to prove:2sin(A+45)+sin(A-45)=sin^2 A-cos^2 A
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2016-06-26T11:16:14+05:30
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)
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)
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