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Question

Question: What is the exact value for \[\cos \left( {\dfrac{\pi }{{12}}} \right)\]....

What is the exact value for cos(π12)\cos \left( {\dfrac{\pi }{{12}}} \right).

Explanation

Solution

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(π12)\cos \left( {\dfrac{\pi }{{12}}} \right).

Complete step by step answer:
Given is the angle, cos(π12)\cos \left( {\dfrac{\pi }{{12}}} \right)
It can be written as, cos(π12)=cos(π4π6)\cos \left( {\dfrac{\pi }{{12}}} \right) = \cos \left( {\dfrac{\pi }{4} - \dfrac{\pi }{6}} \right)
We know that, cos(AB)=cosA.cosB+sinA.sinB\cos (A - B) = \cos A.\cos B + \sin A.\sin B
So we will substitute the value for angle A and B,
cos(π4π6)=cosπ4.cosπ6+sinπ4.sinπ6\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(π4π6)=12.32+12.12\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(π4π6)=3+122\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 2\sqrt 2
cos(π4π6)=(3+122)×22\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(π4π6)=(3×2+222×2)\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(π4π6)=6+22×2\cos \left( {\dfrac{\pi }{4} - \dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 6 + \sqrt 2 }}{{2 \times 2}}
The denominator will be,
cos(π4π6)=6+24\cos \left( {\dfrac{\pi }{4} - \dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 6 + \sqrt 2 }}{4}

Hence the correct answer is cos(π12)=6+24\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.