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this question is giving me some issue because i know sin and cos is the same as $45$ degrees but there is a $2$ on the right side where cosine is.

So how would i get $\theta$?

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    $\begingroup$ Try squaring both sides and remember the relationship between $\sin^2$ and $\cos^2$ $\endgroup$ – Simon S Oct 6 '15 at 2:01
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Square both sides to get

$\begin{array}{l} {\sin ^2}\theta = 4{\cos ^2}\theta \Rightarrow 1 - {\cos ^2}\theta = 4{\cos ^2}\theta \Rightarrow \\ {\cos ^2}\theta = \frac{1}{5} \end{array}$

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  • $\begingroup$ oh, how simple. $\endgroup$ – Caddy Heron Oct 6 '15 at 2:12
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Let $\theta \in \mathbb{R}$; note that $\cos^{2}\theta = 1 - \sin^{2}\theta$. Then by assumption $\cos^{2}\theta = 1 - 4\cos^{2}\theta$, so $\cos^{2}\theta = 1/5$.

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