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$\cos(\theta)-\tan(\phi)\sin(\theta) = a$

From the above equation, how can we find $\theta$? We know $\phi$ and $a$ but not $\theta$. I tried, but wasn't be able to figure it out.

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2 Answers 2

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$\tan(\phi) = \dfrac {\sin(\phi)}{\cos \phi}$.

and use the formula $\cos (A + B) = \cos A \cos B – \sin A \sin B$.

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There are several strategies to solve this problem. One of them could be to bring $\tan\phi\sin\theta$ to the right, then write $\cos\theta$ as $\sqrt{1-\sin^2\theta}$ and then square both sides to obtain a quadratic in $\sin\theta$. But this technique usually introduced repeated roots. So I recommend avoiding it.

The other strategy is to write $\tan\phi$ as $\frac{\sin\phi}{\cos\phi}$, then multiply $\cos\phi$ on both sides and simplify.

$ \cos\theta - \frac{\sin\phi}{\cos\phi}\sin\theta = a \\ \implies \cos\theta\cos\phi - \sin\theta\sin\phi = a\cos\phi \\ \implies \cos (\theta + \phi) = a\cos\phi $

Let $\cos\delta = a\cos(\theta + \phi)$ where $0\le \delta < \pi$. Thus,

$$ \delta = \cos^{-1} (a\cos\phi). $$

Also,

$$ \cos(\theta + \phi) = \cos\delta \\ \implies \theta + \phi = 2n\pi \pm \delta\\ \implies \theta = 2n\pi \pm \delta - \phi $$

where $n\in \Bbb Z$. Of course while writing $\tan\phi = \frac{\sin\phi}{\cos\phi}$, we assume that $\phi \ne \frac{n\pi}{2}$.

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