Trigonometric Ratios in the Second Quadrant
The question gives us the cosine of an angle (θ) and the range for the angle, which is the second quadrant (\(\frac{\pi}{2} < \theta < \pi\)), and asks us to find the values of the remaining trigonometric ratios.
Given:
\[
\cos \theta = -\frac{\sqrt{3}}{2}
\]
Since the cosine of θ is negative and the angle is in the second quadrant, sine will be positive (as sine is positive in the second quadrant).
Let's use the Pythagorean identity to find sine:
\[
\sin^2 \theta + \cos^2 \theta = 1
\]
Substitute the given value of cos θ:
\[
\sin^2 \theta + \left(-\frac{\sqrt{3}}{2}\right)^2 = 1
\]
\[
\sin^2 \theta + \frac{3}{4} = 1
\]
Solving for \(\sin^2 \theta\):
\[
\sin^2 \theta = 1 - \frac{3}{4} = \frac{1}{4}
\]
Taking the square root to find sin θ yields two possible values, +1/2 and -1/2. Since we are in the second quadrant and sine is positive:
\[
\sin \theta = \frac{1}{2}
\]
The tangent of θ is the ratio of sine to cosine:
\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\]
\[
\tan \theta = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}
\]
\[
\tan \theta = -\frac{1}{\sqrt{3}}
\]
\[
\tan \theta = -\frac{\sqrt{3}}{3}
\]
For the reciprocals:
\[
\csc \theta = \frac{1}{\sin \theta} = 2
\]
\[
\sec \theta = \frac{1}{\cos \theta} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}
\]
\[
\cot \theta = \frac{1}{\tan \theta} = -\sqrt{3}
\]
So, the six trigonometric ratios for the given angle θ in the second quadrant are:
\[
\sin \theta = \frac{1}{2}, \cos \theta = -\frac{\sqrt{3}}{2}, \tan \theta = -\frac{\sqrt{3}}{3}
\]
\[
\csc \theta = 2, \sec \theta = -\frac{2\sqrt{3}}{3}, \cot \theta = -\sqrt{3}
\]
