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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!

  • 0
    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 $\tan(\alpha)=\frac{1}{2}$.2012-01-05

4 Answers 4

9

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

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

5

As suggested by Jonas:

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

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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".

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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.

4

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

1

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 $(\cos \alpha, \sin \alpha) \in \{(\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}), (-\frac{1}{\sqrt{5}}, -\frac{1}{\sqrt{5}})\}.$