一題三角難題

θ=π/7
1. (tan^2)θ=(sin^2)θ/(cos^2)θ=[1-cos2θ]/ [1+cos2θ]….倍角公式
2.cos2θcos4θcos6θ=1/8……….倍角公式
3. cos2θ+cos4θ+cos6θ=-1/2…..棣美弗或積化和差
4. cos2θcos4θ+cos4θcos6θ+ cos2θcos6θ=-1/2…(由2.3. 倍角公式)
5. [1+cos2θ] [1+cos4θ] [1+cos6θ]= 1/8…….(由2.3.4)
6. [1-cos2θ] [1+cos4θ] [1+cos6θ]
+[1+cos2θ] [1-cos4θ] [1+cos6θ]
+[1+cos2θ] [1+cos4θ] [1-cos6θ]=21/8….. (由2.3.4)
(故)(tan^2)θ+ (tan^2) 2θ+ (tan^2) 3θ
= [1-cos2θ]/ [1+cos2θ]+ [1-cos4θ]/ [1+cos4θ]+ [1-cos6θ]/ [1+cos6θ]
=6./5.
=21

恩，就是C(2n+1,2)

θ=π/(2n+1), n in N, then (tanθ)^2 + (tan 2θ)^2 +...+(tan nθ)^2 = n(2n+1)