If sinθ = -4/5 in the third quadrant, find cos(θ/2)

Trigonometric Problem: Half-Angle Formula

Given: \( \sin \theta = -\frac{4}{5} \) and \(\theta\) lies in the third quadrant.
Find \( \cos \frac{\theta}{2} \).

Select the correct option:

A. \( \frac{1}{\sqrt{5}} \)

B. \( -\frac{1}{\sqrt{5}} \)

C. \( \sqrt{\frac{2}{5}} \)

D. \( -\sqrt{\frac{2}{5}} \)

Detailed Solution

We are given \( \sin \theta = -\frac{4}{5} \) and that \(\theta\) lies in the third quadrant. In the third quadrant, both sine and cosine are negative.

Step 1: Use the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1. \] Substitute \( \sin \theta = -\frac{4}{5} \): \[ \left(-\frac{4}{5}\right)^2 + \cos^2 \theta = 1 \quad \Rightarrow \quad \frac{16}{25} + \cos^2 \theta = 1. \] Solve for \( \cos^2 \theta \): \[ \cos^2 \theta = 1 - \frac{16}{25} = \frac{9}{25}. \] Since \( \theta \) is in the third quadrant, we take the negative square root: \[ \cos \theta = -\frac{3}{5}. \]

Step 2: Use the half-angle formula for cosine: \[ \cos \frac{\theta}{2} = \pm \sqrt{\frac{1+\cos \theta}{2}}. \] Substitute \( \cos \theta = -\frac{3}{5} \): \[ \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 - \frac{3}{5}}{2}} = \pm \sqrt{\frac{\frac{2}{5}}{2}} = \pm \sqrt{\frac{1}{5}} = \pm \frac{1}{\sqrt{5}}. \]

Step 3: Determine the sign of \( \cos \frac{\theta}{2} \). Since \(\theta\) is in the third quadrant (i.e., between \(180^\circ\) and \(270^\circ\)), then \(\frac{\theta}{2}\) is between \(90^\circ\) and \(135^\circ\) (second quadrant) where cosine is negative.

Thus, \[ \cos \frac{\theta}{2} = -\frac{1}{\sqrt{5}}, \] which corresponds to option B.