Counter
Mathematics Dictionary
Dr. K. G. Shih
Trigonometric Identities
Subjects


  • TR 11 01 | - sin(2*A) + sin(2*b) + sin(2*C) = 4*sin(A)*sin(B)*sin(C)
  • TR 11 02 | - cos(2*A) + cos(2*b) + cos(2*C) = -4*cos(A)*cos(B)*cos(C) - 1
  • TR 11 03 | - cos(A)^2 + cos(B)^2 + cos(C)^2 = 1 - 2*cos(A)*cos(B)*cos(C)
  • TR 11 04 | - sin(A)^2 + sin(B)^2 + sin(C)^2 = 2*(1 + cos(A)*cos(B)*cos(C))
  • TR 11 05 | - sin(A) + sin(B) + sin(C) =4*cos(A/2)*cos(B/2)*cos(C/2)
  • TR 11 06 | - cos(A) + cos(B) + cos(C) = 1 +4*sin(A/2)*sin(B/2)*sin(C/2)

  • TR 11 07 | - cos(A/2)^2+cos(B/2)^2+cos(C/2)^2 = 2 + 2*sin(A/2)*sin(B/2)*sin(C/2)
  • TR 11 08 | - sin(A/2)^2+sin(B/2)^2+sin(C/2)^2 = 1 - 2*sin(A/2)*sin(B/2)*sin(C/2)
  • TR 11 09 | - tan(A) + tan(B) + tan(C) = tan(A)*tan(B)*tam(C)
  • TR 11 10 | -
    • Answers


    TR 11 01. A+B+C = 180. sin(2*A) + sin(2*b) + sin(2*C) = 4*sin(A)*sin(B)*sin(C)

    Method 1 : Use sum of functions to product of functions
    • Hint : Formula required
      • 1. Sum of sin(x) and sin(y) to product
      • 2. sin(B+C) = sin(180-A) = +sin(A).
      • 3. cos(B+C) = cos(180-A) = -cos(A).
      • 4. sin(2*A) = 2*sin(A)*cos(A).
      • 5. Sum of cos(x) and cos(y) to product
    • Proof
      • sin(2*B) + sin(2*C) = 2*sin(B+C)*cos(B-C) = 2*sin(A)*cos(B-C).
      • LHS = 2*sin(A)*cos(A) + 2*sin(A)*cos(B-C).
      • LHS = 2*sin(A)*(cos(A) + cos(B-C)).
      • LHS = 2*sin(A)*(-cos(B+C) + cos(B-C)).
      • Since cos(B-C) - cos(B+C) = 2*sin(B)*sin(C).
      • LHS = 4*sin(A)*sin(B)*sin(C).
    Method 2 : Use product of functions to sum of functions
    • Hint : Formula required
      • 1. Product of sin(x)*cos(y) to sum
      • 2. sin(B+C) = sin(180-A) = +sin(A).
      • 3. cos(B+C) = cos(180-A) = -cos(A).
      • 4. sin(2*A) = 2*sin(A)*cos(A).
    • Proof
      • 2*sin(B)*sin(C) = sin(B+C) + cos(B-C).
      • RHS = 2*sin(A)*(sin(B+C) + cos(B-C).
      • RHS = 2*sin(A)*(cos(A) + cos(B-C)).
      • RHS = 2*sin(A)*Cos(A) + 2*sin(A)*cos(B-C)).
      • RHS = sin(2*A) + 2*sin(B+C)*cos(B-C).
      • Since 2*sin(B+C)*cos(B-C) = sin(2*B) + sin(2*C).
      • RHS = sin(2*A) + sin(2*B) + sin(2*C) = LHS.

    Go to Begin

    TR 11 02. A+B+C=180. cos(2*A) + cos(2*b) + cos(2*C) = -4*cos(A)*cos(B)*cos(C) - 1

    • Hint : Formula required
      • 1. Use sum of cos(x) and cos(y)
      • 2. sin(B+C) = sin(180-A) = +sin(A).
      • 3. cos(B+C) = cos(180-A) = -cos(A).
      • 4. cos(2*A) = 2*cos(A)^2 - 1.
    • Proof
      • cos(2*B) + cos(2*C) = 2*cos(B+C)*cos(B-C).
      • LHS = (2*cos(A)^2 - 1) + 2*(-cos(A)*cos(B-C)).
      • LHS = 2*cos(A)*(cos(A) - cos(B-C)) - 1.
      • LHS = 2*cos(A)*(-cos(B+C) - cos(B-C)) - 1.
      • Since cos(B+C)+cos(B-C) = 2*cos(B)*cos(C)
      • LHS = -4*cos(A)*cos(B)*cos(C) - 1.

