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

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Question 551204: Prove that
cos(A+B) + sin(A-B) = 2sin(45°+A)cos(45°+B)
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Prove that
cos(A+B) + sin(A-B) = 2sin(45°+A)cos(45°+B)

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[sqrt%282%29%2F2cos(A) + sqrt%282%29%2F2sin(A)][sqrt%282%29%2F2cos(B)-sqrt%282%29%2F2sin(B)]

2sqrt%282%29%2F2[cos(A) + sin(A)]sqrt%282%29%2F2[cos(B)-sin(B)]

2sqrt%282%29%2F2sqrt%282%29%2F2[cos(A) + sin(A)][cos(B)-sin(B)]

%282%2A2%29%2F%282%2A2%29[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)

Edwin