Question

If $\cos A=\frac{3}{5},\cos B=\frac{4}{5}$ and $\frac{-\pi }{2}<A<0,\frac{-\pi }{2}<B<0$, then the value of $2\sin A+4\sin B=$

• A. $4$
• B. $2$
• C. $-4$
• D. $0$

Answer 1

The correct option is C $-4$

$\cos A=\frac{3}{5}$

& $Aϵ(\frac{-\pi }{2},0)$

$\therefore \sin A=-\sqrt{1-{\left(\frac{3}{5}\right)}^2}$

$=-\frac{4}{5}$ ($\because \sin <0$ when $Aϵ(\frac{-\pi }{2},0)$)

Similarly, $\sin B=-\sqrt{1-{\left(\frac{4}{5}\right)}^2}$

$=\frac{-3}{5}$

$\Rightarrow 2\sin A+4\sin B$

$=\frac{-8}{5}-\frac{12}{5}=\frac{-20}{5}=-4$ is the correct answer