Question

If $\sin A=\frac{4}{5}$ and $\cos B=-\frac{12}{13}$, where $A$ and $B$ lie in the first and third quadrant respectively, then $\cos (A+B)=$

• A. $\frac{56}{65}$
• B. $-\frac{56}{65}$
• C. $\frac{16}{65}$
• D. $-\frac{16}{65}$

Answer 1

The correct option is D

$-\frac{16}{65}$

Explanation for the correct option.

Step 1: Find the value of $\cos A,\sin B$

Given that, $\sin A=\frac{4}{5}$and $\cos B=-\frac{12}{13}$.

We know that, ${\sin }^2\theta +{\cos }^2\theta =1$

$\begin{array}{rcl}\cos A& =& \sqrt{1-{\sin }^{2}A}\\ & =& \sqrt{1-{\left(\frac{4}{5}\right)}^2}\\ & =& \frac{3}{5}\end{array}$

Now the value of $\sin B$is negative because $B$ lies in 3rd quadrant.

$\begin{array}{rcl}\sin B& =& \sqrt{1-{\left(-\frac{12}{13}\right)}^2}\\ & =& \sqrt{1-\frac{144}{169}}\\ & =& \sqrt{\frac{25}{169}}\\ & =& -\frac{5}{13}\end{array}$

Step 2: Find the value of $\cos (A+B)$

We know that, $\cos \left(A+B\right)=\cos A\cos B-\sin A\sin B$.

$\begin{array}{rcl}\cos (A+B)& =& \frac{3}{5}·\left(-\frac{12}{13}\right)-\frac{4}{5}·\frac{-5}{13}\\ & =& -\frac{36}{65}+\frac{20}{65}\\ & =& -\frac{16}{65}\end{array}$

Hence, option D is correct.