© D.J.Dunn www.freestudy.co.uk 1

EDEXCEL NATIONAL CERTIFICATE

UNIT 28 – FURTHER MATHEMATICS FOR TECHNICIANS

OUTCOME 3

TUTORIAL 2 - TRIGONOMETRICAL EQUATIONS

CONTENTS

3 Be able to understand how to manipulate trigonometric expressions and apply trigonometric

techniques

Trigonometrical graphs: amplitude, period and frequency, graph sketching e.g. sin x, 2 sin x,

½ sin x, sin 2x, sin ½ x for values of x between 0 and 360°; phase angle, phase difference;

combination of two waves of the same frequency

Trigonometrical formulae and equations: the compound angle formulae for the addition of sine

and cosine functions e.g. sin (A ± B); expansion of R sin (ωt + α) in the form a cos ωt + b sin ωt

and vice versa

It is assumed that the student has completed the module MATHEMATICS FOR

TECHNICIANS.

© D.J.Dunn www.freestudy.co.uk 2

1. REVIEW OF TRIGONOMETRIC RATIOS

This is the work covered so far in previous tutorials.

SINE sin(A) = b/c cosec(A) = c/b

COSINE cos(A) = a/c sec(θ) = c/a

TANGENT tan(A) = b/a cot (A) = a/b

This is also useful to know. tan (A) = sin(A) / cos(A)

sin2(A) = b

2/c

2 and cos

2(A) = a

2/c

2 and sin

2(A) + cos

2(A) = b

2/c

2 + a

2/c

2

2

2222

c

ab(A)cos(A)sin

c

2 = a

2 + b

2 (Pythagoras)

1ba

ab(A)cos(A)sin

22

2222

SINE RULE sinC

c

sinB

b

sinA

a

COSINE RULE

2bc

acbcos(A)

222

2ca

baccos(B)

222

2ab

cbacos(C)

222

2. COMPOUND ANGLES

SUMS and DIFFERENCES

Prove that cos(A + B) = cos(A) cos(B) – sin(A) sin(B)

Refer to the diagram opposite.

OP cos(A+B) = OR cos(A) + RP cos(A)

substitute cos(A) = - sin(A)

OP cos(A+B) = OR cos(A) - RP sin(A)

OR = OP cos(B) and RP = OP sin(B)

OP cos(A+B) = OP cos(B)cos(A) - OP sin(A) sin(B)

cos(A+B) = cos(A) cos(B) - sin(A) sin(B)

Prove that sin(A + B) = sin(A) cos(B) + cos(A) sin(B)

Refer to the diagram opposite.

OP sin(A+B) = OR sin(A) – RP sin(A)

-sin(A) = cos(A)

OP sin(A+B) = OR sin(A) + RP sin(A)

OR = OP cos(B) and RP = OP sin(B)

OP sin(A+B) = OP cos(B) sin(A) + OP sin(B) cos(A)

sin(A+B) = sin(A) cos(B) + sin(B) cos(A)

By a similar manner to before it can be shown that:

cos(A - B) = cos(A) cos(B) + sin(A) sin(B) and

sin(A - B) = sin(A) cos(B) - cos(A) sin(B)

© D.J.Dunn www.freestudy.co.uk 3

DOUBLE ANGLES

cos(A+B) = cos(A) cos(B) - sin(A) sin(B) put B = A:

cos(2A) = cos(A)cos(A) – sin(A) sin(A)

cos(2A) = cos2(A) – sin

2(A) and since sin

2(A) + cos

2(A) = 1

cos(2A) = 1 – 2sin2(A) = 2cos

2(A) – 1

2

cos(2A)1(A)sin2

and 2

1-cos(2A)(A)cos2

Similarly sin(A+B) = sin(A) cos(B) + sin(B) cos(A)

sin(2A) = 2sin(A) cos(A)

or sin(A) cos(A) = ½ sin(2A)

HALF ANGLES

In the identity cos(2A) = cos2(A) – sin

2(A) if we substitute A = C/2 for A we get

cos(C) = cos2(C/2) – sin

2(C/2)

cos(C) = cos2(C/2) – {1 - cos

2(C/2)}

cos(C) = 2cos2(C/2) – 1

And sin(C) = 2sin(C/2) cos(C/2)

3. PRODUCTS AND SUMS

CHANGING PRODUCTS TO SUMS

We have already shown that: sin(A + B) = sin(A) cos(B) + cos(A) sin(B)

sin(A - B) = sin(A) cos(B) - cos(A) sin(B)

If we add the two lines we get : sin(A + B) + sin(A - B) = 2sin(A) cos(B)

Rearrange, sin (A) cos (B) = ½ {sin (A + B) + sin (A − B)}

If A = B then as before sin (A) cos (A) = ½ {sin (2A) }

CHANGING SUMS TO PRODUCTS

Using sin (A) cos (B) = ½ {sin (A+B) + sin (A−B)}

Change sides ½ {sin (A+B) + sin (A−B)} = sin (A)cos (B)

