Trig3

From Exampleproblems

If $\sin(x)=\frac{-4}{5}\,$ and $x\,$ is in the third quadrant, find $\cos(x)\,$ .

1 Short Version
2 Long Version

Short Version

Since x is in the third quadrant, cos(x) is negative.

So, $\cos (x) = -\sqrt{1-\sin^2(x)} = -\sqrt{1- \left ( \frac{4}{5} \right )^2} = -\frac{3}{5}.$

Long Version

First, if $\sin(x) = y\,$ , then $x = \sin^{-1}(y)\,$ . Note that in Trigonometry, this notation does not mean the reciprocal of $\sin(y)\,$ , which would be denoted $(\sin(y))^{-1} = \frac{1}{\sin(y)}$ . Instead, this is the inverse that takes ratios back to angles, and is often referred to as the arcsin to prevent confusion.

So, for this problem, calculate $\cos\left(\sin^{-1}\left(\frac{-4}{5}\right)\right)$ .

Alternatively, using right triangles (and ignoring the sign for the time being)

$\sin(x) = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}} = \frac{4}{5}$

Thus, by the Pythagorean Thereom

$\mathrm{hypotenuse}^2\,$	$= \mathrm{opposite}^2 + \mathrm{adjacent}^2\,$
$5^2\,$	$= 4^2 + \mathrm{adjacent}^2\,$
$25\,$	$= 16 + \mathrm{adjacent}^2\,$
$25 - 16\,$	$= \mathrm{adjacent}^2\,$
$9\,$	$= \mathrm{adjacent}^2\,$
$3\,$	$= \mathrm{adjacent}\,$

$\cos(x) = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} = \frac{3}{5}$

Therefore, since $\cos(x)\,$ is negative in the third quadrant, the solution is $-\frac{3}{5}$ . This could also be seen by inspection, since this is a 3-4-5 triangle.