Is it possible to find the sine or cos from a tangent? I have a value of a tangent. Is it possible to find the sine and/or cossine from that value? How?
 A: Yes you can. Try drawing a right angle triangle with the angle you're interested in (it doesn't need to be to scale). What side lengths could you use as the opposite and adjacent given your value of tan of the angle? Now you can you Pythagoras's Theorem to determine the remaining side length and hence you can find what sin and cos of the angle would be.
A: Yes, but you must know the range of the angle:
$$1+\tan^2x=\frac1{\cos^2x}\;\ldots$$
A: Hint:$$1+\tan^2(x)=\frac{1}{\cos^2(x)}$$$$1+\cot^2(x)=\frac{1}{\sin^2(x)}$$$$\tan(x)=\frac{1}{\cot(x)}$$
A: If $\tan\theta=\frac ab$
$$\frac{\sin\theta}a=\frac{\cos\theta}b=\pm\sqrt{\frac{\sin^2\theta+\cos^2\theta}{a^2+b^2}}=\pm\frac1{\sqrt{a^2+b^2}}$$
If $b=1,$ 
$$\tan\theta=a,\frac{\sin\theta}a=\frac{\cos\theta}1=\pm\frac1{\sqrt{a^2+1}}$$
A: Not completely: There's a "$\pm$" issue.
We have the standard identity $\sec^2\theta=1+\tan^2\theta$.
So $\cos^2\theta=\dfrac{1}{1+\tan^2\theta}$, and so $\cos\theta=\pm\dfrac{1}{\sqrt{1+\tan^2\theta}}$.
Once you know $\cos^2\theta$, you have $\sin^2\theta= 1-\cos^2\theta$, and then $\sin\theta=\pm\sqrt{1-\cos^2\theta}$.
Notice that if $\theta$ is in the first quadrant, then $\pi+\theta$ is in the third quadrant and both of those angles have the same tangent, but for one of them the sine and cosine are positive, whereas for the other they're both negative.  If $\theta$ is in the second quadrant, then $\theta$ and $\pi+\theta$ again have the same tangent but for $\theta$ the sine is positive and the cosine is negative, whereas for $\pi+\theta$ the sine is negative and the cosine is positive.
