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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!

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There is not a unique solution unless you make some additional assumption on $\alpha$, such as that $\alpha$ is acute. In that case, I suggest sketching a right triangle having an angle $\alpha$ with sides chosen such that $\tan(\alpha)=\frac{1}{2}$. – Jonas Meyer Jan 5 '12 at 11:37
up vote 5 down vote accepted

Coming from an algebraic perspective, consider the well-known identity, $\sin^2 \theta + \cos^2 \theta \equiv 1\, (*)$.

We also know that $\tan \theta \equiv \frac{\sin \theta}{\cos \theta}\, (**)$. So squaring both sides and using $(*)$ on the denominator gives

$$\tan^2 \theta = \frac{\sin^2 \theta}{1-\sin^2 \theta}$$

This rearranges to

$$\sin^2 \theta = \frac{\tan^2 \theta}{1+\tan^2 \theta}$$

This is the best we can do, because $(**)$ above holds if we replace $(\sin \theta, \cos \theta)$ by $(-\sin \theta, -\cos \theta)$, so it's impossible to obtain the sign of $\sin \theta$ from only knowing the value of $\tan \theta$.

We can then obtain an expression for $\cos^2 \theta$ by using $(*)$ again, namely

$$\cos^2 \theta = \frac{1}{1+\tan^2 \theta}$$

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As suggested by Jonas:

1) Draw a right triangle and label one of the (non $90^\circ$) angles $\alpha$.

enter image description here

2) You know that the tangent of $\alpha$ is ${1\over2}$. Since $\tan={\text{opposite}\over \text{adjacent}}$, you can label the side of the triangle adjacent to $\alpha$ "1" and the opposite side "2".

enter image description here

3) By the Pythagorean theorem, you can find the length of the hypotenuse of the triangle.

enter image description here

4) Now you can read $\sin(\alpha)$ from the completed triangle. And remember sin from the angle is opposite/hypotenuse - 1/radical 5

This will give you one solution. There is another solution, given when $\alpha$ is a third quadrant angle.

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You can obtain the cosine as follows:

$$ \begin{align*} cos^2 x + \sin^2 x &= 1 \\ \cos^2 x + \cos^2x \tan^2 x &= 1 \\ \cos^2 x(1 + \tan^2 x) &= 1 \\ \cos x &= \pm \frac{1}{\sqrt{1 + \tan^2 x}} \end{align*} $$

Then $\sin x = \pm \sqrt{1 - \cos^2 x}$

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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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