Question

How do you find the value of $\sin \left( \dfrac{7\pi }{12} \right)$?

Answer 1

Hint: In the given question, we have been asked to find the value of a trigonometric function. Now, the argument of the given trigonometric function is not in the range of the known values of the trigonometric functions as given in the standard table, in which values lie from $0$ to $\pi /2$. But we can calculate that by using the formula of the periodicity of the given trigonometric function and then solving it.

Formula Used:
We are going to use the formula of $a+b$ of sine function, which is:
\[sin\left( A\text{ }+\text{ }B \right)\text{ }=\text{ }sinA\text{ }cosB\text{ }+\text{ }cosA\text{ }sinB\].

Complete step by step answer:
Here, we have to calculate the value of $\sin \left( \dfrac{7\pi }{12} \right)$.
Hence, $\sin \left( \dfrac{7\pi }{12} \right)=\sin \left( \dfrac{\pi }{2}+\dfrac{\pi }{12} \right)=\sin \left( \dfrac{\pi }{2} \right)\cos \left( x \right)+\cos \left( \dfrac{\pi }{2} \right)\sin \left( x \right)$
$\Rightarrow \left( 1 \right)\cos \left( \dfrac{\pi }{12} \right)+\left( 0 \right)\sin \left( \dfrac{\pi }{12} \right)$
$\Rightarrow \cos (\dfrac{\pi }{12})$
Now, we know that
$\cos \left( \dfrac{\pi }{12} \right)=\dfrac{\sqrt{2}+\sqrt{6}}{4}$

Hence, $\cos \left( \dfrac{\pi }{12} \right)=\dfrac{\sqrt{2}+\sqrt{6}}{4}$

Note:
In the given question, we applied the concept of addition of $2$ angles of the sine function, so it is necessary that we know all the formulas by heart for completing the numerical easily.