Chapter 7



CHAPTER 6 SOLUTIONS TO Reinforcement exercises IN TRIGONOMETRY

6.3.1 Radian measure and the circle

6.3.1A.

Express as radians

i) 36o ii) 101o iii) 120o iv) 250o

v) 340o vi) – 45o vii) – 110o viii) 15o

ix) 27o x) 273o

Solution

All we have to remember in converting between degrees and radians is that

2π radians = 360 degrees

Then:

i) 36o = = radians = radians

Note that it is usual to leave radian measure in such form, rather than convert it to decimal form. Once you get the hang of this sort of problem it is little more than practice in cancelling down fractions.

ii) Since 2π radians = 360 degrees, 1o = radians and so 101o = radians

iii) 120o = radians = radians

iv) 250o = = radians

v) 340o = = radians

vi) – 45o = – = – radians

vii) – 110o = – = – radians

viii) 15o = = radians

ix) 27o = = radians

x) 273o= = radians

6.3.1B.

Express the following radian measures in degrees in the range 0 ≤ θ < 360o.

i) ii) 14( iii) – iv)

v) vi) vii) viii)

ix) – x)

Solution

Whereas to convert degrees to radians we have to multiply by to go in the opposite direction and convert radians to degrees we have to multiply by .

i) radians = ( = 120o

ii) 14( radians = 14( ( = 7 ( 360o = 360o on reduction to the 360 range

iii) – radians = – ( = – 90o = 270o

iv) radians = ( = 60o

v) radians = ( = 30o

vi) radians = ( = 450o = 90o on reduction to 360 range

vii) radians = ( = 40o

viii) radians = ( = 225o

ix) – radians = – ( = – 72o = 308o

x) radians = ( = 15o

6.3.1C.

Determine the length of arc and the area of the sector subtended by the following angles in a circle of radius 4cm.

i) 15o ii) 30o iii) 45o iv) 60o

v) 90o vi) 120o vii) 160o viii) 180o

Solution

The length of arc of a circle of radius r subtended by an angle θ RADIANS at the centre is given by s = rθ, and the corresponding area of sector is A = r2θ. since the radius is the same, 4cm in each case, we can write these as

s = 4θ and A = 8θ

and the calculations are then just an exercise in changing degrees to radians.

i) For θ ’ 15o = radians we have s = 4 = cm and A = 8 = cm2

ii) For θ ’ 30o = radians we have s = 4 = cm and A = 8 = cm2

iii) For θ ’ 45o = radians we have s = 4 = π cm and A = 8 = 2π cm2

iv) For θ ’ 60o = radians we have s = 4 = cm and A = 8 = cm2

v) For θ ’ 90o = radians we have s = 4 = 2π cm and A = 8 = 4π cm2

vi) For θ ’ 120o = radians we have s = 4 = cm and A = 8 = cm2

vii) For θ ’ 160o = radians we have s = 4= cm and A = 8 = cm2

viii) For θ ’ 180o = π radians we have s = 4π cm and A = 8π cm2

6.3.2 Definitions of the trig ratios

6.3.2A.

Write down the exact values of sine, tan, sec, and complementary ratios for all 'special' angles 0, π/2, π/3, π/4, π/6.

Solution

This may seem like a lot to remember, but with a few aids it is not too bad. It is also very useful to be able to recall these results easily (particularly those for sine and cosine), as they occur frequently (pun intended!) in such topics as AC circuit theory and signal analysis. All you really need to remember are the 30-60 and 45 triangles given in Figure 6.4, and then all of the results are just a matter of definition. Try to get used to using the surds, such as and (See Section 1.2.7). It is much better to leave them as they are until the end of calculation, rather than immediately replacing them by decimal approximations. All you have to remember is, for example ()2 = 2, which is something like how we handle the imaginary j in complex numbers (Chapter 12), by using j2 = ( 1 whenever possible. If you didn’t get all the results right first time, use Figure 6.4 to check the following answers.

| |θ |0 |/2 |/3 |/4 |/6 |

| |sin θ |0 |1 | | | |

| |tan θ |0 |nd | |1 | |

| |sec θ |1 |nd |2 | | |

| |cos θ |1 |0 | | | |

| |cot θ |nd |0 | |1 | |

| |cosec θ |nd |1 | | |2 |

(nd means not defined, because in principle we are trying to divide by zero – in practice, however, it is common to regard, for example, tan(/2) as ‘infinity’).

