Prove that

Cos (A+B) + Sin (A-B) = 2sin (45+A). cos (45+B)

Question

Walsh

Mathematics

3 October, 08:10

Prove that

Cos (A+B) + Sin (A-B) = 2sin (45+A). cos (45+B)

+1

Answers (1)

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Addisyn Alvarez · Answer 1

Step-by-step explanation:

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

= 2 (sin45°cosA + cos45°sinA) (cos45°cosB - sin45°sinB)

But sin45=cos45 = (sqrt2) / 2

= 2 ((sqrt2) / 2 * cosA + (sqrt2) / 2 * sinA) ((sqrt2) / 2 * cosB - (sqrt2) / 2 * sinB)

= 2 ((sqrt2) / 2 * (cosA + sinA)) * ((sqrt2) / 2 * (cosB - sinB))

= 2 * (sqrt2) / 2 * (sqrt2) / 2 * (cosA + sinA) * (cosB - sinB)

= (cosA + sinA) * (cosB - sinB)

= cosAcosB + sinAcosB - cosAsinB - sinAsinB

Regrouping:

= (cosAcosB - sinAsinB) + (sinAcosB - cosAsinB)

= cos (A+B) + sin (A-B)

Prove thatCos (A+B) + Sin (A-B) = 2sin (45+A). cos (45+B)

Prove that

Cos (A+B) + Sin (A-B) = 2sin (45+A). cos (45+B)