Trig. Identity

2013-06-25 2:55 pm
Prove the identity :
[2(sin 50 + cos 50 tan 70)]/(sqrt 3 tan 70 - 1) = sin 60/sin 40.
Note : Angles are in degrees.
更新1:

Correction : Disregard the '2' on the LHS. Sorry for the error.

更新2:

Correction : the '2' should be there !! The question is correct. Sorry again !!!

回答 (2)

2013-06-25 4:38 pm
✔ 最佳答案
[2(sin 50 + cos 50 tan 70)] / (sqrt 3 tan 70 - 1)
= [2(sin 50 + cos 50 tan 70)] / (sqrt 3 tan 70 - 1)
= [2(cos 40 + sin 40 (1 / tan 20)] / [(sqrt 3 / tan 20) - 1]
= [2(cos 40 tan 20 + sin 40] / (sqrt 3 - tan 20)
= [2(cos 40 sin 20 + sin 40 cos 20] / (sqrt 3 cos 20 - sin 20)
= 2(sin 60) / (sqrt 3 cos 20 - sin 20)
= sin 60 / [(sqrt 3 / 2) cos 20 - (1 / 2) sin 20]
= sin 60 / (sin 60 cos 20 - cos 60 sin 20)
= sin 60 / sin 40
參考: knowledge
2013-06-26 3:04 am
LHS
= 2(sin50° + cos50° tan70°)]/( sqrt3 tan70° - 1)
= 2(sin50°cos70° + cos50° sin70°)]/ [(sqrt3 tan70° - 1)cos70°] (∵cos70°≠0)
= 2( sin120°) / (sqrt3 sin70° - cos70°) (Note: sin120°= sin(180°-60°) = sin60°)
= sin60° / [(sqrt3)/2 * sin70° - (1/2)cos70°)
= sin60° / (sin70°cos30° - sin30°cos70°)
= sin60° / sin40°
= RHS


收錄日期: 2021-04-28 14:41:30
原文連結 [永久失效]:
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