Find $\sin(\theta)$ when $\sin(\theta) > 0$, $\tan(\theta)=7/24$.

Question

Find $\sin(\theta)$ when $\sin(\theta) > 0$, $\tan(\theta)=7/24$.

I have no idea what to do.

Answer 1

Hint

use $$\sin \theta ={\tan\theta\over \sqrt{1+\tan^2\theta}}$$whenever $\sin \theta , \tan \theta \ge 0$.

Answer 2

Use $\cos^2\theta = 1 - \sin^2\theta$ and $\tan\theta = \frac{\sin\theta}{\cos\theta}=c$. Then $\sin\theta = c\sqrt{1-\sin^2\theta}$, then $\sin^2\theta = c^2-c^2\sin^2\theta$, and finally $\sin\theta = \frac{c}{\sqrt{1+c^2}}$. Note that when taking the roots I have taken the positive such since $\sin\theta>0$ and also $\tan\theta >0$, which implies $\cos\theta > 0$.

Answer 3

Draw a right triangle that shows $\tan \theta = 7/24.$ There are infinitely many, but choosing the one with legs $7$ and $24$ is a swell choice. Now use Pythagorean Theorem to find the length of the hypotenuse. Finding the values of any of the other trig functions should be easy now.

The only snag is that you have to think about which quadrant you're in. But you have $\sin \theta$ and $\tan \theta$ both positive, so you're in the first quadrant.