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Dr. K. G. Shih
Half angle formula


  • Q01 | - Sin(A) = 2*Sqr(s*(s-a)*(s-b)*(s-c))/(b*c)
  • Q02 | - Area of triangle = Sqr(s(s-a)*(s-b)*(s-c)/(b*c))
  • Q03 | - tan(A/2) = Sqr((s-b)*(s-c)/(s*(s-a))
  • Q04 | - cos(A/2) = Sqr(s*(s-a)/(b*c))
  • Q05 | - sin(A/2) = Sqr((s-b)*(s-c)/(b*c))

    • Q01. Sin(A) = 2*Sqr(s*(s-a)*(s-b)*(s-c))/(b*c)

    Definition
    • Triangle ABC with 3 sides a, b, c
    • s = (a + b + c)/2
    Method 1
    • Hint : Use cosine law a^2 = b^2 + c^2 - 2*b*c*cos(A)
    Proof
    • cos(A) = (b^2 + c^2 - a^2)/(2*b*c).
    • sin(A)^2 = 1 - cos(A)^2.
    • sin(A)^2 = 1 - (b^2 +c^2 - a^2)^2/(4*b^2*c^2).
    • sin(A)^2 = (1 - (b^2 +c^2 - a^2)/(2*b*c))*(1 + (b^2 +c^2 - a^2)/(2*b*c)).
    • sin(A)^2 = ((b+c)^2 - a^2)*(a^2 - (b+c)^2)/(4*b^2*c^2).
    • sin(A)^2 = (a+b+c)*(-a+b+c)*(a-b+c)*(a+b-c)/(4*b^2*c^2).
    • Let s = (a+b+c)/2.
    • Hence a+b+c = 2*s, -a+b+c = 2*(s-a), a-b+c = 2*(s-b0 and a+b-c = 2*(s-c)
    • Hence sin(A)^2 = 16*s*(s-a)(s-b)*(s-c)/(4*b^2*c^2)
    • Hence sin(A)^2 = 4*s*(s-a)*(s-b)*(s-c)/(b^2*c^2)
    • Hence sin(A) = 2*Sqr(s(s-a)*(s-b)*(s-c)/(b*c))/(b*c).
    Method 2
    • Hint : Use sin(A/2) = Sqr((s-b)*(s-c)/(b*c)).
    • Use cos(A/2) = Sqr(s*(s-a)/(b*c)).
    • Use sin(A) = 2*sin(A/2)*cos(A/2).
    Proof
    • Hence sin(A) = 2*Sqr(s*(s-a)(s-b)*(s-c))/(b*c).

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    Q02. Heron formula : Sqr(s(s-a)*(s-b)*(s-c)/(b*c))

    Proof
    • Since area of triangle = b*c*sin(A)/2 .............. (1)
    • Since sin(A) = 2*Sqr(s*(s-a)*(s-b)*(s-c))/(b*c) .... (2)
    • Substitute (2) into (1)
    • Hence area of triangle = Sqr(s*(s-a)*(s-b)*(s-c)/(b*c))
    • This is called Heron formula (MD 2002 program 20 13 or 20 16).

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    Q03. tan(A/2) = Sqr((s-b)*(s-c)/(s*(s-a))

    Method 1
    • Hint : Use Area of triangle = r*s where r is radius of incircle.
    • Use r = (s-a)*tan(A/2).
    Proof
    • tan(A/2) = r/(s-a).
    • r = Area/s.
    • r = Sqr(s*(s-a)*(s-b)*(s-c))/s.
    • Hence tan(A/2) = Sqr((s-b)*(s-c)/(s*(s-a)).
    Method 2
    • Hint : Hence tan(A/2) = Sqr((s-b)*(s-c)/(s*(s-a)).
    Identities
    • tan(A/2) = Sqr((s-b)*(s-c)/(s*(s-a)).
    • tan(B/2) = Sqr((s-c)*(s-a)/(s*(s-b)).
    • tan(C/2) = Sqr((s-a)*(s-b)/(s*(s-c)).