6.3.2B.

Classify as odd or even functions: cos, sin, tan, sec, cosec, cot.

Solution

Again, you either know or don't know that cosine is even and sine is odd. So tan = sin/cos is odd. sec = 1/cos is even. cosec = 1/sin is odd. cot = cos/sin is odd. Remember such results as Odd ( Even = Odd, O/E = O, O ( O = E, etc.

6.3.2C.

Express as trig functions of x (n is an integer) :-

i) sin(x + n() ii) cos(x + n() iii) tan(x + n()

iv) sin v) cos vi) tan

where n is an integer.

Solution

There is a bit more meat to this question! The safest approach is to use the corresponding compound angle formulae. Also note that we will be frequently using such results as cos n( = (( 1)n, sin n( = 0, etc, so make sure that you understand these results.

i) From the compound angle formula for sin(A + B) we have

sin(x + n() = sin x cos nπ + cos x sinnπ

= sin x cos nπ = (– 1)n sin x

ii) From the compound angle formula for cos(A + B) we have

cos(x + n() = cos x cos nπ + sin x sinnπ

= cos x cos nπ = (– 1)n cos x

iii) tan(x + n() = = = tan x

iv) sin= sin x cos + sin cos x

The form of this depends on whether n is odd or even. the easiest way to find the general result is to try a few specific values of n and then see what the general pattern is. We then find that:

If n is odd then cos = 0 and sin = (– 1)(n–1)/2

So sin= (– 1)(n–1)/2 cos x if n is odd.

On the other hand if n is even then sin = 0 and cos = (– 1)n/2

So sin= (– 1)n/2 sin x if n is even.

v) cos= cos x cos – sin sin x

= (– 1)n/2 cos x if n is even.

– (– 1)(n–1)/2 sin x if n is odd

= (– 1)(n +1)/2 sin x if n is odd.

vi) tan=

Again we get different answers depending on whether n is odd or even. Using the results of iv) and v) we get

If n is even:

tan= = tan x

and if n is odd

tan= = – cot x

6.3.3 Sine and cosine rules and the solution of triangles

With the standard notation solve the following triangles.

i) A = 70o , C = 60o , b = 6

ii) B = 40o , b = 8, c = 10

iii) A = 40o , a = 5, c = 2

iv) a = 5 b = 6, c = 7

v) A = 40o , b = 5, c = 6

vi) A = 120o , b = 3, c = 5

Solution

i) In this case we are given two angles, A = 70o and C = 60o, and one side, b = 6. From A = 70o and C = 60o we know that the third angle must be B = 180o – 70o – 60o = 50o.

We can now use the sine rule to find the two other sides:

= =

so

= =

Hence

a = = 7.3601 to 4dp

c = = 6.7831 to 4dp

ii) This is the case of two sides b = 8 and c = 10, and a non-included angle B = 40o. In this case we can get two possible triangles to fit the information given - it is the 'ambiguous case'. Firstly, apply the sine rule to attempt to find 'sin C'. We have

= = =

so

sin C = = 0.8035 to 4dp

Now although this looks like one answer, we have to remember that

sin (180o – C) = sin C

So if we take C = C1 to be the acute angle solution, then there is another solution with an obtuse angle C2 = 180o – C1. Both solutions will produce triangles satisfying the given condition. For the acute angle result we find

C1 = 53.46o to 2 dp

The obtuse angle result is then

C2 = 180o – 53.46o = 126.54o to 2 dp

So we have two possible triangles for which to find A and a.

