教育/科学

已知cos(αβ)=0,求证sin(α2β)=sinα

2005-03-02 20:16:09新***

已知cos(αβ)=0,求证sin(α2β)=sinα：cos(α + β)= 0 α + β = kπ + π/2 (k∈N) α = kπ +?

cos(α + β)= 0 α + β = kπ + π/2 (k∈N) α = kπ + π/2 -β ,sinα=cosβ, sin(α + 2β)= sin(kπ + π/2 + β)= cosβ = sinα

2005-03-02 21:51:40

因为: cos(α + β)= 0 所以: sin(α + 2β)= sin[2(α+β)-α] = sin2(α+β)*cosα - cos2(α+β)*sinα = [2sin(α+β)*cos(α+β)]*cosα - {2*[cos(α+β)]^2 - 1}*sinα = sinα

2005-03-02 21:52:06

∵cos(α + β)= 0 ∴α + β = kπ + π/2 (k∈N) ∴sin(α + 2β)= sin(kπ + π/2 + β)= cosβ = sinα

2005-03-02 20:23:50

证明：因为cos(α+β)=0 所以 α+β=kπ+π/2 因此sin(α+2β)=sin[2（α+β)-α]=sin(2kπ+π-α)=sin(π-α)=sinα

2005-03-02 20:22:28

很赞哦！ (5)