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Prove that

cos(A+B) + sin(A-B) = 2sin(45°+A)cos(45°+B)

2 answers

LS
= cosAcosB - sinAsinB + sinAcosB - cosAsinB

RS
= 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
= LS
Which formula did you use to derive the second last step ?
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