C1 = 53.46o

Then

A = 180o – 53.46o – 40o = 180o – 53.46o = 86.54o

and

=

or

a = = 12.42 to 2dp

C2 = 126.54o

Then

A = 180o – 126.54o – 40o = 13.46o

Note: If it had happened that the larger value of C produced a negative answer here then the answer would not be a valid solution - we would effectively only have one triangle.

So

=

or

a = = 2.9 to 2dp

iii) For A = 40o, a = 5, c = 2 we again have two sides and a non-included angle. The sine rule gives

=

so

sin C = = 0.2571 to 4dp

This gives two values of C, as before:

C1 = 14.9o to 2 dp

C2 = 180o – 14.9o = 165.1o to 2 dp

In the acute case we obtain for B:

B = 180o – A – C1 = 180o – 40o – 14.9o = 125.1o

But in the obtuse case we see that A and C2 add to give 165.1o + 40o = 205.1o. Since this is greater than 180o no triangle exists for this value of C which must therefore be rejected.

For the allowed value of C = 14.9o we find the remaining side from

=

or

b = = 6.36 to 2dp

iv) a = 5, b = 6, c = 7 is the case of all three sides and finding the angles. We can use the cosine rule to find any of the angles.

From a2 = b2 + c2 – 2bc cos A we have

cos A = = = 0.7143

and

A = 44.42o

Similarly for B we find

cos B = = 0.543

and so

B = 57.11o

We can then find C from 180o – A – B = 180o – 44.42o – 57.11o = 78.47o

v) A = 40o, b = 5, c = 6 is the case of two sides and the included angle

We have

a2 = b2 + c2 – 2bc cos A = 52 + 62 – 2(5)(6) cos 40o

= 25 + 36 – 60 cos 40o = 15.04373

and so

a = 3.88

Now the sine rule gives

= =

so

sin C =

from which

C = 84.02o

Similarly we get for B:

B = sin– 1 = 55.98o

It pays to check our answer by confirming that the angles add up to 180o (or alternatively of course we could forego the check and determine, say, B by subtraction from 180o). So, the final answer is

a = 3.88, B = 55.98o, C = 84.02o

vi) A = 120o, b = 3, c = 5 is as for v) but now we have an obtuse angle – but the calculations are basically the same, just remembering that cos 120o = – .

We have

a2 = 32 + 52 – 2(3)(5) cos 120o

= 9 + 25 + 30 (0.5) = 49

and so

a = 7

Now the sine rule gives

= =

So

sin B =

from which

B = 21.77o

Similarly we get for C:

C = sin– 1 = 38.21o

So the answer is

a = 7, B = 21.77o, C = 38.21o

6.3.4 Graphs of trigonometric functions

Sketch the graphs of

i) 2 sin (2t + ) ii) 3 cos (3t – )

Solution

Both sin and cosine are of course just continuous waves, and in sketching them we only have to size and locate them. In the general form x = A sin(or cos) ((t + () A gives the amplitude, or the ‘height’ of the wave above the mean level, ( determines the number of oscillations in a given period of t and ( fixes the location of the graph by ensuring that at t = 0, x = A sin (, ie it passes through A sin ( on the vertical axis, with t plotted horizontally. Using these ideas you should obtain the graphs in the figures.

[pic]

6.3.5 Inverse trigonometric functions

6.3.5A.

Find the principal value and general solutions for sin–1x and cos–1x for the following values.

i) 0 ii) iii) –

iv) v) vi) –

Solution

The inverse sine and cosine can be evaluated from the general results (given in radian form)

sin–1x = nπ + (–1)n PV

cos–1 x = 2nπ ± PV

where PV denotes the corresponding principle values, which take the ranges

– [pic] sin ( 1 x [pic]

0 [pic] cos ( 1 x [pic] (

Applying these results, with the appropriate principal values we obtain the table below

| x |sin–1x |sin–1x |cos–1x |cos–1x |

| |PV |GS |PV |GS |

|i) 0 |0 |n( | |2nπ ± |

|ii) | |n( + (–1)n | |2nπ ± |

|iii) – |– |n( +(–1)n+1 | |2nπ ± |

|iv) | |n( + (–1)n | |2nπ ± |

|v) | |n( + (–1)n | |2nπ ± |

|vi) – |– |n( +(–1)n+1 | |2nπ ± |

where n is an integer

6.3.5B.

