Complex Variables

Question

I am studying complex variables but I seem to have forgotten basic trig!! I have sin($\theta$) = $\sqrt3$/$2$ and cos($\theta$)=$-1$/$2$. I need to find $\theta$ in order to find arg(z). In my notes $\theta$=$2\pi$/$3$. How did they find this without using a calculator? Was it through triangles? No working was shown for this.

Answer 1

If $\sin(\theta) = \frac{\sqrt{3}}{2}$ and $\cos(\theta) = -\frac{1}{2}$, then $\frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta) = -\sqrt{3}$. Recall that if $\tan(x) = y$, then $y = \begin{cases} \arctan(y)+2\pi k\\ \pi+\arctan(y)+2\pi k\end{cases} = \arctan(y)+\pi k$ for $k \in \mathbb{Z}$.

Notice that sine is positive and cosine is negative, so the angle must be in quadrant $2$. However, arctangent returns an angle in quadrant $1$ (for positive arguments) and $4$ (for negative arguments) only. This is because the range of $\arctan(x)$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Here, taking $\arctan\left(-\sqrt{3}\right)$ gives us $-\frac{\pi}{3}$, so we add $\pi$ in order to get the angle in the second quadrant - $\frac{2\pi}{3}$.

Similarly, if both sine and cosine are negative, then the angle desired is in quadrant $3$, but since tangent is positive, the angle given by taking the actangent would be the quadrant $1$ angle, so you'd have to add $\pi$ to get the quadrant $3$ angle.

Answer 2

Since $\cos\theta=\frac12$ has acute root $\frac{\pi}{3}$, $\cos\theta=-\frac12$ has obtuse root $\frac{2\pi}{3}$. By $\sin(\pi-x)=\sin x$, we've also satisfied $\sin\theta=\frac{\sqrt{3}}{2}$.