SUM OF SINE AND COSINE FUNCTIONS
(AUXILLIARY ANGLES)
An important technique students are required to master is to replace the sum, or difference,
of sine and cosine functions, that share the same period, with a single sinusoidal function.
This enables equations involving multiple sine and cosine functions to be solved and is
particularly important when solving many ______________________ questions.
sin θ +
cos θ = sin(θ − α)
It is important to become fluent in this technique known as auxilliary equations.
In physics this is the concept of SUPERPOSITION of waves.
The technique requires the compound angle formulae introduced earlier and results in a pair
of simultaneous equations which need to be solved to find the two unknowns R and α.
Generally the required form will be given in the question. If not, experience will tell you if it is
better to use a sine form or a cosine form and whether to use the compound angle as a sum
or difference (so you need to do lots of practice)!!!!
EXAMPLE 5
Express √3 sin θ − cos θ in the form
sin( − ).
Solution
Step 1: Expand the given form and equate it to the original expression.
√3 sin θ − cos θ =
sin θ cos α − cos θ sinα
Step 2: Equate the coefficients of the corresponding terms:
The coefficients of sin θ :
The coefficients of cos θ:
√3 = cos α
1 = sin
Step 3: Solve the two equations simultaneously to obtain values for R and α:
To eliminate α, square both equations and add:
Hence:
3+1= (
4=
=2
+
)
(ignore the negative root as we are really calculating a length, the amplitude of the wave.)
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To eliminate
divide the sin
equation by the cos α equation:
1
√3
=
=
α = 30° (=
π
)
6
Hence: √3 sin θ − cos θ = 2 sin( − 30°)
GENERALISED RESULT
Generalising the result for:
sin θ +
Is a little difficult since the equation for
used.
depends which of the four forms is being
(13)
(14)
tan
=
=
QUESTION 7
Express each of the following in the form
(a)
sin + cos
(b)
3sin + cos
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cos θ
+
or(− ) or
or (− )
sin( + ):
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QUESTION 8
Express each of the following in the form
(a)
sin + cos
(b)
7sin + 24 cos
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cos( − ):
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The auxiliary angle technique can be used to solve equations.
EXAMPLE 6
Solve: sin + cos
= 1 for 0 ≤
≤ 360°
From a previous question we know: sin
+ cos
= √2 sin( + 45°)
∴ √2 sin( + 45°) = 1
1
∴ sin( + 45°) =
√2
We need to change the domain to solve this equation so:
45° ≤ ( + 45°) ≤ 405°
∴
+ 45° = 45°, 135° or 405°
∴
= 0°, 90°
360°
Notice these final three solutions lie within the original domain.
QUESTION 9
Solve the equation 2sin
+ 3cos
= √13 for 0 ≤
≤ 360°
Solution
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QUESTION 10
Solve the equation 3sin
+ 4cos
= 2 for −180 ≤
≤ 180°.
Solution
QUESTION 11 (HSC 2013 Question 12a)
(a)
Write √3 cos x − sin x in the form 2 cos(x + α), where 0 < α < .
(b)
Hence, or otherwise, solve √3 cos
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= 1 + sin , where 0 <
<2 .
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GENERAL SOLUTIONS
Since trigonometric functions are periodic, unless we restrict the domain then most
trigonometric equations will have an infinite set of solutions.
e.g.
cos
=
=. . . −420°, −300°, −60°, 60°, 300°, 420° …
has solutions
We can express these sets of solutions using a general solution, rather than using a
restricted domain.
Yes there are (in some texts) formulae to remember. I would suggest it is easier to develop a
feeling for what the solution is like and write down that solution rather than simply learning
formulae.
The key idea is to express angles in different quadrant as a movement from the x-axis, i.e.
second quadrant is subtracting an acute angle from the negative x-axis
The positive x-axis can be considered as 0°, 360°, 720° … that is _________ multiples of 180°
or
radians. We can express this as 2 ×180° where
is an integer or 2
. Similarly the
negative x-axis with values … − 180°, 180°, 540°, … is ______multiples of , which can be
expressed as (2 − 1)× .
TANGENT EQUATIONS
EXAMPLE 7
Take a simple example: tan
=1
This equation has two solutions for 0o ≤ x ≤ 360o (now to be written as 0 ≤ x ≤ 2π).
They are 45° or 225° (
4
and
).
To obtain these two values, we begin at the positive x-axis and add
move to the negative x-axis and add
radians. We then
radians and we continue this process for every full
rotation. Hence we can express the general solution in words as adding
multiple (odd or even) of
.
Mathematically, this general solution is expressed as:
Note: If
i.e.
radians to any
=
+4
where
is an integer.
is negative that corresponds to a rotation clockwise from the positive x-axis.
= (−1) corresponds to the solution
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=−
in the third quadrant.
Trial Exam Revision Lectures – Maths Extension 1 – Book 1
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This generalises as a result for tan
=
=
where
+
is the related angle.
The general solutions for cosine and sine and a little more involved since the solutions
belonging to each quadrant are expressed as a separate general set.
COSINE FUNCTIONS
For the first quadrant angles are added from the positive x-axis. Hence even multiples of pi
add the related angle .
=2
+
While the fourth quadrant solutions are ______________from the ______________x-axis.
That is even multiples of pi minus the related angle.
=2
−
=2
∓
Often these are then combined as:
SINE FUNCTIONS
For the first quadrant angles are added from the positive x-axis as before, however the other
quadrant where sine is positive is the second and these solutions are ______________from
the ______________x-axis. That is odd multiples of pi minus the related angle.
=2 +
or
= (2 − 1) −
Note: These are best left separate although sometimes quoted combined with a factor
of (−1) which alternates in sign dependent upon whether is odd or even.
