Proving trigonometrical identitiesusing sum and difference of two anglesVania LundinaGrade 10
Basic knowledge
Say we have this diagram:yIn triangle OPQ, angle POQ = A - BP(cosA, sinA)Q(cosB, sinB)ABx1OP and Q are two points on the circle of radius 1 unitCoordinates of P = (cosA, sinA)Coordinates of Q = (cosB, sinB)
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 OPQPQ2 = 12 + 12 – 2(1)(1)cos(A-B)	 	= 2 – 2cos(A-B)
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
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
Since sin ϑ = COS (900 – ϑ)sin(A+B) = cos[900 – (A+B)]		      = cos[(900 – a) – B]		      = cos(900 – A) cosB + sin(900 – A)sinBSo, sin(A+B) = sinAcosA + cosAsinB
Replace B with (-B) from the previous equation, we have;Sin(A-B) = sinAcos(-B) + cosAsin(-B)Sin(A-B) = sinAcosB-cosAsinB
tan(A+B) = sin(A+B)cos(A+B)		       = sinAcosB + cosAsinBcosAcosB – sinAsinB		       = tanA + tanB			 1 – tanAtanB
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
EXAmples
Simplify the equation sin 370cos 410 + cos 370 sin 410;sinAcosB + cosAsinB = sin(A+B)sin37cos 41 + cos37sin41 = sin(37 + 41)					     = sin 780
Simplify 1 – tan150     1 + tan150tan A – tan B = tan (A-B)1 + tanAtanBtan 45 – tan 15    = tan(45-15) 1 + tan 45 tan 151 – tan 15 = tan 3001 + tan 15

Trigono

  • 1.
    Proving trigonometrical identitiesusingsum and difference of two anglesVania LundinaGrade 10
  • 2.
  • 3.
    Say we havethis diagram:yIn triangle OPQ, angle POQ = A - BP(cosA, sinA)Q(cosB, sinB)ABx1OP and Q are two points on the circle of radius 1 unitCoordinates of P = (cosA, sinA)Coordinates of Q = (cosB, sinB)
  • 4.
    Now we canform 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 OPQPQ2 = 12 + 12 – 2(1)(1)cos(A-B) = 2 – 2cos(A-B)
  • 5.
    From the twoequations, 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)sinBSo, 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 + cosAsinBcosAcosB – 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.
  • 12.
    Simplify the equationsin 370cos 410 + cos 370 sin 410;sinAcosB + cosAsinB = sin(A+B)sin37cos 41 + cos37sin41 = sin(37 + 41) = sin 780
  • 13.
    Simplify 1 –tan150 1 + tan150tan A – tan B = tan (A-B)1 + tanAtanBtan 45 – tan 15 = tan(45-15) 1 + tan 45 tan 151 – tan 15 = tan 3001 + tan 15