Figure 116 : Cos(A - B) = cos(A)*cos(B) - sin(A)*sin(B)
Q01. Diagram
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Cos(A - B) 
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Q02. Prove that cos(A - B) = cos(A)*cos(B) + sin(A)*sin(B)
Keywords
- Distance formula
- Cosine law
- Draw triangle circle with center O
- Draw angle A = POX and angle B = QOX
- Assume OP = OQ = 1
- Polar coordinate of A : xp = cos(A) and yp = sin(A)
- Polar coordinate of B : xq = cos(B) and yq = sin(B)
- Distance between A and B
- d = Sqr((xp - xq)^2 + (yp - yq)^2)
- d = Sqr((cos(A) - cos(B))^2 + (sin(A) - sin(B))^2)
- Since cos(A)^2 + sin(A)^2 = 1 and cos(B)^2 + sin(B)^2 = 1
- Hence d^2 = 2 - 2*cos(A)*cos(B) - 2*sin(A)*sin(B) ..... (1)
- Find PQ using cosine law
- d^2 = PO^2 + QO^2 - 2*PO*QO*cos(A - B)
- d^2 = 2 - 2*cos(A - B) ..... (2)
- From (1) and (2) we hav
- 2 - 2*cos(A)*cos(B) - 2*sin(A)*sin(B) = 2 - 2*cos(A-B)
- Hence cos(A-B) = cos(A)*cos(B) + sin(A)*sin(B)
- Geometric proof : See TR 07 04
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Q03. Related questions
Questions
- 1. cos(A + B) = cos(A)*cos(B) - sin(A)*sin(B)
- 2. sin(A + B) = sin(A)*cos(B) + cos(A)*sin(B)
- 3. sin(A - B) = sin(A)*cos(B) - cos(A)*sin(B)
- Question 1 : Let B = -B.
- cos(B) = +cos(B) and
- sin(B) = -sin(B)
- Hence cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B)
- Question 2 : Let B = pi/2 + B.
- cos(pi/2 + A + B) = -sin(A + B)
- cos(pi/2 + A + B) = cos(A)*cos(pi/2 + B) - sin(A)*sin(pi/2 + B)
- Since cos(pi/2 + A + B) = -sin(A + B)
- cos(pi/2 + B) = -sin(B) and
- sin(pi/2 + B) = cos(A)
- -sin(A + B) = -cos(A)*sin(B) - sin(A)*cos(B)
- sin(A + B) = sin(A)*cos(B) + cos(A)*sin(B)
- Question 3 : Let A = pi - A.
- cos(pi - A + B)) = cos(pi - A)*cos(B) + sin(pi - A)*sin(B)
- Since cos(pi - (A - B) = -sin(A - B)
- cos(pi - A) = -sin(A) and
- sin(pi - B) = +cos(A)
- sin(A - B) = sin(A)*cos(B) - cos(A)*sin(B)
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Q04. Reference
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Trigonometry 07 04 Geometric proof
- 07 01 sin(A + B) = sin(A)*cos(B) + cos(A)*sin(B)
- 07 02 sin(A - B) = sin(A)*cos(B) - cos(A)*sin(B)
- 07 03 cos(A + B) = cos(A)*cos(B) - sin(A)*sin(B)
- 07 04 cos(A + B) = cos(A)*cos(B) - sin(A)*sin(B)
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