<s>[INST] How to get sine / cosine value out of tangens I know that: $\tan(\alpha) = 1/2$.
How can I get clean values for sine / cosine without the calculator?
Is there a relationship?
I know that $\sin(\arctan(1/2))$ is a way ... But I hope you get the point.
Thank you!
 [/INST] Here is an algebraic solution.
Recall that $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$. Now your equation states that
$$\sin \alpha = \frac{1}{2}\cos \alpha.$$
Squaring both sides gives
$$\sin^2 \alpha = \frac{1}{4} \cos^2 \alpha = \frac{1}{4}(1 - \sin^2 \alpha).$$
Or
$$\sin^2 \alpha = \frac{1}{5}.$$
Similarly one sees that
$$\cos^2 \alpha = \frac{4}{5}.$$
Thus the possibilities are $$\sin \alpha = \pm \frac{1}{\sqrt{5}}, \quad \cos \alpha = \pm \frac{2}{\sqrt{5}}.$$
We still have to show that these possibilities are possible. We know that there is $\alpha$ such that $\tan \alpha = \frac{1}{2}$. I.e. the original equation has a solution. Now if $\alpha$ is a solution, so is $\alpha + \pi$. From this we can deduce that both $\sin \alpha = \frac{1}{\sqrt{5}}$ and $\sin \alpha = -\frac{1}{\sqrt{5}}$ are attainable and so are $\cos \alpha = \pm \frac{1}{\sqrt{5}}$. Moreover the signs must be the same for both. Thus the solutions are
$$(\cos \alpha, \sin \alpha) \in \{(\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}), (-\frac{1}{\sqrt{5}}, -\frac{1}{\sqrt{5}})\}.$$
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