Question

cos
(π/4 - 2) cos (π/4-4) - Sin (π/4 - 2) Sin (π/4-y) = Sin (x+y)​

Answer 1

Answer:

Since

cos A cos B − sin A sin B = cos ( A+B )

cos (π\4-x) cos (π\4-y)-sin (π\4-x) sin (π\4-y)

Here A = (π\4-x), B= (π\4-y)

= cos [(π\4-x) +(π\4-y)]

= cos [ (π\4+π\4)-(x+y)]

= cos [ (π\2) - (x+y)]

= sin ( x+y )

Answer 2

=cos [(π/4-x) + (π/4-y)]

=cos [π/4-x +π/4-y]

=cos [π/4 + π/4 -x-y]

=cos [π/2 - (x + y)

Putting π =180°

= cos[180°/2-(x+y)]

=cos[90° - (x+y)] (cos(90-∅)=sin=∅

=sin(x + y)

= R.H.S

Hence proved

L.H.S

We know that

cos(A+B) = cos A cos B - sin A sin B

The equation given in question is of this form

where A = (π/4 -x). B = (π/4-y)

Hence

cos (π/4-x) cos (π/4-y) — sin (π/4-x) sin (π/4-y)

Explanation:

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