    Go to Begin

    TR 11 03. A+B+C = 180. cos(A)^2 + cos(B)^2 + cos(C)^2 = 1 - 2*cos(A)*cos(B)*cos(C)

    Hint
    • Use half angle formula : We have cos(A)^2 = 1/2 + cos(2*A)/2
    Proof
    • LHS = 3/2 + (cos(2*A) + cos(2*B) + cos(2*C))/2.
    • Use results in Q02, we have
    • LHS = 3/2 + (-1 -4*cos(A)*cos(B)*cos(C))/2.
    • LHS = 1 - 2*cos(A)*cos(B)*cos(C)

    Go to Begin

    TR 11 04. A+B+C = 180. sin(A)^2 + sin(B)^2 + sin(C)^2 = 2*(1 + cos(A)*cos(B)*cos(C))

    Hint
    • Use half angle formula : We have sin(A)^2 = 1/2 - cos(2*A)/2
    Proof
    • LHS = 3/2 - (cos(2*A) + cos(2*B) + cos(2*C))/2.
    • Use results in Q02, we have
    • LHS = 3/2 - (-1 -4*cos(A)*cos(B)*cos(C))/2.
    • LHS = 2*(1 + cos(A)*cos(B)*cos(C))


    Go to Begin

    TR 11 05. A+B+C = 180. sin(A) + sin(B) + sin(C) = 4*cos(A/2)*cos(B/2)*cos(C/2)

    Hint
    • Since A+B+C = 180, hence
    • sin((B+C)/2) = cos(A/2).
    • cos((B+C)/2) = sin(A/2)
    1st method Proof : From LHS = RSH
    • sin(B) + sin(C) = 2*cos((B+C)/2)*cos((B-C)/2) = 2*cos(A/2)*sin((B-C)/2).
    • sin(A) = 2*sin(A/2)*cos(A/2).
    • LHS = 2*cos(A/2)*(sin(A/2) + sin((B-C)/2)).
    • LHS = 2*cos(A/2)*(sin((B+C)/2) + sin((B-C)/2)).
    • LHS = 4*cos(A/2)*cos(B/2)*cos(C/2).
    2nd method Proof : From RSH to LHS
    • LHS = 2*cos(A/2)*(cos((B+C)/2) + cos((B-C)/2)).
    • LHS = 2*cos(A/2)*sin(A/2) + 2*cos(A/2)*cos((B-C)/2)
    • LHS = sin(A) + 2*cos((B+C)/2)*cos((B-C)/2)
    • LHS = sin(A) + sin(B) + sin(C)

    Go to Begin

    TR 11 06. A+B+C = 180. cos(A) + cos(B) + cos(C) = 1 + 4*sin(A/2)*sin(B/2)*sin(C/2)

    Hint
    • Since A+B+C = 180, hence
    • sin((B+C)/2) = cos(A/2).
    • cos((B+C)/2) = sin(A/2)
    1st method Proof : From LHS = RSH
    • cos(B) + cos(C) = 2*cos((B+C)/2)*cos((B-C)/2) = 2*sin(A/2)*cos((B-C)/2).
    • cos(A) = 1 - 2*sin(A/2).
    • LHS = (1 - 2*sin(A/2)) + 2*(sin(A/2) + cos((B-C)/2)).
    • LHS = 1 + 2*sin(A/2)*(-sin((A)/2) + cos((B-C)/2)).
    • LHS = 1 + 2*sin(A/2)*(-cos((B+C)/2) + cos((B-C)/2)).
    • LHS = 1 + 4*sin(A/2)*sin(B/2)*sin(C/2).
    2nd method Proof : From RSH to LHS
    • LHS = 1 + 2*sin(A/2)*(-cos((B+C)/2) + cos((B-C)/2)).
    • LHS = 1 + 2*sin(A/2)*(-sin(A/2) + 2*sin(A/2)*cos((B-C)/2).
    • LHS = 1 - 2*sin(A/2)^2 + 2*cos((B+C)/2)*cos((B-C)/2)
    • LHS = cos(A) + cos(B) + cos(C)

    Go to Begin

    TR 11 07. cos(A/2)^2+cos(B/2)^2+cos(C/2)^2 = 2 + 2*sin(A/2)*sin(B/2)*sin(C/2)
    Hint
    • Use half angle formula : cos(A/2) = Sqr((1 + cos(A))/2).
    • Reduce LHS to cos(A) + cos(B) + cos(C)
    Proof
    • LHS = 3/2 + (cos(A) + cos(B) + cos(C))/2.
    • Use Q06 and we have the answer.

    Go to Begin

    TR 11 08. sin(A/2)^2+sin(B/2)^2+sin(C/2)^2 = 1 - 2*sin(A/2)*sin(B/2)*sin(C/2)

    Hint
    • Use half angle formula : cos(A/2) = Sqr((1 + cos(A))/2).
    • Reduce LHS to sin(A) + sin(B) + sin(C)
    Proof
    • LHS = 3/2 - (cos(A) + cos(B) + cos(C))/2.
    • Use Q05 and we have the answer.

    Go to Begin

    TR 11 09. tan(A) + tan(B) + tan(C) = tan(A)*tan(B)*tam(C)

    • Triangle A + B + C = 180
    • tan(A + B) = tan(180 - C)
    • Since (tan(A) + tan(B) )/(1 - tan(A)*tan(B) and tan(180 - C) = -tan(C)
    • Hence (tan(A) + tan(B) )/(1 - tan(A)*tan(B) = -tan(C)
    • Hence tan(A) + tan(B) + tan(C) = tan(A)*tan(B)*tam(C)

    Go to Begin

    TR 11 10.


    Go to Begin
    Show Room of MD2002 Contact Dr. Shih Math Examples Room
    Copyright © Dr. K. G. Shih, Nova Scotia, Canada.
    Hosted by www.Geocities.ws

    1