Question

三角函数证明题？要过程！

见图片

Answer 1

证明见图：

Answer 2

(1)
tanθ=(tan(θ/2))/[1-(tan(θ/2)^2]
-2/(tanθ)=-[1-(tan(θ/2)^2]/(tan(θ/2))
tan(θ/2)-1/[tan(θ/2)]=[(tan(θ/2)^2-1]/(tan(θ/2))
=-[1-(tan(θ/2)^2]/(tan(θ/2))=-2/(tanθ)
(2)
(cosφ+sinφ)^2=(sinφ)^2+(cosφ)^2+2sinφcosφ
=1+sin2φ
所以(1+2sin2φ)/(cosφ+sin φ)=cosφ+sinφ
(3)
1+cos 2θ+2(sinθ)^2
=1+1-2(sinθ)^2+2(sinθ)^2
=2
(4)
(1+sin2θ-cos2θ)/(1+sin2θ+cos2θ)
=(1+2sinθcosθ-(1-2(sinθ)^2))/(1+2sinθcosθ+2(cosθ)^2-1)
=(2sinθcosθ+2(sinθ)^2)/(2sinθcosθ+2(cosθ)^2)
=sinθ/cosθ
=tanθ