Trigonometric questions
回答 (3)
2009-06-05 6:47 am
tanθ=sinθ/cosθ
tan^2θ=sin^2θ/cos^2θ
r^2=sin^2θ/(1-sin^2θ)
r^2-r^2sin^2θ=sin^2θ
r^2=sin^2θ(1+r^2)
sin^2θ=r^2/(1+r^2)
sinθ=+/-r/sqrt(1+r^2)
tan^2θ=sin^2θ/cos^2θ
r^2=sin^2θ/(1-sin^2θ)
r^2-r^2sin^2θ=sin^2θ
r^2=sin^2θ(1+r^2)
sin^2θ=r^2/(1+r^2)
sinθ=+/-r/sqrt(1+r^2)
2009-06-05 6:36 am
(Using a diagram to respresent the question will be better.)
Given that tanθ=r,
i.e. tanθ=r/1
In the graph, we can say that the opp.side(BC) of θ is r, and
the adj. side(AB) is 1.
From the Pyth. thm.,
we have the hyp.(AC) =±√(1+r^2).
Therefore, we have sinθ=BC/AC=± r/√(1+r^2).
ps. the diagram I drew is:
http://i146.photobucket.com/albums/r254/kwanhimshek/hk_diff_trigo_simplify.jpg
Given that tanθ=r,
i.e. tanθ=r/1
In the graph, we can say that the opp.side(BC) of θ is r, and
the adj. side(AB) is 1.
From the Pyth. thm.,
we have the hyp.(AC) =±√(1+r^2).
Therefore, we have sinθ=BC/AC=± r/√(1+r^2).
ps. the diagram I drew is:
http://i146.photobucket.com/albums/r254/kwanhimshek/hk_diff_trigo_simplify.jpg
2009-06-05 5:51 am
Is θ an acute angle?
收錄日期: 2021-04-22 00:39:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090604000051KK01737