shows that sinx tanx maybe written as 1-cos2x / c osx?
回答 (3)
2020-09-30 1:44 pm
Trigonometric identities used:
• tanθ = sinθ / cosθ
• sin²θ + cos²θ = 1 and thus sin²θ = 1 - cos²θ
L.H.S.
= sinx tanx
= sinx (sinx / cosx)
= sin²x / cosx
= (1 - cos²x) / cosx
= R.H.S.
Hence, sinx tanx = (1 - cos²x)/cosx
• tanθ = sinθ / cosθ
• sin²θ + cos²θ = 1 and thus sin²θ = 1 - cos²θ
L.H.S.
= sinx tanx
= sinx (sinx / cosx)
= sin²x / cosx
= (1 - cos²x) / cosx
= R.H.S.
Hence, sinx tanx = (1 - cos²x)/cosx
2020-09-30 7:14 pm
sinx.tanx = sinx(sinx/cosx)
i.e. sin²x/cosx
Now, you say cos2x instead of cos²x....??
Now, cos2x = 1 - 2sin²x
so, 2sin²x = 1 - cos2x
Then, sin²x = (1 - cos2x)/2
Hence, sinx.tanx => (1 - cos2x)/2cosx
Did you intended to say (1 - cos²x)/cosx...?
Well sin²x + cos²x = 1
so, sin²x = 1 - cos²x
i.e. sinx.tanx => (1 - cos²x)/cosx
:)>
i.e. sin²x/cosx
Now, you say cos2x instead of cos²x....??
Now, cos2x = 1 - 2sin²x
so, 2sin²x = 1 - cos2x
Then, sin²x = (1 - cos2x)/2
Hence, sinx.tanx => (1 - cos2x)/2cosx
Did you intended to say (1 - cos²x)/cosx...?
Well sin²x + cos²x = 1
so, sin²x = 1 - cos²x
i.e. sinx.tanx => (1 - cos²x)/cosx
:)>
2020-09-30 7:01 pm
sin(x)*tan(x) = sin(x)*[sin(x)/cos(x)] = sin^2(x)/cos(x) = [1-cos^2(x)]/cos(x)
收錄日期: 2021-04-24 08:05:00
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200930053111AAu8hDd