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