# Math Help - Triganometry , Help please

1. ## Triganometry , Help please

Hi, could someone tell what what i need to do to answer these q's please, thank you

All the angles are in interval -180degrees to 180degrees.
So , given that sina<0 and cos a =0.5 find a.

Im not sure what to do, so any help is great thank you all.

2. Originally Posted by Chez_
Hi, could someone tell what what i need to do to answer these q's please, thank you

All the angles are in interval -180degrees to 180degrees.
So , given that sina<0 and cos a =0.5 find a.

Im not sure what to do, so any help is great thank you all.
The reference angle is $\frac{\pi}{3}$. And a is in quadrant IV. So what is a?

-Dan

3. First we need to find a reference angle. To do this, find $cos \frac {1}{2}$.

Recall the special triangle with sides 1, 2, and $\sqrt {3}$.

Where n is equal to any number. In this case, n is 1.

Cosine is equal to adjacent over hypotenuse, so in this case:

$cos \frac {1}{2}$

The adjacent side must be 1 and the hypotenuse must be 2. What angle in this triangle represents that? The 60 degree angle, or $\frac {\pi}{3}$ in radians.

So, $cos \frac {1}{2}$ is equal to $\frac {\pi}{3}$.

Now that we have our reference angle, think about the quadrants. We know that $sin a$ is less than zero and therefore negative, so which quadrants are $sin$ negative in? The third and fourth quadrants.

Recall that sin is a periodic function, repeating every $2\pi$.

So, the values of a that satisfy this equation are:

$2n\pi + \frac {\pi}{3}$, where n is any integer.

For which values of n does this equation fall into the third or fourth quadrant? The only correct answer is 1.

$2\pi + \frac {\pi}{3}$

$= \frac {7\pi}{3}$