PROVING TRIGONOMETRIC PROBLEMS WORKSHEET WITH ANSWERS

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

Detailed Answer Key

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 - cos2Acos2cos2B - 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) =  0

cosα + cosβ + cosγ  =  0

sinα + sinβ + sinγ  =  0

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