Find the principal value and general solutions for tan–1x for the following values of x

i) 0 ii) 1 iii) iv) –1 v) –

Solution

Using tan–1 x = nπ ± PV where the PV is in the range ( [pic] tan ( 1 x [pic] we obtain the following results.

|tan–1x |0 |1 | |– 1 |– |

|PV |0 | | |– |– |

|GS |nπ |nπ + |nπ + |nπ – |nπ – |

6.3.6 The Pythagorean identities – 'cos2 + sin2 = 1'

6.3.6A.

i) For c = cos θ, simplify

a) b) c)

ii) For s = sin θ, simplify

a) b) c)

iii) For t = tan θ, simplify

a) b) c)

Solution

All these questions are simply variations on the Pythagorean identities:

cos2θ + sin2θ ∫ 1

1 + tan2θ ∫ sec2θ

cot2θ + 1 ∫ cosec2θ

i) For c = cos θ we have

a) = = = sinθ

b) = = = cot θ

c) = = = tan2θ

ii) For s = sin θ we have

a) = = = cos θ

b) =

c) = = = sec θ tan θ

iii) For t = tan θ we have

a) = = = sec θ

b) = = = sin θ cos θ

c) = = = cos θ cot θ

6.3.6B.

Eliminate θ from the equations

i) x = a cos θ, y = b sin θ ii) x = a sin θ, y = b tan θ

Solution

i) With x = a cos θ, y = b sin θ we have = cos θ and = sin θ and so using the Pythagorean identity:

cos2 θ + sin2 θ ’ 2 + 2 = 1

ii) With x = a sin θ, y = b tan θ we have

y = b tan θ = b = b

But sin θ = so

y = b = =

6.3.6C.

For the following values of sin θ, find the corresponding values of cos θ and tan θ without using your calculator, giving your answers in surd form, and assuming that θ is acute.

i) ii) iii) – iv)

Solution

i) If sin θ = then

cos θ = = = = =

So

tan θ =

ii) If sin θ = then

cos θ = = = = = =

So

tan θ =

iii) If sin θ = then

cos θ = = = = = =

So

tan θ =

6.3.6D.

If r cos θ = 3 and r sin θ = 4 determine the positive value of r, and the principal value of θ.

Solution

Square and add to get

(r cos θ)2 + (r sin θ)2 = r2 cos2 θ + r2 sin2 θ = r2 (cos2 θ + sin2 θ)

= r2 = 32 + 42 = 9 + 16 = 25

so r = 5 (r is always taken positive)

From r cos θ = 3 we therefore have 5 cos θ = 3 or cos θ = from which

θ = cos–1 = 53.13∞ to 2dp

6.3.7 Compound angle formulae

6.3.7A.

Prove the following

i) sin 3θ = 3 sin θ – 4 sin3θ ii) cos 3θ = 4cos3θ – 3 cosθ

iii) = cos θ – sin θ iv) cot θ – tan θ = 2 cot 2θ

v) cot 2θ =

Solution

i) By the compound angle formula we have

sin 3θ = sin (2θ + θ) = sin 2θ cos θ + sin θ cos 2θ

Using the double angle formula on the sin 2θ and cos 2θ gives

sin 3θ ’ 2sin θ cos θ cos θ + sin θ ( cos2θ – sin2θ)

