Prove that $\cos {{100}^{\circ }}+\cos {{20}^{\circ }}=\cos {{40}^{\circ }}$.

Hint: In this question, we are given a sum of two cosine values and we have to prove that to be equal to another cosine value. Since, we don't know values of $\cos {{100}^{\circ }},\cos {{20}^{\circ }},\cos {{40}^{\circ }}$ so we will use trigonometric identities only to prove given equation as equal. For this, we will use property of cosine as:
\[\Rightarrow \cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\].
We will also use the value of $\cos {{60}^{\circ }}$ which is equal to $\dfrac{1}{2}$.

Complete step by step answer:
Here, we are given value of two cosine values which are $\cos {{100}^{\circ }}\text{ and }\cos {{20}^{\circ }}$ and we have to prove it to be equal to another cosine value which is $cos{{40}^{\circ }}$.
Since we don't know value of $\cos {{100}^{\circ }},\cos {{20}^{\circ }},\cos {{40}^{\circ }}$ so we need to use some trigonometric identities here.
As we know, some of the cosine values can become equal to product of two other cosine value using the formula: \[\Rightarrow \cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\]
So let us use this to prove the left hand side to be equal to the right hand side.
Applying formula on left hand side we get (taking $A={{100}^{\circ }},B={{20}^{\circ }}$):
\[\Rightarrow \cos {{100}^{\circ }}+\cos {{20}^{\circ }}=2\cos \left( \dfrac{{{100}^{\circ }}+{{20}^{\circ }}}{2} \right)\cos \left( \dfrac{{{100}^{\circ }}-{{20}^{\circ }}}{2} \right)\]
Simplifying the angles of cosine functions we get:
\[\begin{align}
  & \Rightarrow 2\cos \left( \dfrac{{{120}^{\circ }}}{2} \right)\cos \left( \dfrac{{{80}^{\circ }}}{2} \right) \\
 & \Rightarrow 2\cos {{60}^{\circ }}\cos {{40}^{\circ }} \\
\end{align}\]
Now the left hand side of the equation has been simplified to $2\cos {{60}^{\circ }}\cos {{40}^{\circ }}$. As we know, $\cos {{60}^{\circ }}=\dfrac{1}{2}$ so let us use that for further simplification:
\[\begin{align}
  & \Rightarrow 2\times \dfrac{1}{2}\times \cos {{40}^{\circ }} \\
 & \Rightarrow \cos {{40}^{\circ }} \\
\end{align}\]
Hence, the left hand side has become equal to $\Rightarrow \cos {{40}^{\circ }}$ which is equal to the right hand side.
Hence proved.

Note: Students should always remember trigonometric identities such as cosA+cosB, cosA-cosB, sinA+sinB, sinA-sinB. They should not try to find values of $\cos {{100}^{\circ }}\text{ or }\cos {{20}^{\circ }}$ but always try to use trigonometric identities to get the required results. In the formula don't forget to divide the sum or difference by 2.