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Prove the following cos(pi/4-x)cos(pi/4-y) - sin(pi/4-x)sin(pi/4-y) = sin(x+y)

Prove the following:

Q(6) \small \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)
 

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\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 )

Multiply and divide by 2 both cos and sin functions
We get,

\frac{1}{2}\left [2 \cos\left ( \frac{\pi}{4}-x \right )\cos\left ( \frac{\pi}{4}-y \right ) \right ] + \frac{1}{2}\left [- 2\sin\left ( \frac{\pi}{4}-x \right )\sin\left ( \frac{\pi}{4}-y \right ) \right ]

Now, we know that

2cosAcosB = cos(A+B) + cos(A-B)             -(i)
-2sinAsinB = cos(A+B) - cos(A-B)               -(ii)
 We use these two identities

In our question A =   \left (\frac{\pi}{4}-x \right )

                        B =   \left (\frac{\pi}{4}-y \right )
So, 

\frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi}{4}-x \right) +\left ( \frac{\pi}{4}-y \right ) \right \} + \cos \left \{ \left ( \frac{\pi}{4}-x \right) -\left ( \frac{\pi}{4}-y \right ) \right \} \right ] +\\ \\ \frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi}{4}-x \right) +\left ( \frac{\pi}{4}-y \right ) \right \} - \cos \left \{ \left ( \frac{\pi}{4}-x \right) +\left ( \frac{\pi}{4}-y \right ) \right \} \right ]

\Rightarrow 2 \times \frac{1}{2} \left [ \cos \left \{ \left ( \frac{\pi}{4}-x \right )+\left ( \frac{\pi}{4}-y \right ) \right \} \right ]

= \cos \left [ \frac{\pi}{2}-(x+y) \right ]

As we know that

(\cos \left ( \frac{\pi}{2} - A \right ) = \sin A)
By using this

= \cos \left [ \frac{\pi}{2}-(x+y) \right ]      =\sin(x+y)

                                                                         R.H.S
 

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