Question

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

• A. $sin2A=-\frac{336}{625}$
• 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)

The correct options are
A $sin2A=-\frac{336}{625}$
C $tan\frac{A}{2}=-\frac{1}{7}$

Given $270\circ <A<{360}^{\circ }$

$sinA=-\frac{7}{25}\Rightarrow cosA=\frac{24}{25}sin2A=2sinAcosA=2(-\frac{7}{25})\left(\frac{24}{25}\right)=-\frac{336}{625}cosA=2{cos}^2\frac{A}{2}-1\Rightarrow {cos}^2\frac{A}{2}=\frac{cosA+1}{2}=\frac{49}{50}$
Using $270<A<360\Rightarrow 135<\frac{A}{2}<180$
We get
$cos\frac{A}{2}=-\frac{7}{5\sqrt{2}}\Rightarrow sin\frac{A}{2}=\frac{1}{5\sqrt{2}}\Rightarrow tan\frac{A}{2}=-\frac{1}{7}$