SUM OF SINE AND COSINE FUNCTIONS (AUXILLIARY · PDF fileSUM OF SINE AND COSINE FUNCTIONS...

© The School For Excellence 2016 Trial Exam Revision Lectures – Maths Extension 1 – Book 1 1

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 : √3 = cos α The coefficients of cos θ: 1 = sin Step 3: Solve the two equations simultaneously to obtain values for R and α: To eliminate α, square both equations and add: 3 + 1 = ( + ) 4 = Hence: = 2 (ignore the negative root as we are really calculating a length, the amplitude of the wave.)

© The School For Excellence 2016 Trial Exam Revision Lectures – Maths Extension 1 – Book 1 Page 12

To eliminate divide the sin equation by the cos α equation: 1√3 = =

α = 30° (= π6 ) Hence: √3 sin θ − cos θ = 2 sin( − 30°) GENERALISED RESULT Generalising the result for: sin θ + cos θ Is a little difficult since the equation for depends which of the four forms is being used. (13) = + (14) tan = or(− ) or or (− ) QUESTION 7 Express each of the following in the form sin( + ): (a) sin + cos (b) 3sin + cos


QUESTION 8 Express each of the following in the form cos( − ): (a) sin + cos (b) 7sin + 24 cos


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 ∴ sin( + 45°) = 1√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


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 = 1 + sin , where 0 < < 2 .


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 = has solutions =. . . −420°, −300°, −60°, 60°, 300°, 420° … 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 radians. We then

move to the negative x-axis and add radians and we continue this process for every full

rotation. Hence we can express the general solution in words as adding radians to any multiple (odd or even) of .

Mathematically, this general solution is expressed as: = + 4 where is an integer. Note: If is negative that corresponds to a rotation clockwise from the positive x-axis.

i.e. = (−1) corresponds to the solution = − in the third quadrant.


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 − Often these are then combined as: = 2 ∓

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.


EXAMPLE 8 Find the general solution for: sin = 12

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 + and = (2 − 1) − 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 otherwise you lose one set of solutions! ∴ sin = 0 cos = 1√2

The first solution set is saying: we start are either on the positive x-axis or on the negative x-axis. Hence any multiple of (180o). = The second solution set has a related angle of 4 and is in the first and fourth quadrants. Hence even multiples of add or subtract 4. = 2 ∓ 4


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


QUESTION 13 Find in radians the general solution for the equation 3 cos x – sin x = 1. Solution QUESTION 14 Solve cos 2 cosα 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) , is an integer

B = ( ) , is an integer

C = 2 ± , is an integer

D = ± , is an integer


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, = or better as 4 = where is the FOCAL LENGTH. 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 )

x-2 -1 1 2

y

-2

-1

1

2

Angle of turn

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) 2x t= and 2y t= −

(b) 5sinx α= and 2cosy α= (Hint: Use Pythagorean trigonometric identity) (c) 6x t= and 23y t=


PARAMETRISATION OF PARABOLA For a parabola, we use the parameter, usually , which describes how the slope of the curve changes with changing x. Hence if x2 = 4ay

y = x2

4adydx

= x2a

which we let equal p

∴ p = x2a

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 is the focal length so positive, the co-ordinate is _________________ While the co-ordinate can be ____________________________ The vertex occurs when = ________ On the right hand branch of the parabola is ______________ On the left hand branch of the parabola is ______________

Note: The above diagram uses parameter t instead of p.


QUESTION 17 Express each of the following parabolas in standard parametric equations: Hint: Find the focal length first. (a) 2 12x y=

(b) 2y x=

(c) 2

4yx =

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!!!!

dydx

= x2a

which, at x = x1 is x1

2a

∴ y – y1 = x1

2a (x – x1)

y = y1 + x1 x2a

– (x1)2

2a

But (x1)2 = 4ay1

y = y1 + x1 x2a

– 2y1

∴ 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.


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 dpdx dp dx

= × rewritten by replacing multiplication with

division by a reciprocal. So differentiate the parametric equations separately and then divide them to get the derivative. EXAMPLE 10

Find dydx

for 4x t= and 23y t= .

Solution = 4 and = 6 = = 6 4 = 32


QUESTION 19

Find dydx

for:

(a) 2 3x t= + and 2 2y t t= + (b) 8 5x t= − and 2 4 3y t t= − + QUESTION 20 Find the gradient of the normal to the curve 2x t= , 22y t= at 1t = . Solution

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