這題三角函數要怎麼算啊

圖片參考：http://imgcld.yimg.com/8/n/AB00183634/o/151202250262013872144130.jpg

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_(cos³θ-cos3θ)/cosθ+(sin³θ+sin3θ)/sinθ=[cos³θ-cos(2θ+θ)]/cosθ+[sin³θ+sin(2θ+θ)]/sinθ=[cos³θ-(cos2θcosθ-sin2θsinθ)/cosθ+[sin³θ+(sin2θcosθ+sinθcos2θ)/sinθ=[cos³θ-cos2θcosθ+sin2θsinθ)/cosθ+[sin³θ+sin2θcosθ+sinθcos2θ)/sinθ=[cos³θ-(2cos²θ-1)cosθ+2(sinθcosθ)sinθ]/cosθ+[sin³θ+(2sinθcosθ)cosθ+sinθ(1-2sin²θ)]/sinθ=(cos³θ-2cos³θ+cosθ+2sin²θcosθ)/cosθ+(sin³θ+2sinθcos²θ+sinθ-2sin³θ)/sinθ=(-cos³θ+2sin²θcosθ+cosθ)/cosθ+(-sin³θ+2sinθcos²θ+sinθ)/sinθ=(-cos²θ+2sin²θ+1)+(-sin²θ+2cos²θ+1)=cos²θ+sin²θ+2=1+2=3
p.s.重要公式:1) cos²θ+sin²θ=12) cos(x+y)=cosxcosy-sinxsiny3)sin(x+y)=sinxcosy+sinycosx4) cos2θ=cos²θ-sin²θ=2cos²θ-1=1-2sin²θ5) sin2θ=2sinθcosθ

cos(3a)=4[cos(a)]^3-3cos(a)
sin(3a)=3sin(a)-4[sin(a)]^3代入原式

即求{[cos(a)]^3-4[cos(a)]^3+3cos(a)}/cos(a)
+{[sin(a)]^3+3sin(a)-4[sin(a)]^3}/sin(a)
=-3[cos(a)]^2+3-3[sin(a)]^2+3
=6-3{[sin(a)]^2+[cos(a)]^2]
=6-3*1
=3