Find the exact value of arccos(-root 3 / 2).

Step-by-step solution

1. 1

   Recall the range of arccos - this is the whole problem. $\arccos x$ returns the unique angle $\theta$ with

   $\cos\theta=x\qquad\text{and}\qquad 0\le\theta\le\pi$

   Unlike $\arcsin$, whose range is $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$, $\arccos$ never returns a negative angle. So $-\tfrac{\pi}{6}$ is not a valid answer here even though $\cos(-\tfrac{\pi}{6})=+\tfrac{\sqrt3}{2}$.

2. 2

   Find the reference angle from the magnitude. Ignoring the sign, $\tfrac{\sqrt3}{2}$ is a special value:

   $\cos\frac{\pi}{6}=\frac{\sqrt3}{2}\ \Longrightarrow\ \text{reference angle}=\frac{\pi}{6}\ (30^{\circ})$

3. 3

   Place the angle in the correct quadrant. Cosine is negative on $(\tfrac{\pi}{2},\pi)$ - the second quadrant - which is the only part of the arccos range where a negative output value can occur. The second-quadrant angle with reference $\tfrac{\pi}{6}$ is

   $\pi-\frac{\pi}{6}=\frac{5\pi}{6}$

4. 4

   Confirm the range condition. $\tfrac{5\pi}{6}=150^{\circ}$ satisfies $0\le\tfrac{5\pi}{6}\le\pi$ ✓, so it is the principal value.

5. 5

   Verify by taking the cosine back.

   $\cos\frac{5\pi}{6}=-\cos\frac{\pi}{6}=-\frac{\sqrt3}{2}$

   Numerically $\cos(2.617994)=-0.8660254$ and $-\tfrac{\sqrt3}{2}=-0.8660254$ ✓.