Question

Prove the following:
$\cos \left( \dfrac{\pi }{4}-x \right)\cos \left( \dfrac{\pi }{4}-y \right)-\sin \left( \dfrac{\pi }{4}-x \right)\sin \left( \dfrac{\pi }{4}-y \right)=\sin \left( x+y \right)$

Answer 1

Hint: For solving this question, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side. And we will use trigonometric formulas like $\cos A\cos B-\sin A\sin B=\cos \left( A+B \right)$ and $\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta $ for simplifying the term on the left-hand side. After that, we will easily prove the desired result.

Complete step-by-step answer:

Given:
We have to prove the following equation:
$\cos \left( \dfrac{\pi }{4}-x \right)\cos \left( \dfrac{\pi }{4}-y \right)-\sin \left( \dfrac{\pi }{4}-x \right)\sin \left( \dfrac{\pi }{4}-y \right)=\sin \left( x+y \right)$
Now, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side.
Now, before we proceed we should know the following formulas:
$\begin{align}
  & \cos A\cos B-\sin A\sin B=\cos \left( A+B \right).....................\left( 1 \right) \\
 & \cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta ................................................\left( 2 \right) \\
\end{align}$
Now, we will use the above two formulas to simplify the term on the left-hand side.
On the left-hand side, we have $\cos \left( \dfrac{\pi }{4}-x \right)\cos \left( \dfrac{\pi }{4}-y \right)-\sin \left( \dfrac{\pi }{4}-x \right)\sin \left( \dfrac{\pi }{4}-y \right)$ .
Now, let $A=\dfrac{\pi }{4}-x$ and $B=\dfrac{\pi }{4}-y$ . Then,
$\begin{align}
  & \cos \left( \dfrac{\pi }{4}-x \right)\cos \left( \dfrac{\pi }{4}-y \right)-\sin \left( \dfrac{\pi }{4}-x \right)\sin \left( \dfrac{\pi }{4}-y \right) \\
 & \Rightarrow \cos A\cos B-\sin A\sin B \\
\end{align}$
Now, we will use the formula from the equation (1) to write $\cos A\cos B-\sin A\sin B=\cos \left( A+B \right)$ in the above expression. Then,
$\begin{align}
  & \cos A\cos B-\sin A\sin B \\
 & \Rightarrow \cos \left( A+B \right) \\
\end{align}$
Now, as per our assumption $A=\dfrac{\pi }{4}-x$ and $B=\dfrac{\pi }{4}-y$ . So, we can put $A=\dfrac{\pi }{4}-x$ and $B=\dfrac{\pi }{4}-y$ in the above expression. Then,
$\begin{align}
  & \cos \left( A+B \right) \\
 & \Rightarrow \cos \left( \dfrac{\pi }{4}-x+\dfrac{\pi }{4}-y \right) \\
 & \Rightarrow \cos \left( \dfrac{\pi }{4}+\dfrac{\pi }{4}-x-y \right) \\
 & \Rightarrow \cos \left( \dfrac{\pi }{2}-\left( x+y \right) \right) \\
\end{align}$
Now, we will use the formula from the equation (2) to write $\cos \left( \dfrac{\pi }{2}-\left( x+y \right) \right)=\sin \left( x+y \right)$ in the above expression. Then,
$\begin{align}
  & \cos \left( \dfrac{\pi }{2}-\left( x+y \right) \right) \\
 & \Rightarrow \sin \left( x+y \right) \\
\end{align}$
Now, from the above result, we conclude that the value of the expression $\cos \left( \dfrac{\pi }{4}-x \right)\cos \left( \dfrac{\pi }{4}-y \right)-\sin \left( \dfrac{\pi }{4}-x \right)\sin \left( \dfrac{\pi }{4}-y \right)$ will be equal to the value of the expression $\sin \left( x+y \right)$ . Then,
$\cos \left( \dfrac{\pi }{4}-x \right)\cos \left( \dfrac{\pi }{4}-y \right)-\sin \left( \dfrac{\pi }{4}-x \right)\sin \left( \dfrac{\pi }{4}-y \right)=\sin \left( x+y \right)$
Now, from the above result, we conclude that the term on the left-hand side is equal to the term on the right-hand side.
Thus, $\cos \left( \dfrac{\pi }{4}-x \right)\cos \left( \dfrac{\pi }{4}-y \right)-\sin \left( \dfrac{\pi }{4}-x \right)\sin \left( \dfrac{\pi }{4}-y \right)=\sin \left( x+y \right)$ .
Hence, proved.

Note: Here, the student should first understand what we have to prove in the question. After that, we should proceed in a stepwise manner and apply trigonometric formulas like $\cos A\cos B-\sin A\sin B=\cos \left( A+B \right)$ correctly. Moreover, while simplifying we should be aware of the result and avoid calculation mistakes while solving.