It is given that sin2θ+sinθ-1=0. Without solving for the values for θ, prove that
(a) cos4θ+cos2θ-1=0, and
(b) cos8θ+cos6θ+cos2θ-1=0
Question (Trigonometry)
2008-03-22 11:00 pm
回答 (1)
2008-03-22 11:32 pm
✔ 最佳答案
a. cos4θ+cos2θ-1=(1-sin^2(θ))^2-sin^2(θ)
=sin^2(θ)-sin^2(θ)=0
b.cos8θ+cos6θ+cos2θ-1
=cos^8(θ)+cos^6(θ)-cos^4(θ)
=cos^4(θ)(cos4θ+cos2θ-1)=0
參考: me
收錄日期: 2021-04-22 00:14:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080322000051KK01654