Question

三角函数题 求解！！！

Answer 1

(1) 已cos(π/6+α)=√3/3.求cos(5π/6-α)的值。
解：cos(5π/6-α)=cos(π-π/6-α).
=cos[(π-(π/6+α)].
=-cos(π/6+α).
∴cos(5π/6-α) =-√3/3.

(2) 已知： π<α<2π, cos(α-7π)=-3/5, 求sin(3π+α)的值，tan(α-7π/2)的值。
解： cos[-(7π-α)=cos[6π+(π-α)]=cos(π-α)=-cosα
∴ cosα=3/5.
sin(3π+α)=-sinα.
=-√(1-cos^2α).
=-4/5.
∴sin(3π+α)=-4/5
tan(α-7π/2)=-tan(7π/2-α).
=-tan(4π-π/2-α).
=-tan[4π-(π/2+α)].
=tan(π/2+α)
=-cotα.
=cosα/sinα.
=(3/5)/(-4/5).
∴tan(α-7π/2)=-3/4.