Prove that:

$
\cos\left( \frac{\pi}{4} - x \right)\cos\left( \frac{\pi}{4} - y \right) - \sin\left( \frac{\pi}{4} - x \right)\sin\left( \frac{\pi}{4} - y \right) = \sin(x + y)
$

Q: Prove that: $ \cos\left( \frac{\pi}{4} - x \right)\cos\left( \frac{\pi}{4} - y \right) - \sin\left( \frac{\pi}{4} - x \right)\sin\left( \frac{\pi}{4} - y \right) = \sin(x + y) $

Explanation: \begin{align*} \text{LHS} &= \cos\left( \frac{\pi}{4} - x \right)\cos\left( \frac{\pi}{4} - y \right) - \sin\left( \frac{\pi}{4} - x \right)\sin\left( \frac{\pi}{4} - y \right) \\ &= \cos\left[ \left( \frac{\pi}{4} - x \right) + \left( \frac{\pi}{4} - y \right) \right] \\ &= \cos\left( \frac{\pi}{2} - (x + y) \right) \\ &= \sin(x + y) = \text{RHS} \qquad(\textbf{Hence proved).} \end{align*}

Question

Prove that:

$
\cos\left( \frac{\pi}{4} - x \right)\cos\left( \frac{\pi}{4} - y \right) - \sin\left( \frac{\pi}{4} - x \right)\sin\left( \frac{\pi}{4} - y \right) = \sin(x + y)
$

Accepted Answer

Explanation: 
\begin{align*}
	ext{LHS} &= \cos\left( \frac{\pi}{4} - x ight)\cos\left( \frac{\pi}{4} - y ight) - \sin\left( \frac{\pi}{4} - x ight)\sin\left( \frac{\pi}{4} - y ight) \
&= \cos\left[ \left( \frac{\pi}{4} - x ight) + \left( \frac{\pi}{4} - y ight) ight] \
&= \cos\left( \frac{\pi}{2} - (x + y) ight) \
&= \sin(x + y) = 	ext{RHS} \qquad(	extbf{Hence proved).}
\end{align*}

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Question:

Solution