Trig identities help!?
回答 (1)
2016-07-20 7:08 pm
L.H.S.
= sin²(2x) / sin²(x)
= [2 sin(x) cos(x)]² / sin²(x)
= 4 sin²(x) cos²(x) / sin²(x)
= 4 cos²(x)
= 4 [1 - sin²(x)]
= 4 - 4sin²(x)
= R.H.S.
Hence, sin²(2x) / sin²(x) = 4 - 4sin²(x)
= sin²(2x) / sin²(x)
= [2 sin(x) cos(x)]² / sin²(x)
= 4 sin²(x) cos²(x) / sin²(x)
= 4 cos²(x)
= 4 [1 - sin²(x)]
= 4 - 4sin²(x)
= R.H.S.
Hence, sin²(2x) / sin²(x) = 4 - 4sin²(x)
收錄日期: 2021-04-18 15:21:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160720110408AACoH0Q