三角函数证明题1)1-(1/2sin(2θ))=(sin^3(θ)+cos^3(θ)/sinθ+cosθ)2)tan(3θ)=3tanθ-tan^3(θ)/1-3tan^2(θ)

Question

三角函数证明题
1) 1- (1/2sin(2θ)) = (sin^3(θ)+cos^3(θ)/sinθ+cosθ)
2) tan(3θ)= 3tanθ-tan^3(θ)/1-3tan^2(θ)

Answer 1

证明：
1) 1- (1/2sin(2θ)) =1-sinθcosθ=sin^2(θ)+cos^2(θ)-sinθcosθ
令x=sinθ, y=cosθ
∵x^3+y^3=(x+y)(x^2-xy+y^2)
∴x^2-xy+y^2=(x^3+y^3)/(x+y)
即sin^2(θ)+cos^2(θ)-sinθcosθ=(sin^3(θ)+cos^3(θ)/sinθ+cosθ)
∴1- (1/2sin(2θ)) = (sin^3(θ)+cos^3(θ)/sinθ+cosθ)
2) 这是正切函数的三倍角公式
tan3θ=(tan2θ+tanθ)/(1-tan2θtanθ) .和角公式
=[2tanθ/(1-tan^2(θ))+tanθ]/{1-[2tan^2θ/(1-tan^2(θ))]}.二倍角公式
=[3tanθ-tan^3(θ)]/[1-3tan^2(θ)]