已知函数f(x)=2sin(1/3x-π/6),x属于R.设a,B属[0,兀/2],f(3a+兀/2)=10/13,f(3B+2兀)=6/5求COS(a+B)的值

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=4/5cos(α+β)=cosαcosβ-sinαsinβ=12/...