Question

How do you show that $2\sin x\cos x=\sin 2x$ ? Is it true for $\dfrac{5\pi }{6}$ ?

Answer 1

Hint: Here in this question we have been asked to show that $2\sin x\cos x=\sin 2x$ and verify it for the value of $x$ equal to $\dfrac{5\pi }{6}$ . For that sake we will use the formula $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$.

Complete step by step solution:
Now considering from the question we have been asked to show that $2\sin x\cos x=\sin 2x$ and verify it for the value of $x$ equal to $\dfrac{5\pi }{6}$ .
For that sake we will use the formula $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ which we have learnt during trigonometric basics and we also know that $\dfrac{5\pi }{6}={{150}^{\circ }}$ .
By using $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ if we keep $A=B=x$ then we will have $\sin 2x=2\sin x\cos x$ .
From the trigonometric table which specifies values for different angles we have $\sin {{30}^{\circ }}=\dfrac{1}{2}$ and $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ . We know that ${{150}^{\circ }}$ lies in the second quadrant. In which only sine and cosecant functions are positive remaining are negative. We also know that $\sin \left( {{180}^{\circ }}-x \right)=\cos x$ and $\cos \left( {{180}^{\circ }}-x \right)=-\sin x$ .
Now we can say that
 $\begin{align}
  & \Rightarrow \sin {{150}^{\circ }}=\sin \left( {{180}^{\circ }}-{{30}^{\circ }} \right) \\
 & \Rightarrow \cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2} \\
\end{align}$
Similarly we can also say that
$\begin{align}
  & \Rightarrow \cos {{150}^{\circ }}=\cos \left( {{180}^{\circ }}-{{30}^{\circ }} \right) \\
 & \Rightarrow -\sin {{30}^{\circ }}=\dfrac{-1}{2} \\
\end{align}$
Now we can say that
$\begin{align}
  & 2\sin x\cos x=2\left( \dfrac{\sqrt{3}}{2} \right)\left( \dfrac{-1}{2} \right) \\
 & \Rightarrow \dfrac{-\sqrt{3}}{2} \\
\end{align}$
For $x=\dfrac{5\pi }{6}$ and now we need to check for $\sin 2x$ by doing that we will have $\begin{align}
  & \Rightarrow \sin {{300}^{\circ }}=\sin \left( {{360}^{\circ }}-{{60}^{\circ }} \right) \\
 & \Rightarrow -\sin {{60}^{\circ }}=\dfrac{-\sqrt{3}}{2} \\
\end{align}$ .
Therefore we can conclude that the expression is verified and valid for any value.

Note: During the process of answering questions of this type we should be sure with our trigonometric concepts that we are going to apply in between. This is a very simple and easy question and can be answered accurately in a short span of time. It is completely theory based question and can be answered by applying accurate trigonometric concepts. We can also write
 $\begin{align}
  & \sin {{150}^{\circ }}=\sin \left( {{90}^{\circ }}+{{60}^{\circ }} \right) \\
 & \Rightarrow \sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2} \\
\end{align}$
And similarly the other simplifications can be done.