Question

已知函数f(x)=2sin(1/3 -π/6)(x∈R),

已知函数f(x)=2sin(1/3 -π/6)(x∈R),设α，β∈[0,π/2],f(3α+π/2)=10/13,f(3α+2π)=6/5，求cosαcosβ-sinαsinβ的值

Answer 1

题目是否有误，应该是f(3β+2π)=6/5吧
f(3α+π/2)=2sin(1/3*(3a+π/2)-π/6)=2sina=10/13
sina=5/13 所以cosa=12/13
f(3β+2π)=2sin(1/3*(3β+2π)-π/6)=2sin(β+1/2π)=2cosβ=6/5
cosβ=3/5 所以sinβ=4/5
cosαcosβ-sinαsinβ=12/13*3/5-5/13*4/5=16/65

Answer 2

f(3α+π/2)=2sin(α+π/6-π/6)=2sinα=10/13 sinα=5/13
f(3β+2π）=2sin（β+π/2）=6/5 cosβ=3/5
因为sin^2(α)+cos^2(α)=1且α，β∈[0,π/2],
所以cosα=12/13 sinβ=4/5
cos(α+β)=cosαcosβ-sinαsinβ=16/65