Question

How do you find the exact value of $\cos {62^\circ }$ using the sum and difference, double angle or half angle formulas?

Answer 1

Hint:
In this question, we are asked to find the value of $\cos {62^ \circ }$. Since, ${62^ \circ }$ is in degree, we will convert it into radians using the formula, ${\text{Radians}}\,{\text{ = }}\,{\text{Degrees}} \times \dfrac{\pi }{{180}}$. Then for the obtained answer, we will split and write it as $\cos (a + b)$ and we use the formula, $\cos (a + b) = \cos a\cos b - \sin a\sin b$
Then we obtained the value which is required.

Complete step by step solution:
Let us solve the given problem.
We are asked to find the value of $\cos {62^ \circ }$ using the sum and difference, double angle or half angle formulas.
We use any one of the formula to solve the given problem.
It seems sum and difference formula of cosine is easy compare to other two formulas, we use this and solve the question.
Firstly, we will convert the angle measure from degrees to radians.
We do this using the formula,
${\text{Radians}}\,{\text{ = }}\,{\text{Degrees}} \times \dfrac{\pi }{{180}}$
Here we have ${62^ \circ }$. Hence we get,
${\text{Radians}}\,{\text{ = }}\,62 \times \dfrac{\pi }{{180}}$
$ \Rightarrow {\text{Radians}}\,{\text{ = }}\,\dfrac{{62\pi }}{{180}}$
$ \Rightarrow {\text{Radians}}\,{\text{ = }}\,\dfrac{{7\pi }}{{12}}$
Hence, we get, $\cos ({62^\circ }) = \cos \left( {\dfrac{{7\pi }}{{12}}} \right)$
Now $\dfrac{{7\pi }}{{12}}$ can be written as,
$\dfrac{{7\pi }}{{12}} = \dfrac{\pi }{3} + \dfrac{\pi }{4}$
Hence we have, $\cos \left( {\dfrac{{7\pi }}{{12}}} \right) = \cos \left( {\dfrac{\pi }{3} + \dfrac{\pi }{4}} \right)$
Since, it is in the form of $\cos (a + b)$, we use the trigonometric identity to solve this.
We use angle sum identity of cosine which is given by,
$\cos (a + b) = \cos a\cos b - \sin a\sin b$
Here we have, $a = \dfrac{\pi }{3}$ and $b = \dfrac{\pi }{4}$.
Substituting this we get,
$\cos \left( {\dfrac{\pi }{3} + \dfrac{\pi }{4}} \right) = \cos \left( {\dfrac{\pi }{3}} \right)\cos \left( {\dfrac{\pi }{4}} \right) - \sin \left( {\dfrac{\pi }{3}} \right)\sin \left( {\dfrac{\pi }{4}} \right)$
Now we substitute the values of cosine and sine, we get,
$ \Rightarrow \cos \left( {\dfrac{\pi }{3} + \dfrac{\pi }{4}} \right) = \left( {\dfrac{1}{2}} \right)\left( {\dfrac{{\sqrt 2 }}{2}} \right) - \left( {\dfrac{{\sqrt 3 }}{2}} \right)\left( {\dfrac{{\sqrt 2 }}{2}} \right)$
$ \Rightarrow \cos \left( {\dfrac{\pi }{3} + \dfrac{\pi }{4}} \right) = \left( {\dfrac{{\sqrt 2 }}{4}} \right) - \left( {\dfrac{{\sqrt 6 }}{4}} \right)$
Taking 4 as a common denominator in the R.H.S. we get,
$ \Rightarrow \cos \left( {\dfrac{\pi }{3} + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 - \sqrt 6 }}{4}$
Thus, we have $\cos ({62^\circ }) = \cos \left( {\dfrac{{7\pi }}{{12}}} \right) = \dfrac{{\sqrt 2 - \sqrt 6 }}{4}$.

Hence, the value of $\cos {62^\circ }$ using sum identity of cosine is given by $\dfrac{{\sqrt 2 - \sqrt 6 }}{4}$

Note:
Students must know how to convert angle measure from degrees to radians and vice versa.
We convert degrees to radians using the formula,
${\text{Radians}}\,{\text{ = }}\,{\text{Degrees}} \times \dfrac{\pi }{{180}}$
Students must remember the formulas related to sine and cosine.
Some of them are given below.
(1) $\cos 2x = {\cos ^2}x - {\sin ^2}x$
(2) $\cos 2x = 2{\cos ^2}x - 1$
(3) $\cos 2x = 1 - 2{\sin ^2}x$
(4) $\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$
(5) $\sin 2x = 2\sin x\cos x$