Find the exact value of $\sin (\theta)$ and $\cos (\theta)$ when $\tan (\theta)=\frac{12}{5}$ So I've been asked to find $\sin(\theta)$ and $\cos(\theta)$ when $\tan(\theta)=\cfrac{12}{5}$; my question is if $\tan (\theta)=\cfrac{\sin (\theta) }{\cos (\theta)}$ does this mean that because  $\tan (\theta)=\cfrac{12}{5}$ then $\sin (\theta) =12$ and $\cos(\theta)=5$?
It doesn't seem to be the case in this example, because $\sin (\theta)\ne 12 $ and $\cos (\theta)\ne 12 $. 
Can someone tell me where the error is in my thinking?

Most of the above solutions assume, wrongly, that $\theta$ is acute, i.e. $0^{\circ} < \theta < 90^{\circ}.$ Here I give all the cases for $0^{\circ}<\theta <360^{\circ}.$ I like to use simple identities to answer these questions. We know that $1+\tan^2\theta \equiv \sec^2\theta$. If $\tan\theta = \frac{12}{5}$ then we see that:
$$\begin{eqnarray*}
1+\tan^2\theta &\equiv& \sec^2\theta \\ \\
1+\left(\frac{12}{5}\right)^{\! 2} &=& \sec^2\theta \\ \\
\frac{169}{25} &=& \sec^2\theta
\end{eqnarray*}$$
It follows that $\sec\theta = \pm \frac{13}{5}$, and hence $\cos\theta = \pm \frac{5}{13}$. The question is: which is it, plus or minus? Sadly, there is not enough information in the question. 
Looking at the graph $y=\tan\theta$ we see that $\tan \theta > 0$ if $0^{\circ} < \theta < 90^{\circ}$ or $180^{\circ} < \theta < 270^{\circ}$. That means $\tan \theta = \frac{12}{5} > 0$ has solutions in the range $0^{\circ} < \theta < 90^{\circ}$ or $180^{\circ} < \theta < 270^{\circ}$. However, $\cos \theta > 0$ when $0^{\circ} < \theta < 90^{\circ}$ and $\cos \theta < 0$ when $180^{\circ} <\theta < 270^{\circ}$.
To conclude: $\cos\theta = \frac{5}{13}$ if $0^{\circ} < \theta < 90^{\circ}$ and $\cos\theta = -\frac{5}{13}$ if $180^{\circ} < \theta < 270^{\circ}$.
There are many ways to find $\sin \theta$. You could use the identity $1+\cot^2\theta \equiv \operatorname{cosec}^2\theta$, noting that $\cot$ is the reciprocal of $\tan$ and $\tan\theta = \frac{12}{5} \iff \cot\theta = \frac{5}{12}$. Alternatively, you could use your value of $\cos\theta$ along with the identity $\sin^2\theta + \cos^2\theta \equiv 1$. It's up to you; I'd use the second one.