Let C = A + B and D = A - B ½ {sin (C) + sin (D)} = sin (A) cos (B)

A = C – B and A = D + B add them together and 2A = C + D and subtracting 2B = C – D

½ {sin (C) + sin (D)} = sin {½ (C+D)} cos {½ (C - D)}

{sin (C) + sin (D)} = 2 sin {½ (C+D)} cos {½ (C - D)}

Since C and D are only symbols for angles it must be true that:

sin A + sin B = 2 sin {½ (A + B)} cos {½ (A − B)}

Likewise we can show: sin A − sin B = 2 sin {½ (A − B)} cos {½ (A + B) }

cos A + cos B = 2 cos {½ (A + B)} cos {½ (A − B)}

cos A − cos B = −2 sin {½ (A + B)} sin {½ (A − B)}

© D.J.Dunn www.freestudy.co.uk 4

4. TO SHOW THAT:- R sin(ωt + a) = a cos(ωt) + b sin (ωt)

From the right angle triangle shown we have a/c = sin(A) and b/c = cos(A)

Start with the expression c sin(A + θ)

From the trig identity we can substitute

sin(A + θ) = sin(A) cos(θ) + cos(A) sin(θ)

c sin(A + θ) = c sin(A) cos(θ) + c cos(A) sin(θ)

Substitute a/c = sin(A) and b/c = cos(A)

c sin(A + θ) = c (a/c) cos(θ) + c (b/c) sin(θ)

c sin(A + θ) = a cos(θ) + b sin(θ)

When applied to a phasor or vector where θ = ωt this becomes

c sin(A + ωt) = a cos(ωt) + b sin(ωt)

If the phasor is the hypotenuse we may change c to radius R and

if we change A to α then R sin(α + ωt) = a cos( ωt) + b sin(ωt)

The diagram illustrates the vector.

WORKED EXAMPLE No.1

Two alternating voltages are expressed as v1 = 100 sin(300t) and v2 = 150 cos(300t)

When the voltage is summed the result is v = V sin(4t + ). Determine the value of V and

SOLUTION

R sin(ωt + α) = a cos(ωt) + b sin (ωt)

Change the symbols V sin(ωt + ) = a cos(ωt) + b sin (ωt) It is apparent that ω = 300, a = 150 and b = 100

V sin(300t + ) = 150 cos(300t) + 100 sin (300t)

α = tan-1

(a/b) = tan-1

(150/100) = 56.3o

180.3100150V 22

© D.J.Dunn www.freestudy.co.uk 5

5. CHANGING COS TO SIN AND SIN TO COS

For the same right angle triangle as shown above we know that:

sin(A) = a/c cos(A) = b/c

Consider the expression b cos (θ) + a sin(θ) (noting θ is not A)

This is the same as the expression θsinc

acθcos

c

bc

b cos (θ) + a sin(θ) =

θsinc

aθcos

c

bc

b cos (θ) + a sin(θ) = c{cos(A) cos(θ) + sin(A) sin(θ)}

Now use the identity cos(A - B) = cos(A) cos(B) + sin(A) sin(B)

Only in this case it is cos(A - θ) = cos(A) cos(θ) + sin(A) sin(θ)

b cos (θ) + a sin(θ) = c{cos(A – θ)}

If a = b = 1 then c = √2 A = 45o cos (θ) + sin(θ) = √2 {cos(45

o – θ)}

If we started with the expression a cos (θ) - b sin(θ)

a cos (θ) - b sin(θ) =

θsinc

bθcos

c

ac

a cos (θ) - b sin(θ) = c{sin(A) cos(θ) - cos(A) sin(θ)}

Now use the identity sin(A - θ) = sin(A) cos(θ) - cos(A) sin(θ)

a cos (θ) - b sin(θ) = c{sin(A - θ)}

If a = b = 1 then c = √2 A = 45o cos (θ) - sin(θ) = √2{sin(45

o - θ)}

WORKED EXAMPLE No. 2

Change 3 cos (θ) – 4 sin (θ) into sine form and then cosine form.

SOLUTION

a cos (θ) - b sin(θ) = c{sin(A – θ)}

Let a = 3 and b = 4 hence c = √25 = 5

3 cos (θ) - 4 sin(θ) = 5{sin(A - θ}

cos(A) = b/c = 4/5 sin(A) = a/c = 3/5

A = cos-1

(4/5) = 36.87 o or sin

-1(3/5) = 36.87

o

θ)-87.36sin(5θ) 4sin(θ) 3cos( o

Check put θ = 45o 707.0θ) 4sin(θ) 3cos( and -0.707)45-87.36sin(5 oo

To put the expression into cosine form we only need to know that sin(A) = cos(A - 90o)

Hence sin(36.86o – θ) = cos(36.87

o - 90

o – θ) = cos (- 53.13

o- θ)

3cos(θ) – 4sin(θ) = 5 cos(-53.13 o - θ)

