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

How do you find the value of sin⁡(11π12)\sin \left( {\dfrac{{11\pi }}{{12}}} \right) ?

Explanation

Solution

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⁡(11π12)\sin \left( {\dfrac{{11\pi }}{{12}}} \right)
This term can be written as
⇒sin⁡(π−π12)\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⁡(π−θ)=sin⁡θ\sin \left( {\pi - \theta } \right) = \sin \theta , by using this condition the above equation is written as
⇒sin⁡(π12)\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⁡(θ2)=±1−cos⁡θ2\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
⇒sin⁡(π12)=±1−cos⁡(π6)2\Rightarrow \sin \left( {\dfrac{\pi }{{12}}} \right) = \pm \sqrt {\dfrac{{1 - \cos \left( {\dfrac{\pi }{6}} \right)}}{2}}
On squaring both sides we have
⇒sin⁡2(π12)=1−cos⁡π62\Rightarrow {\sin ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{1 - \cos \dfrac{\pi }{6}}}{2}
On multiplying 2 both sides we get
⇒2sin⁡2(π12)=1−cos⁡π6\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⁡π6=32\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2} , on substituting this we have
⇒2sin⁡2(π12)=1−32\Rightarrow 2{\sin ^2}\left( {\dfrac{\pi }{{12}}} \right) = 1 - \dfrac{{\sqrt 3 }}{2}
On further simplification we have
⇒2sin⁡2(π12)=2−32\Rightarrow 2{\sin ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{2 - \sqrt 3 }}{2}
Dividing the equation by 2 we have
⇒sin⁡2(π12)=2−34\Rightarrow {\sin ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{2 - \sqrt 3 }}{4}
Taking the square on both sides we have
⇒sin⁡(π12)=2−32\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 “2−32\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)=2sin⁡xcos⁡x\sin (2x) = 2\sin x\cos x and the half angle formula is defined as sin⁡(x2)=±1−cos⁡x2\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.