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Trigono
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Trigono

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Transcript

  • 1. Proving trigonometrical identitiesusing sum and difference of two angles
    Vania Lundina
    Grade 10
  • 2. Basic knowledge
  • 3. Say we have this diagram:
    y
    In triangle OPQ,
    angle POQ = A - B
    P(cosA, sinA)
    Q(cosB, sinB)
    A
    B
    x
    1
    O
    P and Q are two points on the circle of radius 1 unit
    Coordinates of P = (cosA, sinA)
    Coordinates of Q = (cosB, sinB)
  • 4. Now we can form an equation using:
    The distance formula
    PQ2 = (cosA-cosB)2 + (sinA-sinB)2
    = cos2A – 2cosAcosB + cos2B + sin2A – 2sinAsinB + sinB2
    = 2 – 2(cosAcosB + sinAsinB)
    The cosine formula from triangle OPQ
    PQ2 = 12 + 12 – 2(1)(1)cos(A-B)
    = 2 – 2cos(A-B)
  • 5. From the two equations, we get:
    2 – 2(cosAcosB + sinAsinB) = 2 – 2cos(A-B)
    We can cross out the “ 2 – 2 “ from the equation;
    2 – 2(cosAcosB + sinAsinB) = 2 – 2cos(A-B)
    cos(A-B) = cosAcosB + sinAsinB
  • 6. Replace B with (-B) into;
    cos[A-(-B) = cosAcos(-B) + sinAsin(-B)
    since cos(-B) = cosB and sin(-B) = -sinB,
    the equation can be simplified into;
    cos(A+B) = cosAcosB - sinAsinB
  • 7. Since sin ϑ = COS (900 – ϑ)
    sin(A+B) = cos[900 – (A+B)]
    = cos[(900 – a) – B]
    = cos(900 – A) cosB + sin(900 – A)sinB
    So, sin(A+B) = sinAcosA + cosAsinB
  • 8. Replace B with (-B) from the previous equation, we have;
    Sin(A-B) = sinAcos(-B) + cosAsin(-B)
    Sin(A-B) = sinAcosB-cosAsinB
  • 9. tan(A+B) = sin(A+B)
    cos(A+B)
    = sinAcosB + cosAsinB
    cosAcosB – sinAsinB
    = tanA + tanB
    1 – tanAtanB
  • 10. Replace b with (-B), we have;
    tan(A-B) = tan A + tan (-B)
    1 – tan A tan(-B)
    tan(A-B) = tanA – tanB
    1 + tanAtanB
  • 11. EXAmples
  • 12. Simplify the equation sin 370cos 410 + cos 370 sin 410;
    sinAcosB + cosAsinB = sin(A+B)
    sin37cos 41 + cos37sin41 = sin(37 + 41)
    = sin 780
  • 13. Simplify 1 – tan150
    1 + tan150
    tan A – tan B = tan (A-B)
    1 + tanAtanB
    tan 45 – tan 15 = tan(45-15)
    1 + tan 45 tan 15
    1 – tan 15 = tan 300
    1 + tan 15

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