cos[pie/4-x]cos[pie/4-y] - sin [pie/4-x]sin [pie/4-y] = sin [x+y]

1.	cos[pie/4-x]cos[pie/4-y] - sin [pie/4-x]sin [pie/4-y] = sin [x+y]
Answer» LHS = cos(π/4 - x).cos(π/4-y)-sin(π/4 -x).sin(π/4-y) Let ( π/4 - x) = A (π/4 - y) = B Then, LHS = cosA.cosB -sinA.sinB But we know, cos(A + B) = cosA.cosB - sinA.sinB use this, cos(A + B) = cos{(π/4 -x) + (π/4 -y)}=cos(π/2 - (x +y)} We know, Cos(π/2 - ∅) = sin∅ use this , = sin(x + y) Hence proved

Answer»

LHS = cos(π/4 - x).cos(π/4-y)-sin(π/4 -x).sin(π/4-y)

Let ( π/4 - x) = A (π/4 - y) = B

Then, LHS = cosA.cosB -sinA.sinB But we know, cos(A + B) = cosA.cosB - sinA.sinB

use this, cos(A + B) = cos{(π/4 -x) + (π/4 -y)}=cos(π/2 - (x +y)}

We know, Cos(π/2 - ∅) = sin∅ use this , = sin(x + y)

Hence proved

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