= 3 sin θ cos2θ – sin3θ = 3 sin θ (1 – sin2θ) – sin3θ

= 3 sin θ – 4 sin3θ

ii) By the compound angle formula we have

cos 3θ = cos (2θ + θ) = cos 2θ cos θ – sin θ sin 2θ

Using the double angle formula on the sin 2θ and cos 2θ gives

sin 3θ ’ ( cos2θ – sin2θ)cos θ – sin θ 2sin θ cos θ

= cos3θ – sin2θ cos θ – 2 (1 – cos2θ) cosθ

cos3θ – (1 – cos 2θ) cos θ – 2 (1 – cos2θ) cosθ

= 4cos3θ – 3 cosθ

iii) =

= = cos θ – sin θ

iv) cot θ – tan θ = –

= = 2

on cross mulitplying and using double angle formulae

= 2 cot 2θ

v) cot 2θ =

6.3.7B.

Without using a calculator or tables evaluate

i) sin 15( cos 15( ii) sin 15( iii) tan(π/12) iv) cos(11π/12)

v) tan(7π/12) vi) cos 75(

Solution

In this and the next question you need to be adept at dealing with surds, so if you are a bit rusty on this, go back to Section 1.2.7 for a bit of revision.

i) sin 15( cos 15( = 2 sin 15( cos 15( = sin[ 2 (15() ] = sin 30(

= . =

ii) sin 15( = sin (45( – 30() = sin 45( cos 30( – sin 30( cos 45(

= – =

=

iii) tan(π/12) = tan 15( = tan (45( – 30() =

= = = =

= = 2 –

iv) cos(11π/12) = cos 165( = cos (120( + 45()

= cos 120( cos 45( – sin 120( sin 45(

= – – = – = –

v) tan(7π/12) = tan = =

= = = – = – (2 + )

vi) cos 75( = cos (30( + 45() = cos 45( cos 30( – sin 30( sin 45(

= – =

which of course is just sin (90( – 75() = sin 15( as I'm sure you will have noticed!

6.3.7C.

Evaluate

i) sin 22.5o ii) cos 22.5o iii) tan 22.5o,

given that cos 45o = 1/.

Solution

This question uses the double angle formulae:

sin 2A ∫ sin A cos A

cos 2A ∫ cos2 – sin2 A

∫ 2 cos2A – 1

∫ 1 – 2 sin2A

tan A ’

i) sin 22.5o = sin (45o/2) = by the double angle formula for cos 2A

= =

=

ii) cos 22.5o = cos (45o/2) =

= =

=

iii) tan 22.5o = = from i) and ii)

= = =

6.3.7D.

Express the following products as sums or differences of sines and/or cosines of multiple angles

i) sin 2x cos 3x ii) sin x sin4x iii) cos 2x sin x iv) cos 4x cos 5x

Solution

These questions use the compound angle formulae 'backwards'.

i) We know that sin 2x cos 3x will occur in the expansion of sin (2x + 3x) and sin (2x – 3x). Specifically, we have

sin 5x = sin(2x + 3x) = sin 2x cos 3x + sin 3x cos 2x

sin (– x) = sin(2x – 3x) = sin 2x cos 3x – sin 3x cos 2x

Adding these gives

sin 5x + sin (– x) = sin 5x – sin x = 2 sin 2x cos 3x

on using the fact that sin x is odd.

So

sin 2x cos 3x =

ii) Remembering 'cos (+) = cos cos – sin sin' we know that sin x sin 4x occurs in cos ( x – 4x) = cos 3x, and in cos (x + 4x) = cos 5x. Thus

cos 5x = cos (x + 4x) = cos x cos 4x – sin x sin 4x

cos 3x = cos (x – 4x) = cos x cos 4x + sin x sin 4x

Subtracting and dividing by 2 then gives

sin x sin 4x =

iii) We have

sin 3x = sin (2x + x) = cos 2x sin x + sin 2x cos x

and

sin (– x) = – sin x = – sin (2x – x) = – sin 2x cos x + sin x cos 2x

from which

cos 2x sin x =

iv) The argument should now be familiar:

cos 9x = cos (4x + 5x) = cos 4x cos 5x – sin 4x sin 5x

cos x = cos (4x – 5x) = cos 4x cos 5x + sin 4x sin 5x

and addition gives

cos 4x cos 5x =

6.3.7E.

