# If sin(theta)=1/3, what is tan(theta)?

Aug 8, 2016

For this problem, a diagram may help.

But before I present to you the diagram, I think it would help to review a bit of terminology.

First of all, recall that $\sin \theta = \text{opposite"/"hypotenuse}$. Also, recall that $\tan \theta = \text{opposite"/"adjacent}$. We're given that $\sin \theta = \frac{1}{3}$, which means that the side opposite $\theta$ measures $1$ unit and the hypotenuse measures $3$.

However, we don't know the measure of the side adjacent. Since we're working with right triangles, we can use pythagorean theorem to determine the missing side.

Let the side adjacent $\theta$ (the one we would like to find) be $b$.

${a}^{2} + {b}^{2} = {c}^{2}$

${b}^{2} = {c}^{2} - {a}^{2}$

$b = \sqrt{{c}^{2} - {a}^{2}}$

$b = \sqrt{{3}^{2} - {1}^{2}}$

$b = \sqrt{9 - 1}$

$b = \sqrt{8}$

$b = 2 \sqrt{2}$

Now here's where the diagram is really useful. Since the problem, doesn't specify a quadrant, we have to determine all quadrants where sine is negative. The question we must ask ourselves is: Where is the side opposite our angle positive? The answer to this question: Quadrants I and II. If you aren't familiar with quadrants, I have included an image that locates them for you at the bottom of the page.

As the diagram above shows, the difference between quadrants I and II is that the x axis is positive in the first quadrant compared to negative in the second quadrant. This means that tangent of $\theta$, abbreviated to $\tan \theta$, will be positive in the first quadrant though negative in the second. In other words, this problem will have two separate answers.

All that is now left to do is to apply the definition of tangent and find our wanted ratio.

As mentioned earlier, $\tan \theta = \text{opposite"/"adjacent}$.

Quadrant 1: The opposite side measures $1$ and the adjacent side measures $2 \sqrt{2}$. Hence, $\tan \theta = \frac{1}{2 \sqrt{2}} = \frac{\sqrt{8}}{8} = \frac{2 \sqrt{2}}{8} = \textcolor{b l u e}{\frac{\sqrt{2}}{4}}$

Quadrant 2: The opposite side measures $1$ and the adjacent side measures $- 2 \sqrt{2}$. Hence, $\tan \theta = - \frac{1}{2 \sqrt{2}} = - \frac{\sqrt{8}}{8} = - \frac{2 \sqrt{2}}{8} = \textcolor{b l u e}{- \frac{\sqrt{2}}{4}}$

Finally, here are the placements of the quadrants, as promised earlier.

Practice exercises:

1. Determine the value of $\cos \alpha$ if $\tan \alpha = - \frac{4}{3}$, if $\sin \alpha < 0$.

$2$. Determine the value of $\csc \beta$ if $\cos \beta = - \frac{19}{7}$, if $\cot \beta > 0$

Hopefully this helps, and good luck!