Originally Posted by

**Blodwynne** It's one of those trigonometric ones that always make my brain go huh?

If cos x= (-4/5) and tan x>0, find the value of Sin x.

I figured that since Cos was negative, it was in either quadrants II or III. Since Tan is positive, it must be in III. After that....I get lost.

~Wynne

Think of a right triangle. The cosine of an angle is the ratio of the adjacent side to the hypotenuse. So, ignoring the negative for now:

Code:

/|
5 / |
/ | b
/ |
/ θ |
----------
4

By the Pythagorean theorem, we have

$\displaystyle b = \sqrt{25 - 16} = \sqrt9 = 3$ (this is a Pythagorean triple: all sides are integers).

Now, $\displaystyle \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$, so we have

$\displaystyle \sin\theta = \frac b5 = \frac35$.

But, our $\displaystyle \theta$ is in quadrant I. Adding $\displaystyle 180^\circ$ to get into quadrant III will make the sine negative:

$\displaystyle \sin x = -\frac35$