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.
Best Answer
HINT:
$$\cos(2\alpha)=\cos\{\theta+\alpha-(\theta-\alpha)\}=\cos(\theta+\alpha)\cos(\theta-\alpha)+\sin(\theta+\alpha)\sin(\theta-\alpha)$$
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.