Question

If $\sin \left(\theta \right)=\frac{12}{13},\left(0<\theta <\frac{\mathrm{\pi }}{2}\right)$ and $\cos \left(\phi \right)=-\frac{3}{5},\left(\mathrm{\pi }<\mathrm{\phi }<\frac{3\mathrm{\pi }}{2}\right)$, then $\sin \left(\theta +\phi \right)$ will be

• A. $\frac{-56}{61}$
• B. $\frac{-56}{65}$
• C. $\frac{1}{65}$
• D. $-56$

Answer 1

The correct option is B

$\frac{-56}{65}$

Explanation for the correct option

Step 1: Given term

$\sin \left(\theta \right)=\frac{12}{13},\left(0<\theta <\frac{\mathrm{\pi }}{2}\right)and\cos \left(\phi \right)=\frac{-3}{5},\left(\mathrm{\pi }<\mathrm{\phi }<\frac{3\mathrm{\pi }}{2}\right)$

Step 2: Simplification and calculate the value of $\sin \left(\theta +\phi \right)$

For $\left(\theta \right)$,

Use the identity ${\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)=1$

Put the value of $\sin \left(\theta \right)=\frac{12}{13}$ in the identity

$\begin{array}{rcl}& \Rightarrow & {\left(\frac{12}{13}\right)}^2+{\cos }^2\left(\theta \right)=1\\ & \Rightarrow & \frac{144}{169}+{\cos }^2\left(\theta \right)=1\\ & \Rightarrow & {\cos }^2\left(\theta \right)=1-\frac{144}{169}\left(\mathrm{take}\mathrm{L}.\mathrm{C}.\mathrm{M}\right)\\ & \Rightarrow & {\cos }^2\left(\theta \right)=\frac{169-144}{169}\\ & \Rightarrow & {\cos }^2\left(\theta \right)=\frac{25}{169}\\ & & \end{array}$

Take square root , then

$\begin{array}{rcl}& \Rightarrow & \cos \left(\theta \right)=\sqrt{\frac{25}{169}}\\ & \Rightarrow & \cos \left(\theta \right)=\frac{5}{13}\end{array}$

Step 3: Same process for the term $\left(\phi \right)$

So ,the value of $\sin \left(\phi \right)$ is $\frac{-4}{5}$. because $\sin \left(\phi \right)$ is negative in quadrant $\left(3\right)$

Step 4: Apply formula

$\begin{array}{rcl}& \Rightarrow & \sin \left(\theta +\phi \right)=\sin \left(\theta \right).\cos \left(\phi \right)+\cos \left(\theta \right).\sin \left(\phi \right)\\ & =& \left(\frac{12}{13}\right).\left(\frac{-3}{5}\right)+\left(\frac{5}{13}\right).\left(\frac{-4}{5}\right)\\ & =& \left(\frac{-36}{65}\right)+\left(\frac{-20}{65}\right)\\ & =& \left(\frac{-36-20}{65}\right)\\ & =& \left(\frac{-56}{65}\right)\\ & & \\ & & \end{array}$

$\therefore$ The value of $\sin \left(\theta +\phi \right)$ is $-\frac{56}{65}.$