What is cosine superscript negative 1 baseline left parenthesis negative start fraction start root 3 end root over 2 end fraction right parenthesis?

Question

(1 point)
Responses

start fraction pi over 6 end fraction
Image with alt text: start fraction pi over 6 end fraction

start fraction pi over 3 end fraction
Image with alt text: start fraction pi over 3 end fraction

start fraction 2 pi over 3 end fraction
Image with alt text: start fraction 2 pi over 3 end fraction

start fraction 5 pi over 6 end fraction

Answer 1

To find \(\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)\), we need to determine the angle whose cosine is \(-\frac{\sqrt{3}}{2}\).

The cosine function is negative in the second quadrant and the third quadrant. The reference angle that gives a cosine of \(\frac{\sqrt{3}}{2}\) is \(\frac{\pi}{6}\). Therefore, the angles we need to consider where cosine is negative are:

1. Second quadrant: \[ \pi - \frac{\pi}{6} = \frac{5\pi}{6} \]

2. Third quadrant: \[ \pi + \frac{\pi}{6} = \frac{7\pi}{6} \]

However, since the range of the inverse cosine function \(\cos^{-1}(x)\) is defined from \(0\) to \(\pi\), we only take the second quadrant angle.

Thus, \(\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6}\).

So, the answer is: \[ \frac{5\pi}{6} \]