已知函数f(x)=2sin(1/3x-π/6),x∈R (1)求f(0)的值; (2)设α,β∈[0,π/2],f(3α+π/2)=10/13,f(3β+π
已知函数f(x)=2sin(1/3x-π/6),x∈R (1)求f(0)的值;
(2)设α,β∈[0,π/2],f(3α+π/2)=10/13,f(3β+π/2）=6/5.求sin(α+β)的值.
1.f(0)=2sin(-π/6)=2×(-1/2)=-1
2.f(3α+π/2)=2sin[1/3(3α+π/2)-π/6]=2sinα=10/13 sinα=5/13 α∈[0,π/2] cosα=12/13
f(3β+π/2)=2sin[1/3(3β+π/2)-π/6]=2sinβ=6/5 sinβ=3/5 β∈[0,π/2] cosβ=4/5
sin(α+β)=sinαcosβ+sinβcosα=5/13×4/
5+12/13×3/5=56/65
请问一下2.f(3α+π/2)=2sin[1/3(3α+π/2)-π/6]=2sinα=10/13 sinα=5/13 α∈[0,π/2] cosα=12/13
f(3β+π/2)=2sin[1/3(3β+π/2)-π/6]=2sinβ=6/5 sinβ=3/5 β∈[0,π/2] cosβ=4/5
这部分是怎么解的,什么1/3