How do you use the sum and difference formula to simplify $\cos \left( {\dfrac{{17\pi }}{{12}}} \right)$ ?

Hint: In the question above, we have a function $\cos \left( {\dfrac{{17\pi }}{{12}}} \right)$ , and we have to solve it using the sum and difference formula, there are a lot of sine and cosine formulas with sum and difference.
One of the formulae is,
$\cos (A + B) = \cos A\cos B - \sin A\sin B$
We are going to solve this simplification with the help of this formula.

Complete step-by-step solution:
For a function $\cos \left( {\dfrac{{17\pi }}{{12}}} \right)$ , we can see that there exists a $\cos $ and therefore we will be using $\cos ine$ formula that exists in the sum and difference formula.
$\cos (A + B) = \cos A\cos B - \sin A\sin B$
Parting the numbers in two halves, in order to divide it according to the requirement,
$A = \dfrac{{8\pi }}{{12}} = 2\dfrac{\pi }{3}$ and $B = \dfrac{{9\pi }}{{12}} = \dfrac{{3\pi }}{4}$,
Then, $A + B = \dfrac{{17\pi }}{{12}}$ since both the halves make a whole.
Now, substituting the values inside the formula for $\cos $,
$\cos A = \cos (\dfrac{{2\pi }}{3}) = \cos \left( {\pi - \dfrac{\pi }{3}} \right) = - \cos \left( {\dfrac{\pi }{3}} \right) = - \dfrac{1}{2}$
Also, substituting the values inside the formula for $\sin $,
$\sin A = \sin (\dfrac{{2\pi }}{3}) = \sin \left( {\pi - \dfrac{\pi }{3}} \right) = \sin \left( {\dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}$
Substituting the values for the part $B$, in $\cos $,
$\cos B = \dfrac{{\cos (3\pi )}}{4} = \cos \left( {\pi - \dfrac{\pi }{4}} \right) = - \dfrac{{\cos \pi }}{4} = - \dfrac{1}{{\sqrt 2 }}$
And,
Substituting the values for the part $B$, in $\sin $,
$\sin B = \sin \left( {\dfrac{{3\pi }}{4}} \right) = \sin \left( {\pi - \dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$
Now that we have the values that are required, we will substitute them in the formulas,
Hence, $\cos \left( {\dfrac{{17\pi }}{{12}}} \right) = \cos (A + B)$
This clearly shows that the sum and difference method can be followed with the formula which is,
$ = \cos A\cos B - \sin A\sin B$
Substituting the values in the formula for simplifying the equation and multiplying the values,
$ \Rightarrow = \left( { - \dfrac{1}{2}} \right) \times \left( { - \dfrac{1}{{\sqrt 2 }}} \right) - \left( {\dfrac{{\sqrt 3 }}{2}} \right) \times \left( {\dfrac{1}{{\sqrt 2 }}} \right)$
Simplifying the equation further after multiplying the fractions,
$ \Rightarrow = \dfrac{1}{{2\sqrt 2 }} - \dfrac{{\sqrt 3 }}{{2\sqrt 2 }}$
Subtracting the fractions in more detail,
$ \Rightarrow = \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$

Therefore, after using the sum and difference formula, the simplification of $\cos \left( {\dfrac{{17\pi }}{{12}}} \right)$ will be forming the solution $\dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$ .

Note: The sum and difference method have formulas for a lot of sine, cosine, or tangent functions of two given angles. In order to use the formula, we have to first break it into numbers and find out what formula suits the numbers well. For example, in the above question the formula in use is $\cos (A + B) = \cos A\cos B - \sin A\sin B$ . With the help of this identity, we can perform the sum and difference operation on the equation.