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[Solved] Trigonometric equations

 
Trigonometry 1 & 2
Last Post by Staff 4 years ago
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June 15, 2021 3:09 pm
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(@armor)
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June 15, 2021 3:09 pm
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

Let      \small \frac{sin\alpha -cos\alpha }{sin\alpha +cos\alpha }=k\Rightarrow (1-k)sin\alpha =(1+k)cos\alpha        (1)

Denote \small sin\alpha =2-3cos\alpha         (2)

Applying (2) / (1), we have \small cos\alpha =\frac{1-k}{2-k}.

Substituting it into (1), we have  \small sin\alpha =\frac{1+k}{2-k} (k\neq 2).

Since \small sin^2\alpha +cos^2\alpha =1,  then 

\small (\frac{1+k}{2-k})^2+(\frac{1-k}{2-k})^2=1.   

It can be written as \small k^2+4k-2=0.

Solving the equation, we have \small k=-2\pm \sqrt{6}.

As a conclusion,  \small \frac{sin\alpha -cos\alpha }{sin\alpha +cos\alpha }=-2\pm \sqrt{6}.

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