Question

Prove that: Cos(A + B) + sin(A - B) = 2sin(45° + A)cos(45° + B)

Answer 1

I will simplify the right side until it becomes the left side: 2sin(45°+A)cos(45°+B) 2[sin(45°)cos(A) + cos(45)sin(A)][cos(45°)cos(B) - sin(45°)sin(B)] 2[cos(A) + sin(A)][cos(B)-sin(B)] 2[cos(A) + sin(A)][cos(B)-sin(B)] 2[cos(A) + sin(A)][cos(B)-sin(B)] [cos(A) + sin(A)][cos(B)-sin(B)] (1)[cos(A) + sin(A)][cos(B)-sin(B)] [cos(A) + sin(A)][cos(B)-sin(B)] cos(A)cos(B) - cos(A)sin(B) + sin(A)cos(B) - sin(A)sin(B) Rearrange terms: cos(A)cos(B) - sin(A)sin(B) + sin(A)cos(B) - cos(A)sin(B) [cos(A)cos(B) - sin(A)sin(B)] + [sin(A)cos(B) - cos(A)sin(B)] cos(A+B) + sin(A-B)

Answer 2

Given Cos(A + B) + sin(A - B) = 2sin(45° + A)cos(45° + B)
lift hand side ( LHS )
Cos(A + B) + sin(A - B)
cosAcosB - sinAsinB + sinAcosB - cosAsinB

RHS
=> 2(sin45°cosA + cos45°sinA)(cos45cosB - sin45sinB)

=>2(√2/2cosA + √2/2sinA)(√2/2cosB - √2/2sinB)

=> 2( (1/2)cosAcosB - (1/2)cosAsinB + (1/2)sinAcosB - (1/2)sinAsinB)

=> cosAcosB - cosAsinB + sinAcosB - sinAsinB

LHS = RHS 

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