If sin θ = -3/5, and θ lies in the third quadrant, then the value of cos (θ/2) is.

In the third quadrant, sin and cos are negative.

Given : $$sin \theta = \frac{3}{5}$$

=> $$cos \theta = \sqrt{1 - sin^2 \theta} = \sqrt{1 - (\frac{3}{5})^2}$$

=> $$cos \theta = \sqrt{\frac{16}{25}} = - \sqrt{\frac{4}{5}}$$

Also, $$cos 2x = 2 cos^2 x - 1$$

Replacing $$x$$ by $$\frac{\theta}{2}$$

=> $$cos \theta = 2 cos^2 \frac{\theta}{2} - 1$$

$$\therefore cos \frac{\theta}{2} = \sqrt{\frac{cos \theta + 1}{2}}$$

= $$\sqrt{\frac{1 - \frac{4}{5}}{2}} = \sqrt{\frac{1}{10}}$$

$$\because \theta$$ lies in third quadrant, => $$cos \frac{\theta}{2} = - \frac{1}{\sqrt{10}}$$