高二數學-三角函數

1.設*=180度

化簡cos2*/7乘cos4*/7乘cos8*/7之值為?2.設t=cos2*

則2(sin^8*-cos^8*)可表示為?(以t表示)
2.設t=cos2θ

則2(sin^8θ-cos^8θ)可表示為?(以t表示)t=cos^2θ – sin^2θ

cos^2θ sin^2θt^2= cos^4θ sin^4θ-2cos^2θsin^2θ---(1)1= cos^4θ sin^4θ 2cos^2θsin^2θ ---(2)(1) (2)得 (t^2 1)=2(cos^4θ sin^4θ)2(sin^8θ-cos^8θ )=2(sin^4 θ cos^4 θ) (sin^4 θ-cos^4θ)=2(sin^4 θ cos^4 θ) (sin^2 θ cos^2θ) (sin^2 θ-cos^2 θ)=2(sin^4 θ cos^4 θ) t=(t^2 1)t=t^3 thttp://tw.myblog.yahoo.com/sincos-heart/article?mid=3661