<s>[INST] 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?
 [/INST] The error is that $\tan\theta$ merely gives the ratio of $\sin\theta$ and $\cos\theta$. If it is given as a fraction, that does not mean that the numerator must be $\sin\theta$ and the denominator must be $\cos\theta$. It is necessary to use trigonometric identities to solve for $\sin\theta$ and $\cos\theta$.
We have 
$$1+\tan^2\theta=\sec^2\theta$$
so
$$\cos\theta=\sqrt[]{\frac{1}{1+\tan^2\theta}}=\sqrt[]{\frac{1}{1+(\frac{12}{5})^2}}=\sqrt[]{\frac{25}{169}}=\pm\frac{5}{13}$$
Then by the trigonometric identity $\sin^2\theta+\cos^2\theta=1$ we have
$$\sin\theta=\sqrt[]{1-\cos^2\theta}=\sqrt[]{1-\frac{25}{169}}=\sqrt[]{\frac{144}{169}}=\pm\frac{12}{13}$$
Now since $\tan\theta>0$, $\theta$ is in either the first or third quadrant, so $\sin\theta$ and $\cos\theta$ have the same sign. So
$$(\sin\theta,\cos\theta)=(\frac{12}{13},\frac{5}{13}) \text{ or } (\sin\theta,\cos\theta)=(-\frac{12}{13},-\frac{5}{13})$$
Verifying our answer,
$$\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\pm\frac{12}{13}}{\pm\frac{5}{13}}=\frac{12}{5}$$
as the problem stated.
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