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  • 已知函數(shù)f(x)=2sin(1/3x-π/6),x∈R (1)求f(0)的值; (2)設(shè)α,β∈[0,π/2],f(3α+π/2)=10/13,f(3β+π

    已知函數(shù)f(x)=2sin(1/3x-π/6),x∈R (1)求f(0)的值; (2)設(shè)α,β∈[0,π/2],f(3α+π/2)=10/13,f(3β+π
    已知函數(shù)f(x)=2sin(1/3x-π/6),x∈R (1)求f(0)的值;
    (2)設(shè)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
    請(qǐng)問一下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
    這部分是怎么解的,什么1/3
    數(shù)學(xué)人氣:921 ℃時(shí)間:2019-08-21 14:46:51
    優(yōu)質(zhì)解答
    f(x)=2sin(1/3x-π/6)
    f(3α+π/2) 中 x=3α+π/2 ,代入函數(shù)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題解已經(jīng)很清楚了 啊。要問什么問題呢?1/3怎么來的,還有除宇6?已知函數(shù)f(x)=2sin(1/3x-π/6)函數(shù)就是這樣的啊,當(dāng)x=3α+π/2 時(shí)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
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