所以最终结果应该是-16/65