inverse functions when solving trig equations. I am in Pre-calc this semester, and I have been given the problem:
$\cos(\theta) = -\frac12$
The answers I get are $30^o$ and $300^o$ but the answers I get from the homework key are $120^o $ and $240^o$. Why do the inverse equations give me answers that arent correct? Is there a better way to solve these problems? My work is as follows:
$\cos(\theta)=x$
$x= -\frac12$
$x^2+y^2=1$
$(-\frac12)^2+y^2=1$
$\frac14+y^2=1$
$y^2=\frac34$
$\sqrt y^2=\pm\sqrt\frac34$
$y=\pm\frac{\sqrt3}{2}$
$\sin(\theta)=y$
$\sin(\theta)=\frac{\sqrt3}{2}$
$\sin(\theta)=-\frac{\sqrt3}{2}$
$\sin^{-1}(\frac{\sqrt3}{2})=\theta$
$\sin^{-1}(-\frac{\sqrt3}{2})=\theta$
$\theta=60^o,300^o$

The formula for $cosine$ is given by :
$$cos\left(\theta \right)=\frac{adjacent}{hypotenuse}$$
Since that :
$$cos\left(\theta \right)=-\frac{1}{2}$$
So, it can be either :
$$-\frac{1}{2}=\frac{-1}{2}$$
or
$$-\frac{1}{2}=\frac{1}{-2}$$
Since $hypotenuse$ cannot be negative, so it should be :
$$cos\left(\theta \right)=-\frac{1}{2}=\frac{-1}{2}$$
where :
$$adjacent=-1$$
$$hypotenuse=2$$
Since the value of adjacent is negative, so it should consist the angle in second quadrant and third quadrant. So, just use the formula :
Second quadrant :
$$cos\left(-\frac{1}{2}\right)=180^{\circ }-cos\left(\frac{1}{2}\right)=180^{\circ \:}-60^{\circ }=120^{\circ }$$
Third quadrant :
$$cos\left(-\frac{1}{2}\right)=180^{\circ }+cos\left(\frac{1}{2}\right)=180^{\circ \:}+60^{\circ }=240^{\circ }$$
So, the answer is :
$$cos\left(-\frac{1}{2}\right)=120^{\circ },\:240^{\circ }\:for\:0^{\circ }\le \theta \le 360^{\circ }$$
