Question

How do you find the value of \[\sin \left( {\dfrac{{11\pi }}{{12}}} \right)\] ?

Answer 1

Hint: Here the question is related to the trigonometry, we use the trigonometry ratios and we are to solve this question. By using the trigonometry properties we are going to solve this problem. To find the value we need the table of trigonometry ratios for standard angles.

Complete step-by-step answer:
The question is related to trigonometry and it includes the trigonometry ratios. The trigonometry ratios are sine, cosine and tan.
Now consider the given question
   \[\sin \left( {\dfrac{{11\pi }}{{12}}} \right)\]
This term can be written as
  \[ \Rightarrow \sin \left( {\pi - \dfrac{\pi }{{12}}} \right)\]
This will lie in the second quadrant. In trigonometry we have ASTC rules for the trigonometry ratios. The above inequality lies in the second quadrant and the sine trigonometry ratio is positive in the second quadrant.
As we know that \[\sin \left( {\pi - \theta } \right) = \sin \theta \] , by using this condition the above equation is written as
  \[ \Rightarrow \sin \left( {\dfrac{\pi }{{12}}} \right)\]
For the further simplification we use the half angle formula for the sine trigonometry ratio. The formula is defined as \[\sin \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos \theta }}{2}} \]
By using the half angle formula the above equation is written as
   \[ \Rightarrow \sin \left( {\dfrac{\pi }{{12}}} \right) = \pm \sqrt {\dfrac{{1 - \cos \left( {\dfrac{\pi }{6}} \right)}}{2}} \]
On squaring both sides we have
  \[ \Rightarrow {\sin ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{1 - \cos \dfrac{\pi }{6}}}{2}\]
On multiplying 2 both sides we get
  \[ \Rightarrow 2{\sin ^2}\left( {\dfrac{\pi }{{12}}} \right) = 1 - \cos \dfrac{\pi }{6}\]
From the table of trigonometry ratios for standard angles the \[\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}\] , on substituting this we have
  \[ \Rightarrow 2{\sin ^2}\left( {\dfrac{\pi }{{12}}} \right) = 1 - \dfrac{{\sqrt 3 }}{2}\]
On further simplification we have
  \[ \Rightarrow 2{\sin ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{2 - \sqrt 3 }}{2}\]
Dividing the equation by 2 we have
  \[ \Rightarrow {\sin ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{2 - \sqrt 3 }}{4}\]
Taking the square on both sides we have
  \[ \Rightarrow \sin \left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{\sqrt {2 - \sqrt 3 } }}{2}\]
Hence we have solved the trigonometry ratio and found the value.

So, the correct answer is “$\dfrac{{\sqrt {2 - \sqrt 3 } }}{2} $”.

Note: For the trigonometry ratios we have double angle formula and half angle formula. By using these formulas, we can solve the trigonometry ratios. The double angle formula for cosine is defined as \[\sin (2x) = 2\sin x\cos x\] and the half angle formula is defined as \[\sin \left( {\dfrac{x}{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos x}}{2}} \] where x represents the angle. Hence we can obtain the required solution for the question.