inverse functions when solving trig equations.

Question

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:

$\cos(\theta)=x$

$x= -\frac12$

$x^2+y^2=1$

$(-\frac12)^2+y^2=1$

$\frac14+y^2=1$

$y^2=\frac34$

$\sqrt y^2=\pm\sqrt\frac34$

$y=\pm\frac{\sqrt3}{2}$

$\sin(\theta)=y$

$\sin(\theta)=\frac{\sqrt3}{2}$

$\sin(\theta)=-\frac{\sqrt3}{2}$

$\sin^{-1}(\frac{\sqrt3}{2})=\theta$

$\sin^{-1}(-\frac{\sqrt3}{2})=\theta$

$\theta=60^o,300^o$

Answer 1

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 : $$cos\left(-\frac{1}{2}\right)=120^{\circ },\:240^{\circ }\:for\:0^{\circ }\le \theta \le 360^{\circ }$$

Answer 2

Recall that, by definition, $\sin^{-1}$ has range in $[-\pi/2,\pi/2]$ that is $[-90^o,90^o]$ therefore, since $\theta$ belongs to the third and fourth quadrant, we need to adjust the values as follows

• $\sin (60^o)=\sin (180^o-60^o)=\sin(120^o)$

• $\sin (300^o)=\sin (180^o-300^o)=\sin(-120^o)=\sin(240^o)$

which are the correct values.

In conclusion, your work is almost fine with this correction. In general, to solve this kind of equation I warmly suggest to refer to the unit circle definition for the trigonometric function, in this way the solution becomes trivial. Just draw a vertical line passing throught the point $(-1/2,0)$ and determine the intersection with the unit circle.

Answer 3

Since $\cos\theta=-\frac{1}{2}$ and $\sin\theta=\pm\frac{\sqrt 3}{2}$ implies that $\theta$ lies in the quadrant second or third, so $\theta$ is $\pi\mp\frac{\pi}{3}=\frac{2\pi}{3}$ or $\frac{4\pi}{3}.$