Question

Find the values of the other five trigonometric functions, if \( \cos x = -\frac{1}{2} \), and \( x \) lies in the third quadrant.

Hint:

In the third quadrant, both sine and cosine are negative, but tangent is positive. Use the Pythagorean identity to find the missing trigonometric functions.

Answer 1

Step 1: Recall the trigonometric identities.
The six trigonometric functions are: \[ \sin x, \cos x, \tan x, \csc x, \sec x, \text{ and } \cot x \]
Step 2: Given information.
We are given that \( \cos x = -\frac{1}{2} \) and that \( x \) lies in the third quadrant. In the third quadrant, sine and cosine are negative, while tangent is positive.
Step 3: Use the Pythagorean identity.
We can use the identity: \[ \sin^2 x + \cos^2 x = 1 \] Substitute \( \cos x = -\frac{1}{2} \) into the identity: \[ \sin^2 x + \left( -\frac{1}{2} \right)^2 = 1 \] \[ \sin^2 x + \frac{1}{4} = 1 \] \[ \sin^2 x = 1 - \frac{1}{4} = \frac{3}{4} \] \[ \sin x = -\frac{\sqrt{3}}{2} \quad (\text{since \( x \) is in the third quadrant, where sine is negative}) \]
Step 4: Find the remaining trigonometric functions.
Now, we can find the other functions: \[ \tan x = \frac{\sin x}{\cos x} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3} \] \[ \csc x = \frac{1}{\sin x} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \] \[ \sec x = \frac{1}{\cos x} = \frac{1}{-\frac{1}{2}} = -2 \] \[ \cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \]
Step 5: Conclusion.
The values of the other five trigonometric functions are: \[ \sin x = -\frac{\sqrt{3}}{2}, \quad \tan x = \sqrt{3}, \quad \csc x = -\frac{2\sqrt{3}}{3}, \quad \sec x = -2, \quad \cot x = \frac{\sqrt{3}}{3} \]