Question

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

Answer 1

In order to solve this question ,convert $\left( {\dfrac{\pi }{{12}}} \right)$ into some (A-B) and apply $\sin \left( {A - B} \right)$ formula to calculate the answer.

Formula used:
$\sin \left( {A - B} \right) = \sin \left( A \right)\cos \left( B \right) - \sin \left( B \right)\cos \left( A \right)$

Complete step-by-step answer:
In trigonometric table ,the value of $\sin \left( {\dfrac{\pi }{{12}}} \right)$ is not given ,so to find this we have use some other means to find the required value .We’ll use other trigonometric values given in the trigonometric table .
First ,we’ll ,convert $\left( {\dfrac{\pi }{{12}}} \right)$ into some (A-B) form
$\sin \left( {\dfrac{\pi }{{12}}} \right) = \sin \left( {\dfrac{\pi }{4} - \dfrac{\pi }{6}} \right)$
Now we will apply formula $\sin \left( {A - B} \right) = \sin \left( A \right)\cos \left( B \right) - \sin \left( B \right)\cos \left( A \right)$
Where , A is equal to $\dfrac{\pi }{4}$ and B is equal to $\dfrac{\pi }{6}$
$\sin \left( {\dfrac{\pi }{4} - \dfrac{\pi }{6}} \right) = \sin \left( {\dfrac{\pi }{4}} \right)\cos \left( {\dfrac{\pi }{6}} \right) - \sin \left( {\dfrac{\pi }{6}} \right)\cos \left( {\dfrac{\pi }{4}} \right)$ -(1)
From the trigonometric table
$\sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}\, = \,\dfrac{{\sqrt 2 }}{2}$ ,
$\sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2}$
$\cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}\, = \,\dfrac{{\sqrt 2 }}{2}$
$\cos \left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2}$
Putting values in equation (1)
$= \dfrac{{\sqrt 1 }}{2} \times \dfrac{{\sqrt 3 }}{2} - \dfrac{1}{2} \times \dfrac{{\sqrt 2 }}{2} \\\ = \dfrac{{\sqrt 6 }}{4} - \dfrac{{\sqrt 2 }}{4} \\\ = \dfrac{{\sqrt 6 - \sqrt 2 }}{4} \\\$
Therefore, the value of $\sin \left( {\dfrac{\pi }{{12}}} \right)$ is equal to $\dfrac{{\sqrt 6 - \sqrt 2 }}{4}$.

Note: 1. Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.
If T is the smallest positive real number such that $f(x + T) = f(x)$ for all x, then T is called the fundamental period of $f(x)$ .
Since $\sin \,(2n\pi + \theta ) = \sin \theta$ for all values of $\theta$ and n$\in$N.
2. Even Function – A function $f(x)$ is said to be an even function ,if $f( - x) = f(x)$ for all x in its domain.
Odd Function – A function $f(x)$ is said to be an even function ,if $f( - x) = - f(x)$ for all x in its domain.
We know that $\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta$
Therefore,$\sin \theta$ and $\tan \theta$ and their reciprocals,$\cos ec\theta$ and $\cot \theta$ are odd functions whereas $\cos \theta$ and its reciprocal $\sec \theta$ are even functions.