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    Q04. cos(A/2) = Sqr(s*(s-a)/(b*c))

    Method 1
    • Hint
      • 1. Use 1 + tan(x)^2 = sec(x)^2.
      • 2. Use cos(x) = 1/sec(x).
    • Proof
      • cos(A/2)^2 = 1/(Sec(A/2)^2).
      • cos(A/2)^2 = 1/(1 + tan(A/2)^2).
      • cos(A/2)^2 = 1/(1 + (s-b)*(s-c)/(s*(s-a))
      • cos(A/2)^2 = s*(s-a)/(s*(s-a) + (s-b)*(s-c)).
      • s*(s-a) + (s-b)*(s-c) = s^2 - s*a + s^2 - s*(b+c) + b*c.
      • s*(s-a) + (s-b)*(s-c) = 2*s^2 - s*(a+b+c) + b*c.
      • s*(s-a) + (s-b)*(s-c) = 2*s^2 - s*(2*s) + bc = b*c.
      • Hence cos(A/2) = Sqr(s*(s-a)/(b*c)).
    Method 2
    • Hint
      • 1. Use cosine law.
      • 2. Use cos(2*x) = 2*cos(x)^2 - 1.
      • 3. Use a^2 + 2*b*c + c^2 = (b+c)^2.
    • Proof
      • Cosine law : cos(A) = (b^2 + c^2 - a^2)/(b*c).
      • Since cos(2*x) = 2*cos(x)^2 -1.
      • Hence cos(A/2)^2 = (1 + cos(A))/2.
      • Hence cos(A/2) = Sqr(1 + (b^2 + c^2 - a^2)/(2*b*c)).
      • Hence cos(A/2) = Sqr(2*b*c + (b^2 + c^2 - a^2))/(2*b*c)).
      • Hence cos(A/2) = Sqr((b + c)^2 - a^2)/(2*b*c)).
      • Hence cos(A/2) = Sqr((b + c + a)*(b + c - a)/(2*b*c)).
      • Let 2*s = a + b + c and then b + c - a = a + b + c - 2*a = 2*s - 2*a.
      • Hence cos(A/2) = Sqr((2*s)*(2*s - 2*a)/(2*b*c)).
      • Hence cos(A/2) = Sqr(s*(s - a)/(b*c)).
    Identities
    • cos(A/2) = Sqr(s*(s-a)/(b*c)).
    • cos(B/2) = Sqr(s*(s-b)/(c*a)).
    • cos(C/2) = Sqr(s*(s-c)/(a*b)).

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    Q05. sin(A/2) = Sqr((s-b)*(s-c)/(b*c)

    Method 1
    • Hint
      • Use sin(x) = tan(x)*cos(x).
      • tan(A/2) = Sqr((s-b)*(s-c)?(s*(s-a)).
      • cos(A/2) = Sqr(s*(s-a)/(b*c)).
    • Proof
      • sin(A/2)^2 = tan(A/2)*cos(A/2).
      • sis(A/2)^2 = Sqr((s-b)*(s-c)/(s*(s-a))*Sqr(s*(s-a)/(b*c)).
      • sin(A/2)^2 = Sqr((s-b)*(s-c)/(b*c)).
      • Hence sin(A/2) = Sqr((s-b)*(s-c)/(b*c)).
    Method 2
    • Hint
      • 1. Use cosine law.
      • 2. Use cos(2*x) = 1 - 2*sin(x)^2.
      • 3. Use a^2 - 2*b*c + c^2 = (b-c)^2.
    • Proof
      • Cosine law : cos(A) = (b^2 + c^2 - a^2)/(b*c).
      • Since cos(2*x) = 1 - 2*sin(x)^2.
      • Hence sin(A/2)^2 = (1 - cos(A))/2.
      • Hence sin(A/2) = Sqr(1 - (b^2 + c^2 - a^2)/(2*b*c)).
      • Hence sin(A/2) = Sqr(2*b*c - (b^2 + c^2 - a^2))/(2*b*c)).
      • Hence sin(A/2) = Sqr(a^2 - (b - c)^2 )/(2*b*c)).
      • Hence sin(A/2) = Sqr((a + b - c)*(a + c - b)/(2*b*c)).
      • Let 2*s = a + b + c and then a + b - c = a + b + c - 2*c = 2*s - 2*c.
      • Hence sin(A/2) = Sqr((2*s)*(2*s - 2*a)/(2*b*c)).
      • Hence sin(A/2) = Sqr((s - b)*(s - c)/(b*c)).
    Identities
    • sin(A/2) = Sqr((s-b)*(s-c)/(b*c)).
    • sin(B/2) = Sqr((s-c)*(s-a)/(c*a)).
    • sin(C/2) = Sqr((s-a)*(s-b)/(a*b)).

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