1. ## Quick trig question

Hi
I need help on how to do this question:

Find the value of $\displaystyle \theta$ between 0 and $\displaystyle 2\pi$ such that $\displaystyle cos\theta = \frac{3}{5}$ and $\displaystyle sin\theta = \frac{-4}{5}$

P.S

2. Originally Posted by Paymemoney
Hi
I need help on how to do this question:

Find the value of $\displaystyle \theta$ between 0 and $\displaystyle 2\pi$ such that $\displaystyle cos\theta = \frac{3}{5}$ and $\displaystyle sin\theta = \frac{-4}{5}$

P.S
This is a 3-4-5 triangle, and the angle has to be in the fourth quadrant since cosine is positive and sine is negative.

So $\displaystyle \theta = 2\pi - \arccos{\frac{3}{5}}$.

You could also do $\displaystyle \theta = 2\pi - \arcsin{\frac{4}{5}}$.

3. Originally Posted by Prove It
This is a 3-4-5 triangle, and the angle has to be in the fourth quadrant since cosine is positive and sine is negative.

So $\displaystyle \theta = 2\pi - \arccos{\frac{3}{5}}$.

You could also do $\displaystyle \theta = 2\pi - \arcsin{\frac{4}{5}}$.

I tried to do both ways and they get different answers.
if you say they are the same shouldn't they get the same answer??