Question

If $A$ lies between ${270}^o$ and ${360}^o$ and $\sin A=\frac{-7}{25}$, then

• A. $\sin 2A=-\frac{336}{625}$
• B. $\cos \frac{A}{2}=\frac{\sqrt{2}}{5}$
• C. $\tan \frac{A}{2}=-\frac{1}{7}$
• D. $\sin \frac{A}{2}=-\frac{\sqrt{2}}{10}$.

Answer 1

Answer: (A), (C)

$\tan \frac{A}{2}=-\frac{1}{7}$
$\sin 2A=-\frac{336}{625}$

In 4 th quadrant, sine function is negative while the cosine function is positive.

$\cos A=\sqrt{1-(\sin A{)}^2}=\sqrt{1-(\frac{-7}{25}{)}^2}=\frac{24}{25}$

$\sin 2A=2\sin A\cos A=2\times \frac{-7}{25}\times \frac{24}{25}=\frac{-336}{625}$
$\tan \left(\frac{A}{2}\right)=-\sqrt{(\frac{1-cosA}{1+cosA})}$
$\tan \left(\frac{A}{2}\right)=-\sqrt{\frac{1}{49}}$
$\tan \frac{A}{2}=-\frac{1}{7}$