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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
v) 340o
vi) – 45o
ix) 27o
x) 273o
iii)
vii)
120o
iv) 250o
– 110o viii) 15o
Solution
All we have to remember in converting between degrees and radians is that
2 radians = 360 degrees
Then:
i)
360o 2

36 = 10 = 10 radians = 5 radians
o
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.
101

ii) Since 2 radians = 360 degrees, 1o = 180 radians and so 101o = 180 radians
120
2
iii) 120o = 180 radians = 3 radians
250 25
iv) 250o = 180 = 18 radians
340
17
v) 340o = 180 = 9 radians
45

vi) – 45o = – 180 = – 4 radians
110
11
vii) – 110o = – 180 = – 18 radians
15

viii) 15o = 180 = 12 radians
-1-
27 3
ix) 27o = 180 = 20 radians
x)
273 31
273o= 180 = 60 radians
6.3.1B.
Express the following radian measures in degrees in the range 0  < 360o.
2


i)
ii)
14
iii)
–
iv)
3
2
3
5
2
5

v) 6
vi)
vii)
viii)
2
9
4
2

ix) – 5
x) 12
Solution

Whereas to convert degrees to radians we have to multiply by 180 to go in the opposite
180
direction and convert radians to degrees we have to multiply by
.

i)
2
2
180
radians
=

= 120o
3
3

180
= 7 360o = 360o on reduction to the 360 range

ii) 14 radians = 14 
180


iii) – 2 radians = – 2 
= – 90o = 270o

iv)
180


radians
=

= 60o
3
3

v)
180


radians
=

= 30o
6
6

vi)
5
2 radians =
vii)
2
2
180
radians
=

= 40o
9
9

5
180
o
o
2   = 450 = 90 on reduction to 360 range
-2-
viii)
5
5
180
radians
=

= 225o
4
4

2
2
180
ix) – 5 radians = – 5 
= – 72o = 308o


x) 12
radians =
180


= 15o
12

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)
v)
15o
90o
ii)
vi)
30o
120o
iii)
vii)
45o
160o
iv)
viii)
60o
180o
Solution
The length of arc of a circle of radius r subtended by an angle  RADIANS at the centre is
1
given by s = r, and the corresponding area of sector is A = 2 r2 since the radius is the
same, 4cm in each case, we can write these as
s = 4 and A = 8
and the calculations are then just an exercise in changing degrees to radians.
2




i) For 15o = 12 radians we have s = 4 12 = 3 cm and A = 8 12 = 3 cm2
2
4



ii) For 30o = 6 radians we have s = 4 6 = 3 cm and A = 8 6 = 3 cm2



iii) For 45o = 4 radians we have s = 4 4 =  cm and A = 8 4 = 2 cm2
4
8



iv) For 60o = 3 radians we have s = 4 3 = 3 cm and A = 8 3 = 3 cm2



v) For 90o = 2 radians we have s = 4 2 = 2 cm and A = 8 2 = 4 cm2
-3-
2
vi) For 120o = 3 radians we have
2
2 16
8
s = 4 3 = 3 cm and A = 8 3 = 3
cm2
8
8
8 64
32
vii) For 160o = 9 radians we have s = 4 9 = 9 cm and A = 8 9 = 9 cm2
viii) For 180o =  radians we have
s = 4cm and A = 8 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, /2, /3, /4, /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 2 and 3 (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 = 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.
-4-

0
 /2
 /3
 /4
 /6
sin 
0
1
3
2
1
2
1
2
tan 
0
nd
3
1
sec 
1
nd
2
2
1
3
2
3
cos 
1
0
1
2
1
2
3
2
cot 
nd
0
1
3
cosec 
nd
1
1
3
2
3
2
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)
n 
n 


iv) sinx + 2 
v) cosx + 2 




iii)
vi)
tan(x + n)
n 

tanx + 2 


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
-5-
sin(x + n) = sin x cos n + cos x sinn
= sin x cos n = (– 1)n sin x
ii) From the compound angle formula for cos(A + B) we have
cos(x + n) = cos x cos n + sin x sinn
= cos x cos n = (– 1)n cos x
iii) tan(x + n) =
sin(x + n) (– 1)n sin x
=
= tan x
cos(x + n) (– 1)n cos x
n 

n 
n 
iv) sinx + 2  = sin x cos  2  + sin  2  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:
n 
n 
If n is odd then cos  2  = 0 and sin  2  = (– 1)(n–1)/2
 
 
n 

So sinx + 2  = (– 1)(n–1)/2 cos x if n is odd.


