Question

How do you find the exact value of $\cos \left( {\dfrac{{13\pi }}{{24}}} \right)\sin \left( {\dfrac{{13\pi }}{{24}}} \right)$ using the half angle formula?

Answer 1

In the above question, we are given a trigonometric function as $\cos \left( {\dfrac{{13\pi }}{{24}}} \right)\sin \left( {\dfrac{{13\pi }}{{24}}} \right)$ . We have to find the exact value of the given trigonometric function using a well known formula of trigonometry called the half angle formula. We can also use the similar formula called the double angle formula. The double and half angle formula respectively, are given as:
$\sin \left( {2x} \right) = 2\sin x\cos x$
and
$\sin \left( {\dfrac{x}{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos x}}{2}}$

Complete step by step answer:
Given trigonometric function is
$\Rightarrow \cos \left( {\dfrac{{13\pi }}{{24}}} \right)\sin \left( {\dfrac{{13\pi }}{{24}}} \right)$
We have to find the exact value of the above given trigonometric function.
Using the double angle identity of trigonometry, since we have we can also write the above function as,
$\Rightarrow \sin \left( {2x} \right) = 2\sin x\cos x$
Hence, we can write it as
$\Rightarrow \sin x\cos x = \dfrac{1}{2}\sin \left( {2x} \right)$
Now putting $x = \dfrac{{13\pi }}{{24}}$ in the above obtained equation, we get
$\Rightarrow \sin \left( {\dfrac{{13\pi }}{{24}}} \right)\cos \left( {\dfrac{{13\pi }}{{24}}} \right) = \dfrac{1}{2}\sin \left( {2 \cdot \dfrac{{13\pi }}{{24}}} \right)$
We can also write the above equation as,
$\Rightarrow \cos \left( {\dfrac{{13\pi }}{{24}}} \right)\sin \left( {\dfrac{{13\pi }}{{24}}} \right) = \dfrac{1}{2}\sin \left( {\dfrac{{13\pi }}{{12}}} \right)$ ...(1)
Now consider $\sin \left( {\dfrac{{13\pi }}{{12}}} \right)$ , we can write it as
$\Rightarrow \sin \left( {\dfrac{{13\pi }}{{12}}} \right) = \sin \left( {\pi + \dfrac{\pi }{{12}}} \right)$
That gives,
$\Rightarrow \sin \left( {\dfrac{{13\pi }}{{12}}} \right) = - \sin \left( {\dfrac{\pi }{{12}}} \right)$
We can also write it as,
$\Rightarrow \sin (\dfrac{13\pi}{12})= -\sin (\dfrac{\pi/6}{2})$ ...(2)
Now from the half angle identity, we have
$\Rightarrow \sin \left( {\dfrac{x}{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos x}}{2}}$
Since, $\dfrac{\pi }{6}$ lies in the first quadrant and sine function is positive in the first quadrant, therefore we can take the positive sign in the half angle identity while putting $x = \dfrac{\pi }{6}$ .
Therefore, we have
$\sin (\dfrac{\pi/6}{2}) = \sqrt{\dfrac{1-\cos (\pi/6)}{2}}$
Now since $cos\left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2}$ , that gives,
$\sin (\dfrac{\pi/6}{2})= \sqrt{\dfrac{1-\dfrac{\sqrt 3}{2}}{2}}$
$\sin (\dfrac{\pi/6}{2}) = \sqrt {\dfrac{{2 - \sqrt 3 }}{4}}$
Therefore, we get
$\sin (\dfrac{\pi/6}{2}) = \dfrac{1}{2} \sqrt {2-\sqrt 3}$
Substituting this value in equation (2), we have
$\sin (\dfrac{13\pi}{12})=-\sin (\dfrac{\pi/6}{2})$
Hence,
$\Rightarrow \sin \left( {\dfrac{{13\pi }}{{12}}} \right) = - \dfrac{1}{2}\sqrt {2 - \sqrt 3 }$
Again, putting $\sin \left( {\dfrac{{13\pi }}{{12}}} \right) = - \dfrac{1}{2}\sqrt {2 - \sqrt 3 }$ in equation (1), we have
$\Rightarrow \cos \left( {\dfrac{{13\pi }}{{24}}} \right)\sin \left( {\dfrac{{13\pi }}{{24}}} \right) = \dfrac{1}{2}\sin \left( {\dfrac{{13\pi }}{{12}}} \right)$
Hence,
$\Rightarrow \cos \left( {\dfrac{{13\pi }}{{24}}} \right)\sin \left( {\dfrac{{13\pi }}{{24}}} \right) = \dfrac{1}{2} \cdot \left( { - \dfrac{1}{2}\sqrt {2 - \sqrt 3 } } \right)$
Therefore, we get
$\Rightarrow \cos \left( {\dfrac{{13\pi }}{{24}}} \right)\sin \left( {\dfrac{{13\pi }}{{24}}} \right) = - \dfrac{1}{4}\sqrt {2 - \sqrt 3 }$
That is the required solution.
Therefore, the exact value of $\cos \left( {\dfrac{{13\pi }}{{24}}} \right)\sin \left( {\dfrac{{13\pi }}{{24}}} \right)$ is $- \dfrac{1}{4}\sqrt {2 - \sqrt 3 }$.

Note:
Sometimes the double angle formula and the half angle formula is also written in their other forms where $2x$ is replaced by its half, i.e. $x$ and $\dfrac{x}{2}$ is replaced by its double, i.e. $x$ .
Their other forms are written as the following identities:
$\Rightarrow \sin \left( {2x} \right) = 2\sin x\cos x$
Hence,
$\Rightarrow \sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$
Similarly,
$\Rightarrow \sin \left( {\dfrac{x}{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos x}}{2}}$
Therefore,
$\Rightarrow \sin x = \pm \sqrt {\dfrac{{1 - \cos 2x}}{2}}$
Ultimately, we have
$\Rightarrow \sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} = \pm \sqrt {\dfrac{{1 - \cos 2x}}{2}}$