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My question is:

If $\cos(\theta -\alpha) = \frac{3}{5}$ and $\sin(\theta +\alpha) =\frac{12}{13}$, find $\cos(2\alpha)$.

Attempt I: \begin{align*} &\cos^2(\theta -\alpha)+\sin^2(\theta +\alpha) = \frac{9}{25} + \frac{144}{169}\\ \Rightarrow &\cos^2(\theta -\alpha)- \cos^2(\theta +\alpha) = \frac{9}{25} + \frac{144}{169} - 1 = \frac{896}{4225}. \end{align*} But I thought it won't work.

Then I tried this:

Attempt II: \begin{align*} &\begin{cases} \cos\theta \cos\alpha + \sin\theta \sin\alpha = \cos(\theta -\alpha) = \frac{3}{5},\\ \sin\theta \cos\alpha + \cos\theta \sin\alpha = \sin(\theta +\alpha) =\frac{12}{13}, \end{cases}\\ &\cos\theta (\sin\alpha + \cos\alpha) + \sin\theta (\sin\alpha + \cos\alpha) = \frac{99}{65},\\ &\sqrt{2}\left[\sin\left(\alpha+\frac{\pi}{4}\right) \cos\left(\alpha -\frac{\pi}{4}\right)\right] = \frac{99}{65}. \end{align*} But I thought here its better to convert both parts to cosines, so I did: \begin{align*} &\sqrt{2}\left[\cos\left(\alpha-\frac{\pi}{4}\right) \cos\left(\alpha -\frac{\pi}{4}\right)\right] = \frac{99}{65},\\ &\sqrt{2} \cos^2\left(\alpha-\frac{\pi}{4}\right) = \frac{99}{65}. \end{align*} But I think it also didn't work ....

Please guide. Thanks.

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As $\cos(\theta-\alpha)=\frac35,$ $\sin(\theta-\alpha)=\pm \sqrt{1-\left(\frac35\right)^2}=\pm\frac45$

Similarly, as $\sin(\theta+\alpha)=\frac{12}{13},$ $\cos(\theta+\alpha)=\pm\sqrt{1-\left(\frac{12}{13}\right)^2}=\pm \frac5{13}$

If we assume $0<\theta \pm \alpha<\frac\pi2, $ we can consider the positive values only.

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Nice. +1. I took five seconds to realise that the 'HONT' is actually a 'HINT', though :-D. – user1551 Jul 6 '13 at 9:19
+1 very nice solution – Ovi Jul 6 '13 at 10:30
@user1551 If your question was answered you can accept this answer by clicking the check mark below the vote count. – Ovi Jul 6 '13 at 10:33
Ovi: Hont: I think that one should have been directed at @sachin – fuglede Jul 6 '13 at 10:37
@user1551 haha sorry about that. But I think sachin will still get a notification of these comments – Ovi Jul 6 '13 at 10:45

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