Question

If $\sin \theta =\frac{24}{25}$ and $0^{\circ }<\theta <90^{\circ }$ then what is the value of $\sin \left(\frac{\theta }{2}\right)?$

• A. $\frac{12}{25}$
• B. $\frac{7}{25}$
• C. $\frac{3}{5}$
• D. $\frac{4}{5}$

Answer 1

Answer: (C)

We have, $\sin \theta =\frac{24}{25},0^{\circ }<\theta <90^{\circ }$
${\cos }^2\theta =1-{\sin }^2\theta =1-{\left(\frac{24}{25}\right)}^2$
Since $\theta$ lies in first quadrant
$\Rightarrow \cos \theta =\frac{7}{25}\Rightarrow \cos \theta =1-2{\sin }^2\frac{\theta }{2}$
$\Rightarrow 2{\sin }^2\frac{\theta }{2}=1-\cos \theta =1-\frac{7}{25}\Rightarrow 2{\sin }^2\frac{\theta }{2}=\frac{18}{25}$
$\Rightarrow {\sin }^2\frac{\theta }{2}=\frac{9}{25}\Rightarrow \sin \frac{\theta }{2}=\pm \frac{3}{5}\Rightarrow \sin \frac{\theta }{2}=\frac{3}{5}$
[Negative sign discarded since $\theta$ is in first quadrant]