Question 1 :
Show that :
cos2A + cos2B - 2cosAcosBcos(A + B) = sin2(A + B)
Question 2 :
If cos(α − β) + cos(β − γ) + cos(γ − α) = −3/2, then prove that
cos α + cos β + cos γ = sinα + sinβ + sinγ = 0
Question 1 :
Show that :
cos2A + cos2B - 2cosAcosBcos(A + B) = sin2(A + B)
Answer :
cos2A + cos2B - 2cosAcosBcos(A + B) :
= cos2A + cos2B - 2cosAcosB[cosAcosB - sinAsinB]
= cos2A + cos2B - 2cos2Acos2B + 2cosAcosBsinAsinB
= cos2A - cos2Acos2B + cos2B - cos2Acos2B + 2cosAcosBsinA sin B
Factor out cos2A from the 1st and 2nd terms and factor out cos2B from the 3rd and 4th terms.
= cos2A(1 - cos2B) + cos2B(1 - cos2A) + 2cosAcosBsinAsinB
= cos2A(sin2B) + cos2B(sin2A) + 2cosA cosBsinAsinB
= (cosAsinB)2 + (cosBsinA)2 + 2cosAcosBsinAsin B
= (cosAsinB + cosBsinA)2
= [sin (A + B)]2
= sin2(A + B)
Question 2 :
If cos(α − β) + cos(β − γ) + cos(γ − α) = −3/2, then prove that
cos α + cos β + cos γ = sinα + sinβ + sinγ = 0
Answer :
cos(α − β) = cos α cos β - sin α sin β ----(1)
cos(β − γ) = cos β cos γ - sin β sin γ ----(2)
cos(γ − α) = cos γ cos α - sin γ sin α ----(3)
(1) + (2) + (3) :
cos(α − β) + cos(β − γ) + cos(γ − α)
= cos α cos β + sin α sin β + cos β cos γ + sin β sin γ + cos γ cos α + sin γ sin α
So,
cos α cos β + sin α sin β + cos β cos γ + sin β sin γ + cos γ cos α + sin γ sin α = -3/2
2[cos α cos β + sin α sin β + cos β cos γ + sin β sin γ + cos γ cos α + sin γ sin α] + 3 = 0
2cos α cos β + 2sin α sin β + 2cos β cos γ + 2sin β sin γ + 2cos γ cos α + 2sin γ sin α] + 3 = 0
2cos α cos β + 2sin α sin β + 2cos β cos γ + 2sin β sin γ + 2cos γ cos α + 2sin γ sin α] + (cos2α + sin2α) + (cos2β + sin2β) + (cos2γ + sin2γ) = 0
cos2α + cos2β + cos2γ + 2cos α cos β + 2cos β cos γ + 2cos γ cos α + sin2α + sin2β + sin2γ + 2sin α sin β + 2sin β sin γ + 2sin γ sin α = 0
(cosα + cosβ + cosγ)2 + (sinα + sinβ + sinγ)2 = 0
cosα + cosβ + cosγ = 0
sinα + sinβ + sinγ = 0
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