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# Trigono

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### Transcript

• 1. Proving trigonometrical identitiesusing sum and difference of two angles
Vania Lundina
• 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