三角恒等式的推導

1)sinα sinβ sinγ= (-1/2) ( cos(α + β) - cos(α - β) ) sinγ= (-1/2) ( cos(α + β) sinγ - sinγ cos(α - β) )= (-1/2) ( (1/2)sin(α+β+γ) - (1/2)sin(α+β-γ) - (1/2)sin(γ+α-β) - (1/2)sin(γ-α+β) )= [sin(β+γ-α) + sin(γ+α-β) + sin(α+β-γ) - sin(α+β+γ)] / 4
2)[cos(β+γ-α) + cos(γ+α-β) + cos(α+β-γ) + cos(α+β+γ)] / 4= [ 2 cosγ cos(β - α) + 2 cos(α + β) cosγ ] / 4 = (1/2) cosγ ( cos(β - α) + cos(α + β) )= (1/2) cosγ ( cosβ cosα + sinβ sinα + cosα cosβ - sinα sinβ )= cosα cosβ cosγ