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EXAMPLE 8
Find the general solution for:
sin
1
2
=
Previously using degrees you would have written:
=. . . , 30° , 390° (360° + 30°), 750° (720° + 30°), … in first quadrant and
=. . . , 150°(180° − 30°) , 510° (540° − 30°), … in second quadrant.
Now the general solution is given as :
=2
+
= (2 − 1) −
and
In words this is:
___________ multiples of
________
and ___________ multiples of
EXAMPLE 9
Find the general solution to the equation √2 sin cos
________
.
= sin
Solution
√2 sin cos − sin = 0
sin (√2 cos − 1) = 0
Note: Do NOT divide throughout by sin
∴ sin
otherwise you lose one set of solutions!
=0
cos
=
1
√2
The first solution set is saying: we start are either on the positive x-axis or on the negative xaxis. Hence any multiple of (180o).
=
The second solution set has a related angle of
Hence even multiples of
add or subtract
4
.
=2
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4
and is in the first and fourth quadrants.
∓
4
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QUESTION 12
Find the general solution for the following equations:
(a)
2 cos
= √3.
(b)
√3 tan x + 1 = 0
(c)
2 cos
+ sin − 1 = 0
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QUESTION 13
Find in radians the general solution for the equation 3 cos x – sin x = 1.
Solution
QUESTION 14
Solve cos 2α = cos α giving the general solution(s) in radians.
Solution
QUESTION 15 (HSC 2013 Multiple Choice Question 6)
Let | | ≤ 1. What is the general solution of sin 2 = a ?
A
=
+ (−1)
B
=
(
C
=2
D
=
)
±
±
, is an integer
, is an integer
, is an integer
, is an integer
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GEOMETRY OF THE PARABOLA
PARAMETRIC REPRESENTATION
Cartesian equations usually express
as a function of , e.g. ( ).
You are familiar with the CARTESIAN equation of a parabola,
where is the FOCAL LENGTH.
=
or better as 4
=
Parametric equations introduce a new variable called the parameter and each co-ordinate is
written as a function of this parameter giving two equations = ( ) and = ( ) which is
supposed to make the curve easier to describe.
For example, points on a circle are often described by using x and y co-ordinates which
satisfy the equation of that particular circle (e.g. x2 + y2 = 4).
A parameter for a circle could however be the angle θ through which a radius turns, hence
we need only one value to describe a given circle not two values
(e.g. = 2 cos and = 2 sin )
y
2
1
Angle of turn
-2
-1
1
2
x
-1
-2
We can easily convert a pair of parametric equation back to the Cartesian equivalent by
solving the two equations simultaneously to ELIMINATE THE PARAMETER.
QUESTION 16
Express each of the following as Cartesian equations by eliminating the parameter:
(a)
x = 2t and y = 2 − t
(b)
x = 5sin α and y = 2 cos α
(c)
x = 6t and y = 3t 2
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(Hint: Use Pythagorean trigonometric identity)
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PARAMETRISATION OF PARABOLA
For a parabola, we use the parameter, usually
curve changes with changing x.
, which describes how the slope of the
2
Hence if
x = 4ay
2
y = x
4a
dy = x
2a
dx
which we let equal p
∴ p = x
2a
So p (i.e. the parameter) is the gradient.
Hence by multiplying, we get x = 2ap and then, by substitution, y = ap2.
We use these co-ordinates (2ap, ap2) on the parabola x2 = 4ay just as we would use normal
(Cartesian) co-ordinates.
ALWAYS REMEMBER:
The parameter p tells us the GRADIENT of the tangent to the parabola at that point.
The parametric equations of a concave up parabola are:
=2
=
Note: Since
While the
is the focal length so positive, the
co-ordinate is _________________
co-ordinate can be ____________________________
The vertex occurs when
= ________
On the right hand branch of the parabola
On the left hand branch of the parabola
is ______________
is ______________
Note: The above diagram uses parameter t instead of p.
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QUESTION 17
Express each of the following parabolas in standard parametric equations:
Hint: Find the focal length first.
(a)
x 2 = 12 y
(b)
y = x2
(c)
x2 =
y
4
QUESTION 18 (Just for fun)
Show the Cartesian equation of the tangent to x2 = 4ay at the point P (x1, y1) is
xx1 = 2a (y + y1).
Solution
HINT – but no peeping until you have finished!!!!
dy = x which, at x = x is x1
1
2a
2a
dx
∴
y – y1 = x1 (x – x1)
2a
y = y1 + x1 x – (x1)
2a
2a
2
2
But (x1) = 4ay1
y = y1 + x1 x – 2y1
2a
∴
x x1 = 2a (y + y1)
The tricky step to remember is the fourth line where since the point lies on the parabola you
can substitute it into the equation of the parabola and hence simplify the ( ) term.
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DERIVATIVE OF PARAMETRIC PARABOLA
We can also determine the derivative directly from the two parametric equations without
forming the Cartesian equation. This strategy requires the special result:
=
Note: this is really just the chain rule
dy dy dp
=
×
rewritten by replacing multiplication with
dx dp dx
division by a reciprocal.
So differentiate the parametric equations separately and then divide them to get the
derivative.
EXAMPLE 10
Find
dy
for x = 4t and y = 3t 2 .
dx
Solution
= 4 and
=6
=
=6
=
4
3
2
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QUESTION 19
Find
dy
for:
dx
(a)
x = 2t + 3 and y = t 2 + 2t
(b)
x = 8t − 5 and y = t 2 − 4t + 3
QUESTION 20
Find the gradient of the normal to the curve x = 2t , y = 2t 2 at t = 1 .
Solution
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