# Why do we omit the negative sign when finding basic angle when solving trigonometric equations?

Suppose I'm asked to solve for $\cos \theta = -0.5$, for $\theta$ between $0^\circ$ and $360^\circ$ inclusive.

I am told that the first step would be to buy the basic angle, $\alpha$. The way I am told to do this is by omitting the negative sign from -0.5 and hence finding sine inverse of 0.5 to get $60^\circ$. I understand the next steps of identifying the quadrant $\theta$ lies in etc: what I don't understand is why I need to omit the negative sign when finding the basic angle. I know the basic angle itself is acute and positive but why must the trigonometric function of the angle be positive for me to find the basic angle too?

Omission of the negative sign is not necessary.

The technique you described in your question involves finding a reference angle in the first quadrant (the reason you drop the negative sign), determining the quadrants in which the cosine is negative, and then finding the angles in those quadrants with that reference angle.

In this case, the reference angle is $60^\circ$, the quadrants in which the cosine is negative are the second and third quadrants (since the $x$-coordinate is negative), and the angles in those quadrants with a reference angle of $60^\circ$ are, respectively, $180^\circ - 60^\circ = 120^\circ$ and $180^\circ + 60^\circ = 240^\circ$. While this method works, it is not necessary.

The arccosine function (inverse cosine function) is defined by $$\arccos x = \theta \iff x = \cos\theta, 0 \leq \theta \leq \pi$$ where $\theta$ is measured in radians. Since $2\pi/3$ is the unique angle $\theta$ in the interval $[0, \pi]$ such that $\cos\theta = -1/2$, $$\arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$$ Thus, one solution of the equation $\cos\theta = -1/2$ is $2\pi/3$, which is $120^\circ$. Using symmetry, we obtain $$\cos\theta = \cos(2\pi - \theta)$$ since the $x$-coordinates of the points where the terminal sides of the angles $\theta$ and $2\pi - \theta$ intersect the unit circle are equal. (The angle $\theta$ in the diagram is not the same as the one in the problem.) Hence, another solution is $$2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}$$ which is $240^\circ$.