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What is the exact value for $\cos \left( {\dfrac{\pi }{{12}}} \right)$.

Last updated date: 13th Jul 2024
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Hint:We will split the full in in two angles like the formula of addition and subtraction of angles. Then we will apply the formula of addition and subtraction of angles.Putting the values of respective functions for the respective angle we will calculate the value for $\cos \left( {\dfrac{\pi }{{12}}} \right)$.

Given is the angle, $\cos \left( {\dfrac{\pi }{{12}}} \right)$
It can be written as, $\cos \left( {\dfrac{\pi }{{12}}} \right) = \cos \left( {\dfrac{\pi }{4} - \dfrac{\pi }{6}} \right)$
We know that, $\cos (A - B) = \cos A.\cos B + \sin A.\sin B$
So we will substitute the value for angle A and B,
$\cos \left( {\dfrac{\pi }{4} - \dfrac{\pi }{6}} \right) = \cos \dfrac{\pi }{4}.\cos \dfrac{\pi }{6} + \sin \dfrac{\pi }{4}.\sin \dfrac{\pi }{6}$
We will place the values of the respective function and the angle,
$\cos \left( {\dfrac{\pi }{4} - \dfrac{\pi }{6}} \right) = \dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }}.\dfrac{1}{2}$
Here since the denominator of both the terms is same we can directly add the numerator,
$\cos \left( {\dfrac{\pi }{4} - \dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$
Now we will multiply the numerator and denominator by $\sqrt 2$
$\cos \left( {\dfrac{\pi }{4} - \dfrac{\pi }{6}} \right) = \left( {\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right) \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}$
In the numerator there will be separate multiplication and in denominator the product of two roots with same under root will be the number itself,
$\cos \left( {\dfrac{\pi }{4} - \dfrac{\pi }{6}} \right) = \left( {\dfrac{{\sqrt 3 \times \sqrt 2 + \sqrt 2 }}{{2\sqrt 2 \times \sqrt 2 }}} \right)$
Now,
$\cos \left( {\dfrac{\pi }{4} - \dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 6 + \sqrt 2 }}{{2 \times 2}}$
The denominator will be,
$\cos \left( {\dfrac{\pi }{4} - \dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 6 + \sqrt 2 }}{4}$

Hence the correct answer is $\cos \left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{\sqrt 6 + \sqrt 2 }}{4}$.

Note: If the value for any trigonometric function is not available directly, we at such times use these types of formulas. That includes either double angles, triple angles, sum and difference formulae, factorization and defactorization formulae. These are used as per the requirement of the problem.