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
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
f(x)=2sin(1/3x-π/6)
f(3α+π/2) 中 x=3α+π/2 ,代入函数式：
f(3α+π/2)=2sin[1/3(3α+π/2)-π/6]=2sin（α+π/6-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题解已经很清楚了 啊。要问什么问题呢？1/3怎么来的，还有除宇6？已知函数f(x)=2sin(1/3x-π/6)函数就是这样的啊，当x=3α+π/2 时1/3 x= 1/3(3α+π/2)=α+π/6 所以 ：f(3α+π/2)=2sin[1/3(3α+π/2)-π/6]=2sin（α+π/6-π/6）=2sinα=10/13sinα=5/13