Question

If ${180}^0<\theta <{270}^0,sin\theta =-\frac{3}{5},thencos\frac{\theta }{2}=$

• A. $\frac{1}{\sqrt{10}}$
• B. $\frac{1}{32}$
• C. $\frac{1}{16}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$\frac{1}{\sqrt{10}}$

As ${180}^{\circ }<\theta <{270}^{\circ }⟹{90}^{\circ }<\frac{\theta }{2}<{135}^{\circ }$

Given $\sin \theta =-\frac{3}{5}⟹\cos \theta =-\sqrt{1-(\frac{-3}{5}{)}^2}=-\frac{4}{5}$

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