Question

1.已知cos(α-π/3)=12/13 π/3<α<π/2 求cosα 2.已知sinα+sinβ=3/5 cosα+cosβ=4/5 求cos(α-β)

详细过程

Answer 1

(1)π/3<α<π/2. ∴α-π/3∈(0,π/6).
∵cos(α-π/3)=12/13 .∴sin(α-π/3)=5/13.
∴cosα=cos[(α-π/3)+π/3]=cos(α-π/3)cosπ/3-sin(α-π/3)sinπ/3=(12-5√3)/26.
(2) 1=(sinα+sinβ)²+(cosα+cosβ)²=(sin²α+sin²β+2sinαsinβ)+(cos²α+cos²β+2cosαcosβ)
=sin²α+cos²+sin²β+cos²β+2(sinαsinβ+cosαcosβ)=2+2cos(α-β).
∴cos(α-β)= -1/2.