Question

What is the value of $\sin \left( {\dfrac{{17\pi }}{{12}}} \right)$ ?

Answer 1

Hint : Sine function is periodic with the period of $2n\pi$ and so we will convert the given degrees of angle of sine in the form of the $2n\pi$ finding the correlation then will identify the location of the angle in the quadrant then will apply All STC rule for the resultant required value.

Complete step-by-step answer :
Take the given expression: $\sin \left( {\dfrac{{17\pi }}{{12}}} \right)$
The above expression can be re-written: $\sin \left( {\dfrac{{17\pi }}{{12}}} \right) = \sin \left( {\dfrac{{5\pi }}{{12}} + \pi } \right)$
Using the trigonometric table and the unit circle, also sine is negative in third quadrant
$\sin \left( {\dfrac{{17\pi }}{{12}}} \right) = - \sin \left( {\dfrac{{5\pi }}{{12}}} \right)$
Now, using the identity –
$2{\sin ^2}\theta = 1 - \cos 2\theta$
$2{\sin ^2}\left( {\dfrac{{5\pi }}{{12}}} \right) = 1 - \cos 2\left( {\dfrac{{5\pi }}{{12}}} \right)$
Simplify the above expression-
$2{\sin ^2}\left( {\dfrac{{5\pi }}{{12}}} \right) = 1 - \cos \left( {\dfrac{{5\pi }}{6}} \right)$ …. (A)
Now, $\cos \left( {\dfrac{{5\pi }}{6}} \right) = \cos \left( {\pi - \dfrac{\pi }{6}} \right)$
Cosine is negative in second quadrant –
$\cos \left( {\dfrac{{5\pi }}{6}} \right) = \cos \left( {\pi - \dfrac{\pi }{6}} \right) = - \dfrac{{\sqrt 3 }}{2}$
Place the above value in equation (A)
$2{\sin ^2}\left( {\dfrac{{5\pi }}{{12}}} \right) = 1 + \dfrac{{\sqrt 3 }}{2}$
Simplify the above expression –
$2{\sin ^2}\left( {\dfrac{{5\pi }}{{12}}} \right) = \dfrac{{2 + \sqrt 3 }}{2} \\\ {\sin ^2}\left( {\dfrac{{5\pi }}{{12}}} \right) = \dfrac{{2 + \sqrt 3 }}{4} \;$
Take square root on both the sides of the equation –
$2{\sin ^2}\left( {\dfrac{{5\pi }}{{12}}} \right) = \dfrac{{2 + \sqrt 3 }}{2} \\\ \sin \left( {\dfrac{{5\pi }}{{12}}} \right) = \sqrt {\dfrac{{2 + \sqrt 3 }}{4}} \;$
Simplify –
$\sin \left( {\dfrac{{5\pi }}{{12}}} \right) = \pm \dfrac{{\sqrt {2 + \sqrt 3 } }}{2}$
This is the required solution.
So, the correct answer is “$\pm \dfrac{{\sqrt {2 + \sqrt 3 } }}{2}$ ”.

Note : Remember the All STC rule, it is also known as ASTC rule in geometry. It states that all the trigonometric ratios in the first quadrant ( $0^\circ \;{\text{to 90}}^\circ$ ) are positive, sine and cosec are positive in the second quadrant ( $90^\circ {\text{ to 180}}^\circ$ ), tan and cot are positive in the third quadrant ( $180^\circ \;{\text{to 270}}^\circ$ ) and sin and cosec are positive in the fourth quadrant ( $270^\circ {\text{ to 360}}^\circ$ ).