
How do you find $ \cos \dfrac{{11\pi }}{{12}} $ ?
Answer
563.1k+ views
Hint: All the trigonometric functions have different values for different angles but there is a pattern of the values obtained which gets repeated after a certain interval of the angles. This interval is different for different trigonometric functions. Thus they are periodic functions. We know the value of the cosine function when the angle lies between 0 and $ \dfrac{\pi }{2} $ . So to find the value of the cosine of the angles greater than $ \dfrac{\pi }{2} $ or smaller than 0, we use the periodic property of these functions. This way we can simplify the given function and calculate its value.
Complete step-by-step answer:
$ \cos \dfrac{{11\pi }}{{12}} $ can be written as $ \cos (\pi - \dfrac{\pi }{{12}}) $ .
We know that $ \cos (\pi - \dfrac{\pi }{{12}}) = - \cos \dfrac{\pi }{{12}} $
Now, we know that –
$
\cos 2x = 2{\cos ^2}x - 1 \\
\Rightarrow \cos (2 \times \dfrac{\pi }{{12}}) = 2{\cos ^2}\dfrac{\pi }{{12}} - 1 \\
\Rightarrow 2{\cos ^2}\dfrac{\pi }{{12}} = \cos \dfrac{\pi }{6} + 1 \\
\Rightarrow 2{\cos ^2}\dfrac{\pi }{{12}} = \dfrac{{\sqrt 3 }}{2} + 1 \\
\Rightarrow {\cos ^2}\dfrac{\pi }{{12}} = \dfrac{{2 + \sqrt 3 }}{4} \\
\cos \dfrac{\pi }{{12}} = \pm \dfrac{{\sqrt {2 + \sqrt 3 } }}{2} \;
$
As $ \cos \dfrac{\pi }{{12}} $ lies in the first quadrant, so it cannot be negative, that’s why its negative value is rejected. So, $ \cos \dfrac{\pi }{{12}} = \dfrac{{\sqrt {2 + \sqrt 3 } }}{2} $
Hence $ \cos \dfrac{{11\pi }}{{12}} = - \dfrac{{\sqrt {2 + \sqrt 3 } }}{2} $ .
So, the correct answer is “ $ \cos \dfrac{{11\pi }}{{12}} = - \dfrac{{\sqrt {2 + \sqrt 3 } }}{2} $ ”.
Note: To find out the relation between the sides and the angles of a right-angled triangle, we use trigonometry. We know the value of the trigonometric functions of some basic angles $ 0,\dfrac{\pi }{6},\dfrac{\pi }{4},\dfrac{\pi }{3}\,and\,\dfrac{\pi }{2} $ . All the trigonometric functions are related to each other and one can be converted into another using the trigonometric identities like we have used the identity $ \cos 2x = 2{\cos ^2}x - 1 $ in the given question. We know the value of $ \cos \dfrac{\pi }{6} $ , so to find out the value of $ \cos \dfrac{\pi }{{12}} $ we used this identity. Trigonometric functions can be plotted on the graph and the signs of the trigonometric functions are different in different quadrants of the graph, in the first quadrant all the trigonometric functions are positive.
Complete step-by-step answer:
$ \cos \dfrac{{11\pi }}{{12}} $ can be written as $ \cos (\pi - \dfrac{\pi }{{12}}) $ .
We know that $ \cos (\pi - \dfrac{\pi }{{12}}) = - \cos \dfrac{\pi }{{12}} $
Now, we know that –
$
\cos 2x = 2{\cos ^2}x - 1 \\
\Rightarrow \cos (2 \times \dfrac{\pi }{{12}}) = 2{\cos ^2}\dfrac{\pi }{{12}} - 1 \\
\Rightarrow 2{\cos ^2}\dfrac{\pi }{{12}} = \cos \dfrac{\pi }{6} + 1 \\
\Rightarrow 2{\cos ^2}\dfrac{\pi }{{12}} = \dfrac{{\sqrt 3 }}{2} + 1 \\
\Rightarrow {\cos ^2}\dfrac{\pi }{{12}} = \dfrac{{2 + \sqrt 3 }}{4} \\
\cos \dfrac{\pi }{{12}} = \pm \dfrac{{\sqrt {2 + \sqrt 3 } }}{2} \;
$
As $ \cos \dfrac{\pi }{{12}} $ lies in the first quadrant, so it cannot be negative, that’s why its negative value is rejected. So, $ \cos \dfrac{\pi }{{12}} = \dfrac{{\sqrt {2 + \sqrt 3 } }}{2} $
Hence $ \cos \dfrac{{11\pi }}{{12}} = - \dfrac{{\sqrt {2 + \sqrt 3 } }}{2} $ .
So, the correct answer is “ $ \cos \dfrac{{11\pi }}{{12}} = - \dfrac{{\sqrt {2 + \sqrt 3 } }}{2} $ ”.
Note: To find out the relation between the sides and the angles of a right-angled triangle, we use trigonometry. We know the value of the trigonometric functions of some basic angles $ 0,\dfrac{\pi }{6},\dfrac{\pi }{4},\dfrac{\pi }{3}\,and\,\dfrac{\pi }{2} $ . All the trigonometric functions are related to each other and one can be converted into another using the trigonometric identities like we have used the identity $ \cos 2x = 2{\cos ^2}x - 1 $ in the given question. We know the value of $ \cos \dfrac{\pi }{6} $ , so to find out the value of $ \cos \dfrac{\pi }{{12}} $ we used this identity. Trigonometric functions can be plotted on the graph and the signs of the trigonometric functions are different in different quadrants of the graph, in the first quadrant all the trigonometric functions are positive.
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