Check put θ = 45o 5 cos(-53.13

o – 45

o) = 5 cos(-98.13

o) = -0.707

© D.J.Dunn www.freestudy.co.uk 6

WORKED EXAMPLE No. 3

Show that

θtan2θcos1

2θsin

SOLUTION

Substitute sin(2θ) = 2sin(θ) cos(θ)

2θcos1

θcosθ2sin

2θcos1

2θsin

Substitute cos(2θ) = 2cos2(θ) – 1

θcos2

θcosθ2sin

1θcos21

θcosθ2sin

2θcos1

2θsin22

Simplify and

θtanθcos

θsin

2θcos1

2θsin

WORKED EXAMPLE No. 4

Given 2sin2(θ) + 3cos(θ) = 1/2 Solve θ

SOLUTION

Substitute sin2(θ) = 1 - cos

2(θ)

2{1 - cos2(θ)} + 3cos(θ) = 7/2 2 - 2cos

2(θ)} + 3cos(θ) = 7/2

- 2cos2(θ)} + 3cos(θ) – 3/2 = 0 2cos

2(θ) - 3cos(θ) - 3/2 = 0

4

213

4

1293

)2)(2(

)2/3)(2)(4(33θ) cos(

2

cos(θ) = 1.895 or -0.396

Since the cosine value can not exceed 1 the only answer must be cos(θ) = -0.396

θ = 113.3o

WORKED EXAMPLE No. 5

A generator produces a voltage v = 200 sin(2πft) and a current i = 5 sin(2πft) into a purely

resistive load. The electric power is P = vi. Express this as a single term and show that the

power varies from zero to 1000 Watts at double the frequency.

SOLUTION

P = {200 sin(2πft)} {5 sin(2πft)} = 1000 sin2(2πft)

Use the double angle formula 2

cos(2A)1(A)sin2

tf π4cos5000052

tf π22cos11000P

The frequency is doubled and P varies 0 to 1000

(see the plot)

© D.J.Dunn www.freestudy.co.uk 7

Trigonometric function

sin(-A) = - sin(A)

cos(-A) = cos(A)

sin2θ + cos

2θ = 1

sin(2A) = 2sin(A) cos(A)

cos(2A) = cos2(A)- sin

2(A)

cos(A) = 2cos2(A/2) – 1

sin(A) = 2sin(A/2) cos(A/2)

sin (A) cos (B) = ½ {sin (A+B) + sin (A−B)}

cos (A) sin (B) = ½ {sin (A+B) - sin (A−B)}

cos (A) cos (B) = ½ {cos (A+B) + cos (A−B)}

sin (A) sin (B) = -½ {cos (A+B) - cos (A−B)}

cos (θ) - sin(θ) = √2{sin(45o - θ)}

cos (θ) + sin(θ) = √2 {cos(45o – θ)}

sin(A ± B) = sin(A) cos(B) ± cos(A) sin(B)

cos(A ± B) = cos(A) cos(B) sin(A) sin(B)

cos(x)dx

dy then sin(x)y

sin(x)dx

dy then cos(x) y If

Ccosh(x)- sin(x)dx

Csin(x)cos(x)dx

b cos (θ) + a sin(θ) = c{cos(A – θ)}

a cos (θ) - b sin(θ) = c{sin(A - θ)}

© D.J.Dunn www.freestudy.co.uk 8

SELF ASSESSMENT EXERCISE No.1

1. Given sin(3x) = 0.5 solve the two smallest positive values of x.

(Answer 10o or 50

o)

2. The velocity at which a vehicle overturns on an inclined bend is given by μtanθ1

tanθμRgv2

Make θ the subject of the formula

Determine the angle if the vehicle overturns at 13 m/s when R = 30 m and μ = 0.2

(Answer 18.6o)

3. Given cos2(θ) + 2sin

2(θ) = 1.5 Solve the smallest positive value of θ

(Answer 45o)

4. Prove the following by using standard trigonometric identities.

i.

θtanθ) sin(2

2θcos1

ii.

θtan

θ) cos(2 1

2θcos1 2

5. Show that tanA tanB1

tanBtanABAtan

6. Change 5cos(α) + 2 sin(α) into cosine form

(Answer 5.385 cos(α -21.8o)

7. Change θ) cos(θ) sin( 3 into sine form.

(Answer 2 sin(θ -30o)

8. Two alternating voltages are expressed as v1 = 3 sin(4t) and v2 = 4 cos(4t)

When the voltage is summed the result is v = V sin(4t + ). Determine the value of V and

(5 and 53o)

9. Given that 2cos2(θ) + sin(θ) = x find a formula with θ as the subject.

Given x = 2.065 solve the smallest positive value of θ.

(Answer 4.4o)

10. In a complex stress situation, the stress on a plane at angle θ to the x plane is given by the

formula:

τsin(2θ)θ) cos(22

σσ

2

σσσ

yxyxθ

Where

The stress on the x plane is σx = 200 MPa

The stress on the y plane is σy = 100 MPa

The shear stress is τ = 50 MPa

Calculate the angle of the plane where the stress is 218.3 MPa.

(Answer 15o)