Question

Prove that:
$2\sin \left( \dfrac{5\pi }{12} \right)\sin \left( \dfrac{\pi }{12} \right)=\dfrac{1}{2}$

Answer 1

Hint:In this given question, we can use the transformation formula for the product of trigonometric ratios of two angles to sum of trigonometric ratios of the two angles. The formula is as follows: $2\sin A\sin B=\cos (A-B)-\cos (A+B)$, where we can take A equal to $\dfrac{5\pi }{12}$ and B equal to $\dfrac{\pi }{12}$. This way we can simplify the Left Hand Side (LHS) and put the required values of the trigonometric ratios to get the asked proof.

Complete step-by-step answer:
In this solution to the provided question, we will use the following transformation formula to simplify the LHS and get it equal to the RHS:
$2\sin A\sin B=\cos (A-B)-\cos (A+B)........................\left( 1.1 \right)$
So, let us get to the RHS from the LHS as follows:
$LHS=2\sin \left( \dfrac{5\pi }{12} \right)\sin \left( \dfrac{\pi }{12} \right)$
Using equation (1.1) in this, where A is equal to $\dfrac{5\pi }{12}$ and B is equal to $\dfrac{\pi }{12}$, we get
$=\cos \left( \dfrac{5\pi }{12}-\dfrac{\pi }{12} \right)-\cos \left( \dfrac{5\pi }{12}+\dfrac{\pi }{12} \right)$
$=\cos \left( \dfrac{4\pi }{12} \right)-\cos \left( \dfrac{6\pi }{12} \right)$
$=\cos \left( \dfrac{\pi }{3} \right)-\cos \left( \dfrac{\pi }{2} \right).....................\left( 1.2 \right)$
Now, as we know that $\pi $ corresponds to ${{180}^{\circ }}$. So, $\dfrac{\pi }{3}=\dfrac{{{180}^{\circ }}}{3}={{60}^{\circ }}$ and $\dfrac{\pi }{2}=\dfrac{{{180}^{\circ }}}{2}={{90}^{\circ }}$.
Putting these values in equation (1.2), we get
$\begin{align}
  & LHS=\cos \left( \dfrac{\pi }{3} \right)-\cos \left( \dfrac{\pi }{2} \right) \\
 & =\cos {{60}^{\circ }}-\cos {{90}^{\circ }} \\
\end{align}$
Now, we know that $\cos {{60}^{\circ }}=\dfrac{1}{2}$ and $\cos {{90}^{\circ }}=0$, so
$LHS=\dfrac{1}{2}-0=\dfrac{1}{2}=RHS$
Therefore, we arrive at the required condition of LHS=RHS.
Hence, we proved that $2\sin \left( \dfrac{5\pi }{12} \right)\sin \left( \dfrac{\pi }{12} \right)=\dfrac{1}{2}$ .

Note: We must be careful while using equation 1.1, that is $2\sin A\sin B=\cos (A-B)-\cos (A+B)$, as here the subtraction of the cosine of the sum of the angles is done from the cosine of the difference of the angles, instead of the case of sine where the sine of the difference of angles is subtracted from the then sine of the sum of the angles to get twice of the product of cosine and sine of the first and the second angles respectively, that is $2\cos A\sin B=\sin \left( A+B \right)-\sin \left( A-B \right)$.So,students should remember the trigonometric transformation formulae for solving these types of questions.