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Question: How do you find \( \cos \dfrac{{11\pi }}{{12}} \) ?...

How do you find cos11π12\cos \dfrac{{11\pi }}{{12}} ?

Explanation

Solution

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 π2\dfrac{\pi }{2} . So to find the value of the cosine of the angles greater than π2\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 :
cos11π12\cos \dfrac{{11\pi }}{{12}} can be written as cos(ππ12)\cos (\pi - \dfrac{\pi }{{12}}) .
We know that cos(ππ12)=cosπ12\cos (\pi - \dfrac{\pi }{{12}}) = - \cos \dfrac{\pi }{{12}}
Now, we know that –
cos2x=2cos2x1 cos(2×π12)=2cos2π121 2cos2π12=cosπ6+1 2cos2π12=32+1 cos2π12=2+34 cosπ12=±2+32   \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 cosπ12\cos \dfrac{\pi }{{12}} lies in the first quadrant, so it cannot be negative, that’s why its negative value is rejected. So, cosπ12=2+32\cos \dfrac{\pi }{{12}} = \dfrac{{\sqrt {2 + \sqrt 3 } }}{2}
Hence cos11π12=2+32\cos \dfrac{{11\pi }}{{12}} = - \dfrac{{\sqrt {2 + \sqrt 3 } }}{2} .
So, the correct answer is “ cos11π12=2+32\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,π6,π4,π3andπ20,\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 cos2x=2cos2x1\cos 2x = 2{\cos ^2}x - 1 in the given question. We know the value of cosπ6\cos \dfrac{\pi }{6} , so to find out the value of cosπ12\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.