Higher Maths
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Revision Notes
The Auxiliary Angle
The Wave Equation
find corresponding values of θ
solve equations
of the form
a cos θ + b sin θ = c
Express
a cos θ + b sin θ
in the form
r cos(θ ± α)
or r sin(θ ± α)
find maximum and minimum
values of expressions of the form
a cos θ + b sin θ
and the corresponding values of θ
To express a cos θ + b sin θ in the form r cos(θ ± α)
Expand the required form
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
(1)the coefficients of cos θ must be equal: a = r cos α
(2)the coefficients of sin θ must be equal: b = ∓ r sin α
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
tan a =
∓b
a
Þ a = tan-1 ( ∓ ba ) + p n
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)
Expand the required form
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
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
Divide one equation by the other to get
tan a = -43 Þ a = tan -1 ( - 43 ) + p n
Þ a = ..., - 0·927, 2·214, 5·356, 8·497, ...
Use the equations to decide in which quadrant α lies.
cos a = 53 Þ a is in 1st or 4th quadrant
sin a = - 45 Þ a is in 3rd or 4th quadrant
So α is in 4th quadrant. Thus α = 5·356 radians
Giving: 3 cos θ + 4 sin = 5cos(θ + 5·356)
Test
Yourself?
6
5
4
3
2
1
0
0
0.524
1.048
1.572
2.096
2.62
3.144
3.668
4.192
4.716
5.24
5.764
-1
-2
-3
-4
-5
-6
y=3cos x
y = 4sin x
y=3cos x + 4 sin x
back
Two swings are set in motion. Their heights
above an arbitrary line can be modelled by:
Swing 1:
h1 = 5 cos x
Swing 2:
h2 = 12 sin x
15
10
5
reveal
Swing 1
0
0
1
2
3
4
5
6
7
Swing 2
-5
-10
-15
Express the difference in their heights in the form r sin (x – a)
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 = 169
so 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 form
a 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) + a
or 2π + cos–1(c/r) + a
etc
Test
Yourself?
When p cos x + q sin x is
expressed as r cos(x + a)
When p cos x + q sin x is
expressed as r sin(x + a)
4
4
3
3
2
2
1
1
0
-1
0
0
2
4
6
8
-1
-2
-2
-3
-3
-4
-4
Max: p cos x + q sin x = r when x + a = 0
Min: p cos x + q sin x = – r when x + a = π
0
2
4
6
8
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?
2
1.5
1
0.5
0
0
1.57
3.14
4.71
6.28
-0.5
-1
-1.5
-2
y=sinx
y=cosx
y= sinx + cosx
For what values of x, 0 ≤ x ≤ 2π,
is sin x + cos x = 0·5
reveal
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 = 0·5
so √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…), etc
x + 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 by
h = 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.
4
3
2
reveal
1
0
0
-1
-2
1
2
3
4
5
6
7
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.
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