n 
n 
On the other hand if n is even then sin  2  = 0 and cos  2  = (– 1)n/2
 
 
n 

So sinx + 2  = (– 1)n/2 sin x if n is even.


n 

n 
n 
v) cosx + 2  = cos x cos  2  – sin  2  sin x


 
 
= (– 1)n/2 cos x if n is even.
– (– 1)(n–1)/2 sin x if n is odd
-6-
= (– 1)(n +1)/2 sin x if n is odd.
n 

sinx + 2 
n 



vi) tanx + 2  =
n 



cosx + 2 


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:
n 
(– 1)n/2 sin x



tan x + 2 =
= tan x

 (– 1)n/2 cos x
and if n is odd
n  (– 1)(n–1)/2 cos x

tanx + 2  =
= – cot x

 (– 1)(n +1)/2 sin 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 ,
ii) B = 40o ,
b = 8,
iii) A = 40o ,
a = 5,
iv) a = 5
b = 6,
v) A = 40o ,
b = 5,
o
vi) A = 120 ,
b = 3,
b
c
c
c
c
c
=
=
=
=
=
=
6
10
2
7
6
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.
-7-
A
70
b=6
c
60
B
C
a
We can now use the sine rule to find the two other sides:
a
b
c
=
=
sin A sin B sin C
so
a
6
c
o =
o =
sin 70
sin 50
sin 60o
Hence
6 sin 70o
a=
= 7.3601 to 4dp
sin 50o
c=
6 sin 60o
= 6.7831 to 4dp
sin 50o
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
b
8
c
10
sin B = sin 40o = sin C = sin C
so
10 sin 40o
sin C =
= 0.8035 to 4dp
8
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
-8-
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.
A
b=8
c = 10
40
B
C1
C2
C1 = 53.46o
Then
A = 180o – 53.46o – 40o = 180o – 53.46o = 86.54o
and
a
8
o =
sin 86.54
sin 40o
or
8 sin 86.54o
a=
= 12.42 to 2dp
sin 40o
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
-9-
a
8
o =
sin 13.46
sin 40o
or
a=
8 sin 13.46o
= 2.9 to 2dp
sin 40o
iii) For A = 40o, a = 5, c = 2 we again have two sides and a non-included angle.
The sine rule gives
A
40
b
c=2
B
a=5
C
2
5
=
sin C
sin 40o
so
2 sin 40o
sin C =
= 0.2571 to 4dp
5
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
- 10 -
b
5
o =
sin 125.1
sin 40o
or
5 sin 125.1o
b=
= 6.36 to 2dp
sin 40o
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.
A
c=7
B
b=6
a=5
C
From a2 = b2 + c2 – 2bc cos A we have
b2 + c2 – a2
62 + 72 – 52
cos A =
= 2(6)(7)
= 0.7143
2bc
and
A = 44.42o
Similarly for B we find
52 + 72 – 62
cos B = 2(5)(7)
= 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
- 11 -
A
40
6
5
C
a
B
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
5
a
6
=
=
o
sin B sin 40
sin C
so
6 sin 40o
sin C = 3.88
from which
C = 84.02o
Similarly we get for B:
 5 sin 40o
B = sin  3.88  = 55.98o


It pays to check our answer by confirming that the angles add up to 180 o (or
alternatively of course we could forego the check and determine, say, B by
subtraction from 180o). So, the final answer is
–1
a = 3.88, B = 55.98o, C = 84.02o
A = 120o, b = 3, c = 5 is as for v) but now we have an obtuse angle – but the
1
calculations are basically the same, just remembering that cos 120o = – 2 .
vi)
- 12 -
A
120
5
3
C
B
a
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
3
a
7
=
=
o
sin B sin 120
sin 120o
So
sin B =
3 sin 120o
7
from which
B = 21.77o
Similarly we get for C:
 5 sin 120o
 = 38.21o
C = sin– 1 
7


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 + 3 ) ii) 3 cos (3t – 2
)
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
- 13 -
= 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.
- 14 -
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.
1
1
i) 0
ii) 2
iii) – 2
iv)
3
2
1
2
v)
vi)
–
1
2
Solution
The inverse sine and cosine can be evaluated from the general results (given in radian
form)
sin–1x = n + (–1)n PV
cos–1 x = 2n PV
where PV denotes the corresponding principle values, which take the ranges


