Higher Maths
Revision Notes The Auxiliary Angle
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The Wave Equation
find corresponding values of θ
Express a cos θ + b sin θ
in the form r cos(θ ± α)
or r sin(θ ± α)solve equations
of the forma cos θ + b sin θ = c
find maximum and minimum values of expressions of the form
a cos θ + b sin θand the corresponding values of θ
r cos(θ ± α) = r cos θ cos α r sinθ sin α
If a cos θ + b sin θ = r cos θ cos α r sinθ sin α for all values of θ, then
•the coefficients of cos θ must be equal: a = r cos α •the coefficients of sin θ must be equal: b = r sin α
To express a cos θ + b sin θ in the form r cos(θ ± α)
Expand the required form
Example
Square both equations and add them:
a2 + b2 = r2 cos2 α + r2 sin2 α =r2 (cos2 α + sin2 α) = r2
So r = √(a2 + b2 )
Example
Divide one equation by the other to get
Use the equations to decide in which quadrant α lies.
Example
Express 3 cos θ + 4 sin θ in the form r cos(θ + α)
r cos(θ + α) = r cos θ cos α – r sinθ sin α
Equate coefficients:(1) 3 = r cos α (2) –4 = r sin α
Expand the required form
Next see graph
Square both equations and add them:
32 + (–4)2 = r2 cos2 α + r2 sin2 α =r2 (cos2 α + sin2 α) = r2
So r = √(32 + (–4)2)= √25
= 5
Next
Test Yourself?
Divide one equation by the other to get
Use the equations to decide in which quadrant α lies.
So α is in 4th quadrant. Thus α = 5·356 radians
Giving: 3 cos θ + 4 sin = 5cos(θ + 5·356)
back
Two swings are set in motion. Their heights above an arbitrary line can be modelled by:Swing 1: h1 = 5 cos xSwing 2: h2 = 12 sin x
Express the difference in their heights in the form r sin (x – a)
revealreveal
h1 – h2 = 5 cos x – 12 sin x
r sin (x – a) = r sin x cos a – r cos x sin a
Equating coefficients: r cos a = –12; –r sin a = 5
Square and add: r2 (cos2a + sin2a) = (–12)2 + (–5)2 = 169so r = 13
Divide: tan a = (–5) ÷ (–12) = 0·41666… , So a = 0·395, 3·536, 6·678, …
Identify quadrant: sin a = –5/13 … 3rd or 4th quadrant.cosa = –12/13 … 2nd or 3rd quadrant.
Choose the 3rd quadrant angle: a = 3·536
h1 – h2 = 13 sin (x – 3·536)
To solve equations of the forma cos θ + b sin θ = c
• Express the LHS in desired form e.g. r cos (x – a)• r cos (x – a) = c:
solve for x – a: x – a = cos–1(c/r)
or 2π – cos–1(c/r)or 2π + cos–1(c/r)etc
• Solve for x:x = cos–1(c/r) + a
or 2π – cos–1(c/r) + aor 2π + cos–1(c/r) + aetc Test
Yourself?
When p cos x + q sin x is expressed as r cos(x + a)
Max: p cos x + q sin x = r when x + a = 0Min: p cos x + q sin x = – r when x + a = π
When p cos x + q sin x is expressed as r sin(x + a)
Max: p cos x + q sin x = r when x + a = π/2
Min: p cos x + q sin x = – r when x + a = 3π/2
Note that the graph is unchanged
Test Yourself?
For what values of x, 0 ≤ x ≤ 2π,is sin x + cos x = 0·5
revealreveal
sin x + cos x = r sin(x + a) = r sin x cos a + r cosx sin a
Equating coefficients: r cos a = 1 and r sin a = 1
This gives: r = √2 and a = π/4
sin x + cos x = 0·5so √2 sin(x + π/4) = 0·5
x + π/4 = sin–1( 0·353553…), or π – sin–1( 0·353553…),
or 2π + sin–1( 0·353553…), etcx + π/4 = 0·361, 2·78, 6·64,…x = –0·424, 1·995, 5·859,…
In the required domain, x = 1·995, 5·859
As the paddle turns, the height of a point on it can be modelled byh = sin x + 2 cos x + 1 where h units is the height above the water.
Find the minimum height and the value of x at which it occurs.
revealreveal
The minimum of sin (x + 1·107) = –1 when (x + 1·107) = 3π/2
So the minimum height is 1 – √5 = –1·24 units when x = 3·605
First express sin x + 2 cos x in the form r sin(x + a)sin x + 2 cos x = r sin(x + a) = r sin x cos a + r cosx sin a
Equating coefficients: r cos a = 1 and r sin a = 2
This gives: r = √5 and a = 1·107 So h = √5 sin(x + 1·107) + 1.