Prove that $ \cos \left( {{60}^{\circ }}+\theta \right)+\cos \left( {{60}^{\circ }}-\theta \right)=\cos \theta $ .

Hint: In this question, we need to prove the trigonometric function $ \cos \left( {{60}^{\circ }}+\theta \right)+\cos \left( {{60}^{\circ }}-\theta \right) $ to be equal to $ \cos \theta $ . For this, we will use the formula of addition and subtraction of angles in the cosine function to evaluate our answer. We will also use a trigonometric ratio table to find the values of $ \cos {{60}^{\circ }} $ . Formula of cosine that we will use are:
\[\begin{align}
  & \left( i \right)\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B \\
 & \left( ii \right)\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B \\
\end{align}\]

Complete step by step answer:
Here we are given the equation as $ \cos \left( {{60}^{\circ }}+\theta \right)+\cos \left( {{60}^{\circ }}-\theta \right)=\cos \theta $ .
We need to prove the left side to be equal to the right side. For this, let us first pick the left side.
We have $ \cos \left( {{60}^{\circ }}+\theta \right)+\cos \left( {{60}^{\circ }}-\theta \right) $ .
As we can see first term is in the form of cos(A+B) and second term is in the form of cos(A-B). So we can apply the formula of addition and subtraction of angles in cosine function given by,
 $ \cos \left( A+B \right)=\cos A\cos B-\sin A\sin B\text{ and }\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B $ .
Here $ A={{60}^{\circ }}\text{ and }B=\theta $ we get,
 $ \cos \left( {{60}^{\circ }}+\theta \right)+\cos \left( {{60}^{\circ }}-\theta \right)=\cos {{60}^{\circ }}\cos \theta -\sin {{60}^{\circ }}\sin \theta +\cos {{60}^{\circ }}\cos \theta +\sin {{60}^{\circ }}\sin \theta $ .
Cancelling $ \sin {{60}^{\circ }}\sin \theta $ we get,
 $ \cos \left( {{60}^{\circ }}+\theta \right)+\cos \left( {{60}^{\circ }}-\theta \right)=2\cos {{60}^{\circ }}\cos \theta $ .
From the trigonometric ratio table, we know that $ \cos {{60}^{\circ }}=\dfrac{1}{2} $ so putting in the values of $ \cos {{60}^{\circ }} $ in above equation we get,
 $ \cos \left( {{60}^{\circ }}+\theta \right)+\cos \left( {{60}^{\circ }}-\theta \right)=2\times \dfrac{1}{2}\times \cos \theta $ .
Cancelling 2 with 2 we get, $ \cos \left( {{60}^{\circ }}+\theta \right)+\cos \left( {{60}^{\circ }}-\theta \right)=\cos \theta $ .
Which is equal to the right hand side of the given equation.
Hence proved.

Note:
Students should take care of the signs while solving this sum. Keep in mind all the trigonometric properties and identities. Student can solve this sum in following way also,
We are given left side as, $ \cos \left( {{60}^{\circ }}+\theta \right)+\cos \left( {{60}^{\circ }}-\theta \right) $ .
As we can see, it is in the form cosC+cosD. So let us apply the formula of cosine angles given by $ \cos C+\cos D=\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{D-C}{2} \right) $ . Here, $ C={{60}^{\circ }}\text{ and }D=\theta $ we get,
 $ \begin{align}
  & \Rightarrow \cos \left( \dfrac{{{60}^{\circ }}+\theta +{{60}^{\circ }}-\theta }{2} \right)\cos \left( \dfrac{{{60}^{\circ }}+\theta -{{60}^{\circ }}+\theta }{2} \right) \\
 & \Rightarrow \cos \left( \dfrac{{{120}^{\circ }}}{2} \right)\cos \left( \dfrac{2\theta }{2} \right) \\
 & \Rightarrow \cos {{60}^{\circ }}\cos \theta \\
\end{align} $
Putting in the values of $ \cos {{60}^{\circ }}=\dfrac{1}{2} $ we get, $ \cos \theta $ .
Which is the right side.
Hence proved.