– 2  sin  1 x  2
0  cos  1 x  
Applying these results, with the appropriate principal values we obtain the table below
x
i)
ii)
iii)
iv)
v)
0
1
2
1
–2
3
2
1
2
sin–1x
sin–1x
cos–1x
PV
GS
0
n
PV

2

3
2
3

6
GS

2n  2

2n  3
2
2n  3

2n  6

4

2n  4

6

–6

3

4
n + (–1)
n +(–1)
n

n+1

n + (–1)
n

n + (–1)
n

- 15 -
cos–1x
vi) –

–4
1
2
n +(–1)
n+1

3
4
3
2n  4
where n is an integer
6.3.5B.
Find the principal value and general solutions for tan–1x for the following values of x
1
i) 0
ii) 1
iii) 3
iv) –1
v) –
3
Solution

Using tan–1 x = n PV where the PV is in the range  2  tan
1

x  2 we obtain the
following results.
tan–1x
0
1
PV
0

4

3

–4
GS
n

n + 4

n + 3

n – 4
3
–1
6.3.6 The Pythagorean identities – 'cos2 + sin2 = 1'
6.3.6A.
i) For c = cos , simplify
a)
ii)
1–c
b)
c
1 – c2
c)
1 – c2
c2
c)
s
1 – s2
For s = sin , simplify
a)
iii)
2
2
1–s
b)
1 – s2
s2
For t = tan , simplify
a)
1 + t2
b)
t
1 + t2
c)
1
t 1 + t2
Solution
All these questions are simply variations on the Pythagorean identities:
- 16 -
1
3

–6

n – 6
–


cos2 + sin2  1
1 + tan2  sec2
cot2 +
1
cosec2

i)
For c = cos  we have
a)
b)
c)
ii)
c
1 – c2
=
1 – cos2
sin2 = sin
=
cos 
=
1 – cos2
cos
= cot 
sin
1 – c2 1 – cos2 sin2
=
=
= tan2
c2
cos2
cos2
For s = sin  we have
a)
1 – s2 =
b)
1 – s2
cos2
cos 
=
=
= cosec  cot 
2
2
s
sin 
sin2
c)
iii)
1 – c2 =
s
1 – s2
=
1 – sin2 =
sin 
1 – sin2
=
cos2 = cos 
sin
= sec  tan 
cos2
For t = tan  we have
a)
1 + t2
=
1 + tan2 =
sec2 = sec 
t
tan 
tan 
= sin  cos 
2 =
2 =
1+t
1 + tan  sec2
1
1
1
c)
=
=
= cos  cot 
2
2
t 1+t
tan  1 + tan  tan  sec 
b)
6.3.6B.
Eliminate  from the equations
i) x = a cos 
y = b sin 
ii)

- 17 -
x = a sin ,
y = b tan 
Solution
x
y
i) With x = a cos  y = b sin  we have a = cos  and b = sin  and so using
the Pythagorean identity:
x 2 y  2
cos2  + sin2 a + b = 1
 
 
ii) With x = a sin , y = b tan  we have
y = b tan  = b
sin 
=b
cos 
sin 
1 – sin2 
x
But sin  = a so
y=b
x/a
x2
1 – a 
 
bx
=
a
1–
x2
a2
=
bx
a2 – x2
6.3.6C.
For the following values of sin , find the corresponding values of cos  and tan  without
using your calculator, giving your answers in surd form, and assuming that  is acute.
2
1
8
7
i) 5
ii) 13
iii) – 17
iv)
25
Solution
2
i) If sin  = 5 then
cos  =
1 – sin2  =
So
tan  =
22
1 – 5 =
 
4
1 – 25 =
21
21
=
25
5
sin 
2/5
2
=
=
cos 
21
21/5
1
ii) If sin  = 13 then
cos  =
1 – sin2  =
 1 2
1 – 13 =
 
1
1 – 169 =
- 18 -
168
168 2 42
169 = 13 = 13
So
tan  =
sin 
1/13
1
=
=
cos 
42/13 2 42
7
iii) If sin  = 25 then
cos  =
1 – sin2  =
 7 2
1 – 25 =
 
