Question

If $2(\sin a+\sin b)x-2\sin (a-b)y=3$ and $2(\cos a+\cos b)x+2\cos (a-b)y=5$ are perpendicular to each other then, $\sin 2a\sin 2b$ equals to

Answer 1

Step $1$: Find slopes of the given equations.

Let $m_1$ and $m_2$ be slopes of equation $\left(1\right)$ and equation $\left(2\right)$ respectively.

$$\begin{gathered} m_1=\frac{\sin \left(a\right)+\sin \left(b\right)}{\sin (a-b)} \\ {m}_2=\frac{\cos \left(a\right)+\cos \left(b\right)}{\cos (a-b)} \end{gathered}$$

Step $2$: Find the required value:

Given that both the lines are perpendicular to each other.

Thus, the product of the slope of the lines will be $-1$.

$$\begin{gathered} \frac{\sin \left(a\right)+\sin \left(b\right)}{\sin (a-b)}\times \frac{\cos \left(a\right)+\cos \left(b\right)}{\cos (a-b)}=-1 \\ \Rightarrow \frac{\sin \left(a\right)\cos \left(a\right)+\sin \left(a\right)\cos \left(b\right)+\sin \left(b\right)\cos \left(a\right)+\sin \left(b\right)\cos \left(b\right)}{\sin (a-b)\cos (a-b)}=-1 \\ \Rightarrow \sin \left(a\right)\cos \left(a\right)+\sin \left(a\right)\cos \left(b\right)+\sin \left(b\right)\cos \left(a\right)+\sin \left(b\right)\cos \left(b\right)=-\sin (a-b)\cos (a-b) \\ \Rightarrow \frac{1}{2}\sin \left(2a\right)+\frac{1}{2}s\mathrm{in}\left(2b\right)=-\sin \left(a\right)\cos \left(b\right)-\sin \left(b\right)\cos \left(a\right)-\sin (a-b)\cos (a-b) \end{gathered}$$ ' $\left[2\sin \left(a\right)\cos \left(a\right)=\sin \left(2a\right)\right]$

$$\begin{gathered} \Rightarrow \sin \left(2a\right)+\sin \left(2b\right)=-2\left[\frac{1}{2}\sin (2a-2b)+\sin (a+b)\right] \\ \Rightarrow \sin \left(2a\right)+\sin \left(2b\right)=-\sin 2(a-b)-2\sin (a+b) \end{gathered}$$ ; $\left[\sin (a+b)=\sin \left(a\right)\cos \left(b\right)+\cos \left(a\right)\sin \left(b\right)\right]$

Hence the value of $\sin \left(2a\right)\sin \left(2b\right)$ is $-\sin 2(a-b)-2\sin (a+b)$.