Prove the following identities (Hint: put P = (A + B)/2, Q = (A – B)/2 in the left-hand sides, expand and simplify and re-express in terms of P and Q)

i) sin P + sin Q ∫ 2sin cos

ii) sin P – sin Q ∫ 2cossin

iii) cos P + cos Q ∫ 2coscos

iv) cos P – cos Q ∫ –2sinsin

Solution

The question hints at the method of proofs of these relations, but there are some shortcuts for the alert. We prove the first in full.

i) Put A = and B = , or P = A + B and Q = A – B.

Then

sin P + sin Q = sin (A + B) + sin (A – B)

= sin A cos B + sin B cos A + sin A cos B – sin B cos A

= 2 sin A cos B

= 2 sin cos

ii) - iv) can be proved similarly, or we can use i) along with such ploys as replacing Q by – Q in i) to get ii) and, for example in iii)

iii) cos P + cos Q ∫ sin (90( – P) + sin (90( – Q)

= 2 sin cos

2 sin cos

= 2 cos cos

6.3.8 Trigonometric equations

6.3.8A.

Give the general solution to each of the equations:

i) cos θ = 0 ii) cos θ = –1 iii) cos θ = –

iv) sin θ = 0 v) sin θ = –1 vi) sin θ =

vii) tan θ = 0 viii) tan θ = –1 ix) tan θ =

Solution

These are really examples in inverse functions, and we have in fact done some of them already in RE 6.3.5. We only need to use the results of Section 6.2.5 for the general forms of the inverse trig functions:

sin–1x = nπ + (–1)n PV

cos–1 x = 2nπ ± PV

tan–1 x = nπ + PV

Where PV is the principal value, and n is an integer.

i) An equation of the form cos θ = a has the general solution θ = 2nπ ± PV, so cos θ = 0 has the general solution (GS)

θ = 2nπ ±

since the PV of cos–1 0 is .

ii) cos θ = –1 has the GS θ = 2nπ ± π = (2n ± 1)π

since the PV of cos–1 (– 1) is π.

iii) The PV of θ for cos θ = – is θ = π – = and so the GS is

θ = 2nπ ±

iv) The GS of sin θ = a is θ = nπ ± (– 1)n PV. Since the PV of θ for sin θ = 0 is θ = 0, the GS of sin θ = 0 is just

θ = nπ

v) The GS of sin θ = –1 is

θ = nπ ± (– 1)n

since the PV of sin–1 (– 1) is – .

vi) No, sin θ = is not a misprint - just a trap for the unwary! Since, for real angles, θ, – 1 [pic] sin θ [pic] 1 it follows that sin θ = has no solutions at all.

vii) The GS of tan θ = a is

θ = nπ + PV

and the PV of θ for tan θ = 0 is θ = 0, so the GS of tan θ = 0 is

θ = nπ

viii) The GS of tan θ = – 1 is

θ = nπ +

since the PV of tan–1 (– 1) is .

ix) For tan θ = the PV of tan–1 () is and so the GS is

θ = nπ +

6.3.8B.

Find the general solutions of the equations

i) cos 2θ = 1 ii) sin 2θ = sinθ

iii) cos 2θ + sinθ = 0 iv) cos 2θ + cos 3θ = 0

v) sec2θ = 3 tan θ – 1

Solution

i) The GS of cos 2θ = 1 for 2θ is, from the previous question

2θ = 2nπ

and so for θ the GS is

θ = nπ

ii) To solve sin 2θ = sinθ we must get everything in terms of simple sines and cosines, for which we use the double angle formula

sin 2θ = 2 sinθ cos θ = sinθ

This can be rewritten as

(2 cos θ – 1) sinθ = 0

which shows that

2 cos θ – 1 = 0 or sinθ = 0

ie either cos θ = or sin θ = 0.

