Question

How do you verify the identity $\cos \left( x+y \right)+\cos \left( x-y \right)=2\cos x\cos y$?

Answer 1

Hint: We have the theorems of the compound angles of sum and difference of ratio cos. We add those theorems and get the mentioned formula of transformation of sums and products. Then to verify the result of the theorem we take two arbitrary values of the angles and verify their values.

Complete step by step answer:
The mentioned theorem is about the transformation of sums and products where the main aim of the formula is to deal with the sum or difference of two cosines only into a product of two cosines.
We have the compound angle theorems of cosines of sum and difference of two angles as
$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ and $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$.
We take the sum of the two theorems.
The sum of the two cosines $\cos \left( A+B \right)$ and $\cos \left( A-B \right)$ gives $\cos \left( A+B \right)+\cos \left( A-B \right)$ on the left side of the equation.
On the right side of the equation, we get
$\begin{align}
  & \left( \cos A\cos B-\sin A\sin B \right)+\left( \cos A\cos B+\sin A\sin B \right) \\
 & =\cos A\cos B-\sin A\sin B+\cos A\cos B+\sin A\sin B \\
 & =2\cos A\cos B \\
\end{align}$
The terms $\sin A\sin B$ gets cancelled out in the values of the cosines.
The final equation becomes $\cos \left( A+B \right)+\cos \left( A-B \right)=2\cos A\cos B$.
We replace the values of A and B with $A=x;B=y$ and get $\cos \left( x+y \right)+\cos \left( x-y \right)=2\cos x\cos y$.
Now we verify the equation with the use of angles of $x=60;y=30$.
We place the values on the left side of the equation $\cos \left( x+y \right)+\cos \left( x-y \right)$.
So, $\cos \left( x+y \right)+\cos \left( x-y \right)=\cos \left( 60+30 \right)+\cos \left( 60-30 \right)$
The value will be $\cos \left( x+y \right)+\cos \left( x-y \right)=\cos 90+\cos 30=0+\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{2}$.
On the right-side of the equation, we have $2\cos 60\cos 30=2\times \dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{2}$.
Thus verified $\cos \left( x+y \right)+\cos \left( x-y \right)=2\cos x\cos y$.

Note: The process of converting sums into products or products into sums can make a difference between an easy solution to a problem and no solution at all. Two sets of identities can be derived from the sum and difference identities that help in this conversion. Another form of this conversion is $\cos C+\cos D=2\cos \dfrac{C+D}{2}\cos \dfrac{C-D}{2}$.