So
tan  =
49
1 – 625 =
576
576 24
=
625
25 = 25
sin  7/25
7
= 24/25 = 24
cos 
6.3.6D.
If r cos  = 3 and r sin  = 4 determine the positive value of r, and the principal value of
.
Solution
Square and add to get
(r cos )2 + (r sin )2 = r2 cos2  + r2 sin2  = r2 (cos2  + sin2 )
= r2 = 32 + 42 = 9 + 16 = 25
so r = 5 (r is always taken positive)
3
From r cos  = 3 we therefore have 5 cos  = 3 or cos  = 5 from which
3
 = cos–1 5 = 53.13 to 2dp
 
6.3.7 Compound angle formulae
6.3.7A.
Prove the following
i) sin 3 = 3 sin  – 4 sin3
ii)
cos 3 = 4cos3– 3 cos
iv)
cot  – tan  = 2 cot 2


iii)
cos 2
= cos  – sin 
cos  + sin 
- 19 -
v)
cot2 – 1
cot 2 =
2 cot
Solution
i) By the compound angle formula we have
sin 3 = sin (2 + ) = sin 2 cos  + sin  cos 2
Using the double angle formula on the sin 2 and cos 2 gives
sin 32sin  cos  cos  + sin  ( cos2 – sin2)
= 3 sin  cos2 – sin3 = 3 sin  (1 – sin2) – sin3 
= 3 sin  – 4 sin3


ii) By the compound angle formula we have
cos 3 = cos (2 + ) = cos 2 cos  – sin  sin 2
Using the double angle formula on the sin 2 and cos 2 gives
sin 3( cos2 – sin2)cos  – sin  2sin  cos 
= cos3 – sin2 cos  – 2 (1 – cos2) cos

 
cos  – (1 – cos ) cos  – 2 (1 – cos2) cos
3
2
= 4cos3– 3 cos




iii)
cos 2
cos2 – sin2
=
cos  + sin 
cos  + sin 
=
iv)
(cos  – sin )(cos  + sin )
cos  + sin 
cot  – tan  =
cos 
sin 
–
sin 
cos 
- 20 -
= cos  – sin 

cos2 – sin2
cos 2
=
=2

sin cos 
sin 2
on cross mulitplying and using double angle formulae
= 2 cot 2
cos2 
–1
cos 2 cos2 – sin2 sin2 
cot2 – 1
v) cot 2 =
=
=
=
sin 2
 sin cos 
cos 
2 cot
2
sin 
6.3.7B.
Without using a calculator or tables evaluate
i) sin 15 cos 15
ii) sin 15
v) tan(7/12)
vi) cos 75
iii)
tan(/12)
iv)
cos(11/12)
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)
1
1
1
sin 15 cos 15 = 2 2 sin 15 cos 15 = 2 sin[ 2 (15) ] = 2 sin 30
1 1 1
=2 .2 =4
ii) sin 15 = sin (45 – 30) = sin 45 cos 30 – sin 30 cos 45
1 3 1 1
1
= 2 2 –2
=
( 3 – 1)
2
2 2
=
iii)
2( 3 – 1)
4
tan(/12) = tan 15 = tan (45 – 30) =
- 21 -
tan 45 – tan 30
1 + tan 45 tan 30
1 – 1/ 3
=
=
1 + 1/ 3
3–1
( 3 – 1)2
( 3)2 – 2 3 + 1
=
=
3–1
3 + 1 ( 3 + 1)( 3 – 1)
=
iv)
4–2 3
=2– 3
2
cos(11/12) = cos 165 = cos (120 + 45)
= cos 120 cos 45 – sin 120 sin 45
1 1
3 1
1
2( 3 + 1)
=–2
– 2
=–
( 3 + 1) = –
4
2
2
2 2
v)


tan 4 + tan3
1+ 3
 
 
 
tan(7/12) = tan 4 + 3 =
=

 1 – 3


1 – tan 4 tan 3
 
 
=
(1 + 3)2 1 + 2 3 + 3
4+2
=
=
–
1–3
–2
2
3
= – (2 + 3 )
vi) cos 75 = cos (30 + 45) = cos 45 cos 30 – sin 30 sin 45
3 1
1 1
= 2
– 2
2
2
=
2( 3 – 1)
4

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)
o
given that cos 45 = 1/
cos 22.5o
2 .
iii)

Solution
This question uses the double angle formulae:
sin 2A  sin A cos A
cos 2A cos2 – sin2 A
 2 cos2A – 1
- 22 -
tan 22.5o,
tan A
 1 – 2 sin2A
2 tan A

1 – tan2A
i) sin 22.5o = sin (45o/2) =
1
o
2(1 – cos 45 ) by the double angle formula for cos
2A
1
1
1
–

 =
2
2
=
=
2– 2
2
1
o
2(1 + cos 45 )
ii) cos 22.5o = cos (45o/2) =
1
1
21 + 2 =
=
=
sin 22.5o
iii) tan 22.5 =
=
cos 22.5o
o
=
2( 2 – 1)
4
2– 2
2+ 2
2+ 2
2
2–
2
2+
2
=
2( 2 + 1)
4
from i) and ii)
2)2
(2 –
6
=
6(2 – 2)
6

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'.
- 23 -
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
1
sin 2x cos 3x = 2 (sin 5x – sin x)
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
1
sin x sin 4x = 2 (cos 3x – cos 5x)
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
1
cos 2x sin x = 2 (sin 3x – sin x)
iv) The argument should now be familiar:
cos 9x = cos (4x + 5x) = cos 4x cos 5x – sin 4x sin 5x
- 24 -
cos x = cos (4x – 5x) = cos 4x cos 5x + sin 4x sin 5x
and addition gives
1
cos 4x cos 5x = 2 (cos 9x + cos x)

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)
P + Q
P – Q
i) sin P + sin Q  2sin  2  cos  2 




P
+
Q
P
–
Q




ii) sin P – sin Q  2cos 2  sin  2 




P
+
Q
P
–
Q




iii) cos P + cos Q  2cos 2  cos  2 




P + Q
P – Q
iv) cos P – cos Q  –2sin 2  sin  2 




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 =
P+Q
P–Q
and
B
=
2
2 , 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
P + Q
P – Q
= 2 sin  2  cos  2 




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)
- 25 -
180– (P + Q)
Q – P
 cos 
= 2 sin 

2


 2 

P + Q

P – Q
2 sin 90– 2  cos  2 




P + Q
P – Q
= 2 cos  2  cos  2  





6.3.8 Trigonometric equations
6.3.8A.
Give the general solution to each of the equations:
i) cos  = 0
ii) cos  = –1
3
iii) cos  = – 2
iv) sin  = 0
vii) tan  = 0
v) sin  = –1
viii) tan  = –1
vi) sin  = 3
ix) tan  = 3
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 = n + (–1)n PV
cos–1 x = 2n PV
tan–1 x = n + PV
Where PV is the principal value, and n is an integer.
i) An equation of the form cos  = a has the general solution  = 2n PV, so cos 
= 0 has the general solution (GS)

 = 2n 2 

since the PV of cos–1 0 is 2 .
ii) cos  = –1 has the GS  = 2n  = (2n 
since the PV of cos–1 (– 1) is 
- 26 -
5

is  =  –6 = 6 and so the GS is
3
iii) The PV of  for cos  = – 2
5
 = 2n  6 
iv) The GS of sin  = a is  = n – 1n PV. Since the PV of  for sin  = 0 is  = 0,
the GS of sin  = 0 is just
 = n
v) The GS of sin  = –1 is

 = n  (– 1)n 2

since the PV of sin–1 (– 1) is – 2 .
vi) No, sin  = 3 is not a misprint - just a trap for the unwary! Since, for real
angles, , – 1  sin   1 it follows that sin  = 3 has no solutions at all.
vii) The GS of tan  = a is
 = n + PV
and the PV of  for tan  = 0 is  = 0, so the GS of tan  = 0 is
 = n
viii) The GS of tan  = – 1 is
3
 = n + 4 
3
since the PV of tan–1 (– 1) is  4 .
ix)
For tan  =

3 the PV of tan–1 ( 3 ) is 3 and so the GS is

 = n + 3 
- 27 -
6.3.8B.
Find the general solutions of the equations
i) cos 2= 1

iii) cos 2 + sin = 0
v) sec2 = 3 tan  – 1
Solution
i)
ii)
iv)
sin 2 = sin
cos 2 + cos 3 = 0
The GS of cos 2= 1 for 2 is, from the previous question
2 = 2n
and so for  the GS is
 = n
ii) To solve sin 2 = sin we must get everything in terms of simple sines and
cosines, for which we use the double angle formula
sin 2 = 2 sin cos  = sin
This can be rewritten as
(2 cos  – 1) sin = 0
which shows that
2 cos  – 1 = 0 or sin = 0
1
ie either cos  = 2 or sin  = 0.
1

The GS of sin  = 0 is  = n The GS of cos  = 2 is  = 2n  3 which together
give the full set of solutions for .
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 = 2n + B or A = (2n + 1) – B. So the GS of sin 2
= sin  will be
2 = 2n +  or (2n + 1) – 
Hence
- 28 -
 = 2n or  =
(2n + 1)
where n is an integer
3
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 2 + sin = 0 we must again get everything in terms of simple
sines and cosines, for which we use the double angle formula
cos 2 = 1 – 2sin2
Then the equation can be rewritten as
1 – 2sin2  sin = 0
or
2 sin2 – sin  – 1 = 0
= (2 sin  + 1)(sin  – 1)
1
Hence either sin  = – 2 or sin  = 1.
1
1


The GS of sin  = 1 is  = 2n + 2 . The GS of sin  = 2 is  = n – (– 1)n 6 . These


results can be combined together in the general solution for :
1
2
 = 3n + 2 


An alternative approach is to use the GS of cos A = cos B given in Section 6.2.8,
after converting the sin  to a complementary cosine.
iv) In cos 2 + cos 3 = 0 use of multiple angle formulae is going to be messy. It is
easier to use the result
P + Q
P – Q
cos P + cos Q  2cos  2  cos  2 




We have
5

cos 2 + cos 3 = 2 cos  2  cos 2 = 0
 
 
- 29 -
So
5

cos  2  = 0 or cos 2 = 0
 
 
Hence
5 2n + 1
2n + 1
=



and
therefore

=
 5  
2
 2 


or
 2n + 1
2 =  2   and therefore  = (2n + 1) 
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)
2n + 1
10m + 5
 5   =  5
  = (2m + 1) 




So the GS is
2n + 1
 =  5   where n is any integer


v) sec2 = 1 + tan2 = 3 tan  – 1 and so
tan2  – 3 tan  + 2 = 0
Factorising this quadratic in tan  gives
(tan  – 1)(tan  – 2) = 0
So
tan  = 1 or tan  = 2
Therefore

 = 4 or  = 1.11 radians for the PV
Hence the GS is

 = n + 4 , n + 1.11
- 30 -
6.3.9 The a cos  + b sin  form
6.3.9A.
Write the following in the form a) r sin( + ) b) rcos( + )
i)
iv)
sin  – cos 
3 cos  + 4 sin 
ii)
3 cos  + sin 
iii)
3 sin – cos 
Solution
i)a) We start by writing
sin  – cos  = r sin( + )
= r sin  cos  + r cos  sin 
So
r cos  = 1 and r sin  = – 1
Squaring and adding to eliminate  gives r =
cos  =
1
2
2 , and then
and sin  = –
1
2
The principal value of  is then

 = – 45 = – 4
So we now have
sin  – cos  =
1
1




2 sin
–
cos = 2 sin – 4


2
2


b) A little more quickly now:
sin  – cos  = r cos( + )
= r cos cos  – r sin sin 
- 31 -
So
r sin  = – 1 and r cos  = 1
and again r =
2 , so
sin  = –
1
2
and cos  = –
1
2
The principal value of  is then
3
 = – 135 = – 4
So
3

sin  – cos  = 2 cos – 4 


ii) a)
3 cos  + sin  = r sin( + )
= r sin cos  + r cos sin 
So
r sin  =
3 and r cos  = 1
from which r = 2 and then
1
3
cos  = 2 and sin  = 2
The principal value of  is then

 = 60 = 3
So we have


3 cos  + sin  = 2 sin + 3


- 32 -
b)
3 cos  + sin  = r cos( + )
= r cos  cos  – r sin  sin 
So
r cos  =
3 and r sin  = 1
from which r = 2 and then
1
3
sin  = – 2 and cos  = 2
The principal value of  is therefore

 = – 30 = – 6
So we have


3 cos  + sin  = 2 cos – 6


iii) a)
3 sin  – cos  = r sin( + )
= r sin  cos  + r cos  sin 
So
r cos  =
3 and r sin  = – 1
from which r = 2 and then
1
3
sin  = – 2 and cos  = 2
The principal value of  is then