The GS of sin θ = 0 is θ = nπ The GS of cos θ = is θ = 2nπ ± which together give the full set of solutions for θ.

An alternative approach to this sort of question is to use the GS of sin A = sin B given in Section 6.2.8, namely A = 2nπ + B or A = (2n + 1)π – B. So the GS of sin 2θ = sin θ will be

2θ = 2nπ + θ or (2n + 1)π – θ

Hence

θ = 2nπ or θ = where n is an integer

I will leave it as an exercise for you to confirm that this solution is equivalent to the one we obtained above (Evaluate both for n = 0, 1, 2, ...)

iii) To solve cos 2θ + sinθ = 0 we must again get everything in terms of simple sines and cosines, for which we use the double angle formula

cos 2θ = 1 – 2sin2θ

Then the equation can be rewritten as

1 – 2sin2 θ + sinθ = 0

or

2 sin2 θ – sin θ – 1 = 0

= (2 sin θ + 1)(sin θ – 1)

Hence either sin θ = – or sin θ = 1.

The GS of sin θ = 1 is θ = π. The GS of sin θ = is θ = nπ – (– 1)n . These results can be combined together in the general solution for θ:

θ = π

An alternative approach is to use the GS of cos A = cos B given in Section 6.2.8, after converting the sin θ to a complementary cosine.

iv) In cos 2θ + cos 3θ = 0 use of multiple angle formulae is going to be messy. It is easier to use the result

cos P + cos Q ∫ 2cos cos

We have

cos 2θ + cos 3θ = 2 cos cos = 0

So

cos = 0 or cos = 0

Hence

= π and therefore θ = π

or

= π and therefore θ = (2n + 1) π

These two apparently different looking results can be combined by noting that when n = 5m + 2 where m is any integer (and of course n can be any integer)

π = π = (2m + 1) π

So the GS is

θ = π where n is any integer

v) sec2θ = 1 + tan2θ = 3 tan θ – 1 and so

tan2 θ – 3 tan θ + 2 = 0

Factorising this quadratic in tan θ gives

(tan θ – 1)(tan θ – 2) = 0

So

tan θ = 1 or tan θ = 2

Therefore

θ = or θ = 1.11 radians for the PV

Hence the GS is

θ = nπ + , nπ + 1.11

6.3.9 The a cos θ + b sin θ form

6.3.9A.

Write the following in the form a) r sinb) rcos

i) sin θ – cos θ ii) cos θ + sin θ iii) sinθ – cos θ

iv) 3 cos θ + 4 sin θ

Solution

i) a) We start by writing

sin θ – cos θ = r sin

= r sin θ cos α + r cos θ sin α

So

r cos α = 1 and r sin α = – 1

Squaring and adding to eliminate α gives r = , and then

cos α = and sin α = –

The principal value of α is then

α = – 45( = –

So we now have

sin θ – cos θ = = sin

b) A little more quickly now:

