Processing math: 100%

Thursday, November 6, 2014

2014/101) If \sin\theta = k \sin(\theta + 2\alpha) prove that \tan (\theta + \alpha) = \dfrac{1+k}{1-k}\tan \alpha

we have
k = \dfrac{sin\theta}{\sin(\theta + 2\alpha)}
using componendo dividendo
 \dfrac{k+1}{k-1} = \dfrac{\sin\theta+\sin(\theta + 2\alpha)}{\sin\theta-\sin(\theta + 2\alpha)}
=\dfrac{ 2 \sin(\theta + \alpha ) \cos \theta}  {-2 \cos (\theta + \alpha) \sin \theta}
= \dfrac{ - \tan(\theta + \alpha )}{\tan \theta}

hence \dfrac{1 + k} {1- k} = \dfrac{\tan(\theta + \alpha )}{\tan \theta}

or  \dfrac{1 + k} {1- k}\tan \theta = \tan(\theta + \alpha) 
proved

No comments: