Question

If $\frac{(\cos A)}{3}=\frac{(\cos B)}{4}=\frac{1}{5},-\pi /2<A<0,-\pi /2<B<0$, then the value of $2\sin A+4\sin B$ is

• A. $4$
• B. $-2$
• C. $-4$
• D. $0$

Answer 1

The correct option is C

$-4$

Explanation for the correct options:

Finding the value of $\mathbf{2}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathbf{4}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{B}$:

Given,

$\begin{array}{rcl}\frac{(\cos A)}{3}& =& \frac{(\cos B)}{4}\\ & =& \frac{1}{5}\\ \cos A& =& \frac{3}{5},\\ \cos B& =& \frac{4}{5}\end{array}$

Also, given that $-\pi /2<A<0,-\pi /2<B<0$

In this interval sin is negative.

Thus, $\sin A=\frac{-4}{5},\sin B=\frac{-3}{5}(u\sin gPythagoreantriple)$

$\begin{array}{rcl}2\sin A+4\sin B& =& 2(-4/5)+4(-3/5)\\ & =& -8/5-12/5\\ & =& -20/5\\ & =& -4\end{array}$

The correct answer is option (C).