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Note 8: Trig Relationships
The algebra of trig is similar to the algebra of real numbers.
Example
3sinθ + 2sinθ = 5sinθ
( just like 3x + 2x = 5x )
For more complicated we often use sin2θ + cos2θ = 1
(From earlier lessons)
so
sin2θ = 1- cos2θ
and
cos2θ = 1 – sin2θ
Examples:
1.
3 – 3sin2θ
= 3( 1 - sin2θ)
= 3 cos2 θ
(because cos2θ = 1 – sin2θ)
Examples:
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2.
sin2θcosθ+ cos3θ
= cosθ(sin2θ+ cos2θ) (Factorising)
= cos θ
Exercises:
1. Expand and Simplify:
(sinθ- 2)2
= sin2θ – 4sinθ+ 4 (normal expanding of brackets)
2. Factorise:
sin2θ + 5sinθ+ 6
This is the same method as x2 + 5x + 6 = (x + 2)(x + 3)
So sin2θ + 5sinθ+ 6 = (sinθ + 2)(sinθ+ 3)
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Page 290 Exercise 13 I
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Note 9: Double Angle Formulae
sin2A = 2sinAcosA
cos2A = cos2A – sin2A
= 2cos2A – 1
= 1 – 2sin2A
Example
If sinA = 2/3 find the value of cos 2A
cos2A = 1 – 2sin2A
= 1 – 2 x (2/3)2
= 1/9
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Example
If θis acute and cos2θ = ½ find the values of cosθ and sinθ
1. cos2θ= 2cos2 θ – 1
½ = 2cos2 θ – 1
3/4 = cos2 θ ( -1 and then divide by 2)
cosθ = ±√3/2
cos θ = √3/2
(θis acute, cos is positive)
2. cos2θ= 1 – 2sin2 θ
½ = 1 – 2sin2 θ
1/2 = sin2 θ
sinθ = ±1/2
sinθ = 1/2
(θis acute, sin is positive)
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OR
sin2θ+ cos 2θ= 1
sin2θ + (√3/2)2 = 1
sin2θ + 3/4
=1
1/2 = sin2 θ
sinθ = ±1/2
sinθ = 1/2
(θis acute, sin is positive)
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Page 293 Exercise 13 J
Questions 1 - 5
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Use appropriate ‘double angle’ formula to simplify:
Examples
1.
10sinθcosθ
= 5 (sinθcosθ)
= 5sin2θ
2.
3 – 6cos2 4A
= 3 (1 – 2cos2 4A)
= -3(2cos2 4A – 1)
= -3cos 2(4A)
= -3cos8A
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Page 293 Exercise 13 J
Questions 6 - 8
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Note 10: Tangent Relationships
Remember
tanθ = sinθ
cosθ
Example: simplify
1. 4tanxcosx = 4 sinx x cosx
cosx
= 4sinx
2. 3tanx = 3sinx x 1
2sinx
cosx
2sinx
= 3
2cosx
1
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Page 300 Exercise 13 L.2
Questions 3 - 5