Question

What is $\cos A-\cos B$?

Answer 1

Hint: We will split the values of angles $A$ and $B$ in a certain way. Then we will use trigonometric identities to expand the given expression. The trigonometric identities that we will use are $\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y$ and $\cos \left( x-y \right)=\cos x\cos y+\sin x\sin y$. Simplifying this modified expression, we will be able to find the value for the given expression.

Complete step-by-step answer:
We have to find the value for the expression $\cos A-\cos B$. Now, we can write angles $A$ and $B$ in the following manner,
$A=\dfrac{A+B}{2}+\dfrac{A-B}{2}$ and $B=\dfrac{A+B}{2}-\dfrac{A-B}{2}$.
Substituting these values in the given expression, we get
$\cos A-\cos B=\cos \left( \dfrac{A+B}{2}+\dfrac{A-B}{2} \right)-\cos \left( \dfrac{A+B}{2}-\dfrac{A-B}{2} \right)$
We know the following trigonometric identities,
$\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y$
$\cos \left( x-y \right)=\cos x\cos y+\sin x\sin y$
We can use these trigonometric identities to expand the terms on the right hand side of the above expression in the following manner,
$\cos \left( \dfrac{A+B}{2}+\dfrac{A-B}{2} \right)=\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)-\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$
$\cos \left( \dfrac{A+B}{2}-\dfrac{A-B}{2} \right)=\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)+\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$
So, now we have the following expression,
$\begin{align}
  & \cos A-\cos B=\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)-\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right) \\
 & -\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)-\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right) \\
\end{align}$
Simplifying the above expression we get,
$\begin{align}
  & \cos A-\cos B=-\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)-\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right) \\
 & =-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)
\end{align}$
Hence, we get the final expression as $\cos A-\cos B=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$.

Note: The final expression is a formula that relates sum or difference to the product. The same method can be used to prove the remaining results of the sum and product identities. The identity for $\cos A+\cos B$ is given by $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$. The identities for the sum or difference of sine functions and of tangent functions are similar to that of the cosine functions'.