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[Solved] Trigonometric functions

 
Trigonometry 1 & 2
Last Post by Staff 4 years ago
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June 15, 2021 12:42 pm
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(@armor)
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June 15, 2021 12:42 pm
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

Since   \small 4tan\frac{\alpha }{2}=1-tan^2\frac{\alpha }{2}\Rightarrow \frac{4tan\frac{\alpha }{2}}{1-tan^2\frac{\alpha }{2}}=1\Rightarrow 2tan\alpha =1\Rightarrow tan\alpha =\frac{1}{2}

Since  \small 3sin\beta =sin(2\alpha +\beta )=sin(\alpha +\beta )cos\alpha +cos(\alpha +\beta )sin\alpha     (1)

\small 3sin\beta =3sin(\alpha +\beta -\alpha )=3sin(\alpha +\beta )cos\alpha -3cos(\alpha +\beta )sin\alpha   (2)

\small (2)-(1)\Rightarrow sin(\alpha +\beta )cos\alpha =2cos(\alpha +\beta )sin\alpha \Rightarrow tan(\alpha +\beta )=2tan\alpha =1.

For \small \alpha ,\beta \in (0,\frac{\Pi }{4})  thus \small \alpha +\beta =\frac{\Pi }{4}.

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