sin θ – cos θ = r cos

= r cosθ cos α – r sinθ sin α

So

r sin α = – 1 and r cos α = 1

and again r = , so

sin α = – and cos α = –

The principal value of α is then

α = – 135( = –

So

sin θ – cos θ = cos

ii) a) cos θ + sin θ = r sin

= r sinθ cos α + r cosθ sin α

So

r sin α = and r cos α = 1

from which r = 2 and then

cos α = and sin α =

The principal value of α is then

α = 60( =

So we have

cos θ + sin θ = 2 sin

b) cos θ + sin θ = r cos

= r cos θ cos α – r sin θ sin α

So

r cos α = and r sin α = (1

from which r = 2 and then

sin α = – and cos α =

The principal value of α is therefore

α = – 30( = –

So we have

cos θ + sin θ = 2 cos

iii) a) sin θ – cos θ = r sin

= r sin θ cos α + r cos θ sin α

So

r cos α = and r sin α = – 1

from which r = 2 and then

sin α = – and cos α =

The principal value of α is then

α = –

So we have

sin θ – cos θ = 2 sin

b) sin θ – cos θ = r cos

= r cos θ cos α – r sin θ sin α

So

r sin α = – and r cos α = – 1

from which r = 2 and then

cos α = – and sin α = –

The principal value of α is therefore

α = – 150( = –

So we have

sin θ – cos θ = 2 cos

iv) a) 3 cos θ + 4 sin θ = r sin

= r sinθ cos α + r cosθ sin α

So

r sin α = 3 and r cos α = 4

From which r = 5, and then

cos α = and sin α =

The principal value of α is then

α = 36.87( = 0.6435 radians

So we now have

3 cos θ + 4 sin θ = 5 sin)

b) 3 cos θ + 4 sin θ = r cos

= r cosθ cos α – r sinθ sin α

So

r sin α = – 4 and r cos α = 3

and again r = 5, so

cos α = and sin α = –

The principal value of α is then

α = – 53.13(

So

3 cos θ + 4 cos θ = 5 cos

6.3.9B.

Determine the maximum and minimum values of each of the expressions in Question A, stating the values where they occur, in the range 0 [pic] θ [pic] 2π.

Solution

i) a) sin θ – cos θ = sinfrom A.

This has a maximum value of when θ – = , or

θ =

It has a minimum value of – at θ – = , or

θ =

So there is a maximum of at θ = and a minimum value of – at θ =

ii) cos θ + sin θ = 2 sin

This has a maximum of 2 at θ + = , ie θ =

It has a minimum of – 2 at θ + = , ie θ =

iii) sin θ – cos θ = 2 sin

This has a maximum of 2 at θ – = , ie θ =

It has a minimum of – 2 at θ – = , ie θ =

iv) 3 cos θ + sin θ = 5 sin

This has a maximum of 5 at θ + 0.64 = , ie θ = 0.93 radians

It has a minimum of – 5 at θ + 0.64 = , ie θ = 4.07 radians

6.3.9C.

Find the solutions, in the range 0 [pic] θ [pic] 2π, of the equations obtained by equating the expressions in Question A to a) 1 b) –1.

Solution

i) a) sin θ – cos θ = sin= 1 from A.

So

sin=

In the range 0 [pic] θ [pic] 2π the solutions are

θ – = ,

and therefore

θ = ,

b) sin θ – cos θ = sin= – 1, gives

sin= –

In the range 0 [pic] θ [pic] 2π the solutions are

θ – = – ,

and therefore

θ = 0, π

ii) a) cos θ + sin θ = 2 sin= 1

So

sin=

In the range 0 [pic] θ [pic] 2π the solutions are

θ + = ,

and therefore

θ = ,

b) cos θ + sin θ = 2 sin= – 1

So

sin= –

In the range 0 [pic] θ [pic] 2π the solutions are

θ + = ,

and therefore

θ = ,

iii) a) sin θ – cos θ = 2 sin= 1

So

sin=

In the range 0 [pic] θ [pic] 2π the solutions are

θ – = ,

and therefore

θ = , π

b) sin θ – cos θ = 2 sin= – 1

So

sin= –

In the range 0 [pic] θ [pic] 2π the solutions are

θ – = ,

and therefore

θ = , 2π

iv) a) 3 cos θ + 4 sin θ = 5 sin(θ + 0.64) = 1

So

sin(θ + 0.64) = = 0.2

In the range 0 [pic] θ [pic] 2π the solutions are, to two decimal places

θ + 0.64 = π – 0.2 = 2.9 and 2π + 0.2 = 6.48

and therefore

θ = 2.3, 5.84 radians

b) 3 cos θ + 4 sin θ = 5 sin= – 1

So

sin= – = – 0.2

In the range 0 [pic] θ [pic] 2π the solutions are

θ + 0.64 = 3.34 and 6.28

and therefore

θ = 2.7, 5.44

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