Higher Maths Revision Notes The Auxiliary Angle Get Started.

Higher Maths

Revision Notes The Auxiliary Angle

Get Started

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.