=–6
So we have
- 33 -


3 sin  – cos  = 2 sin – 6


b)
3 sin  – cos  = r cos( + )
= r cos  cos  – r sin  sin 
So
r sin  = –
3 and r cos  = – 1
from which r = 2 and then
1
3
cos  = – 2 and sin  = – 2
The principal value of  is therefore
5
 = – 150 = – 6
So we have
5

3 sin  – cos  = 2 cos – 6 


iv)
a) 3 cos  + 4 sin  = r sin( + )
= r sin cos  + r cos sin 
So
r sin  = 3 and r cos  = 4
From which r = 5, and then
4
cos  = 5
3
and sin  = 5
The principal value of  is then
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 = 36.87 = 0.6435 radians
So we now have
3 cos  + 4 sin  = 5 sin( + 0.64) )
b) 3 cos  + 4 sin  = r cos( + )
= r cos cos  – r sin sin 
So
r sin  = – 4 and r cos  = 3
and again r = 5, so
3
cos  = 5
4
and sin  = – 5
The principal value of  is then
 = – 53.13
So
3 cos  + 4 cos  = 5 cos( – 53.13)
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    2.
Solution
i)a) sin  – cos  =


2 sin – 4 from A.


This has a maximum value of
 
2 when  – 4 = 2 , or
3
= 4
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It has a minimum value of –
 3
2 at  – 4 = 2 , or
7
= 4
3
7
So there is a maximum of 2 at  = 4 and a minimum value of – 2 at  = 4
ii)


3 cos  + sin  = 2 sin + 3


 

This has a maximum of 2 at  + 3 = 2 , ie  = 6
7
 3
It has a minimum of – 2 at  + 3 = 2 , ie  = 6
iii)


3 sin  – cos  = 2 sin – 6


2
 
This has a maximum of 2 at  – 6 = 2 , ie  = 3
5
 3
It has a minimum of – 2 at  – 6 = 2 , ie  = 3
iv) 3 cos  + sin  = 5 sin( + 0.64)

This has a maximum of 5 at  + 0.64 = 2 , ie  = 0.93 radians
3
It has a minimum of – 5 at  + 0.64 = 2 , ie  = 4.07 radians

6.3.9C.
Find the solutions, in the range 0    2, of the equations obtained by equating the
expressions in Question A to a) 1
b) –1.
Solution
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i) a) sin  – cos  =


2 sin – 4 = 1 from A.


So
1


sin – 4 =


2
In the range 0    2 the solutions are
  3
–4 =4, 4
and therefore
 3
=2, 2
b) sin  – cos  =


2 sin – 4 = – 1, gives


1


sin – 4 = –


2
In the range 0    2 the solutions are

 5
–4 =– 4, 4
and therefore
 = 0, 
ii) a)


3 cos  + sin  = 2 sin + 3 = 1


So
 1

sin + 3 = 2


In the range 0    2 the solutions are
 5 17
+3 = 6 , 6
and therefore
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 5
=2, 3
b)


3 cos  + sin  = 2 sin + 3 = – 1


So
1


sin + 3 = – 2


In the range 0    2 the solutions are
 7 11
+3 = 6 , 6
and therefore
5 3
= 6 , 2
iii) a)


3 sin  – cos  = 2 sin – 6 = 1


So
 1

sin – 6 = 2


In the range 0    2 the solutions are
  5
–6 =6, 6
and therefore

=3, 
b)


3 sin  – cos  = 2 sin – 6 = – 1


So
1


sin – 6 = – 2


In the range 0    2 the solutions are
- 38 -
 7 11
–6 = 6 , 6
and therefore
4
 = 3 , 2
iv) a) 3 cos  + 4 sin  = 5 sin( + 0.64) = 1
So
1
sin( + 0.64) = 5 = 0.2
In the range 0    2 the solutions are, to two decimal places
 + 0.64 =  – 0.2 = 2.9 and 2 + 0.2 = 6.48
and therefore
 = 2.3, 5.84 radians
b) 3 cos  + 4 sin  = 5 sin( + 0.64) = – 1
So
1
sin( + 0.64) = – 5 = – 0.2
In the range 0    2 the solutions are
 + 0.64 = 3.34 and 6.28
and therefore
 = 2.7, 5.44
- 39 -