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 $ ).