Trigonometry - Mathematics · Trigonometry 1.1 Angular Measure Consider the circle in the plane of...

Chapter 1

Trigonometry

1.1 Angular Measure

Consider the circle in the plane of radius 1, centre the origin. Henceforth this will be known asthe unit circle. Let A be the point (1,0) and let P be any point on the circle.

Degree Measure

P

O Aθ

Figure 3.1.1

Then we shall say that ∠AOP has degree measure θ◦ if

lenth of arc AP

circumference=

θ◦

360,

i.e.θ◦ = 360× length of arc AP

circumference. (1.1)

In this definition please note that the length of arc AP is measured in an anti-clockwise directionfrom A. This is, of course, nothing but the usual definition of the degree measure of an angleused in secondary schools. Thus if P = (0, 1) then

length of arc AP =circumference

4=

2π

4=

π

2.

Henceθ◦ =

360◦

2π× π

2=

360◦

4= 90◦.

Evidently if P = (−1, 0) then θ◦ = 180◦, while if P = (0,−1) then θ◦ = 270◦ (see Figure 3.1.2).

1

2 CHAPTER 1. TRIGONOMETRY

O Az

270◦

Figure 3.1.2

In this way any point P on the unit circle determines a degree measure, and if we take anydegree measure from 0◦ to 360◦ we shall obtain a point P on the unit circle. But why stopthere? Suppose we wish to consider angles of more than 360◦. For example take 380◦. Startingat A = (1, 0), go once around the unit circle, giving 360◦, and then a further 20◦ (Figure 3.1.3).

AO 6

380◦

Figure 3.1.3

Notice that we end up at the same point P as is obtained by considering 20◦.

THIS DOES NOT MEAN 380◦ = 20◦

As a second example consider 1437◦. In this case go three times around the circle, using up3 × 360◦ = 1080◦. There remains 357◦. So we proceed a further 357◦, almost getting back towhere we started.

Now we can see what θ◦ means for any non-negative θ. Can we not attach some meaningto negative angles? What could −45◦ mean? (We could define −45◦ to mean anything wechoose, but we wish to make a sensible and useful definition.) What we should like is that45◦ + (−45◦) = 0◦. Now 45◦ is going to mean moving to a point P anti-clockwise from Athrough 1/8 of a circumference. 0◦ is going to mean remaining at A. Thus it seems sense todefine −45◦ to mean moving to a point Q, say, from A clockwise through 1/8 of a circumference(Figure 3.1.4).

A

Q

O−45◦

Figure 3.1.4

In general if θ◦ is any angle, −θ◦ is the same angle, but the rotation is in the opposite direction.

It may have occurred to you that the number 360 in formula (1.1) was a little curious. Why360 and not 240 or any other number? The reason for the choice of 360 is historical and datesback to Babylonian times. Clearly the number 360 is quite arbitrary and (1.1) would be simplerwithout it. This will give a different kind of angular measure.

1.1. ANGULAR MEASURE 3

Radian Measure

We shall say that ∠AOP has radian measure θ if

θ = length of arc AP (1.2)

O Aθ

P

?

k θ

Figure 3.1.5

In Figure 3.1.5 the arc length AP is θ, so ∠AOP has radian measure θ. Notice that if ∠AOPis a right angle then

θ =circumference

4=

π

2.

Thus

90◦ =π

2radians, 180◦ = π radians,

and

1 radian =180◦

π≈ 57.296◦.

Indeed if θ = 1 radian, then arc AP will have length 1.

P

O A

Y

?1

1 1

Figure 3.1.6

The following are useful to remember:

30◦ = π/6 radians, 45◦ = π/4 radians, 60◦ = π/3 radians, 90◦ = π/2 radians,

180◦ = π radians, 360◦ = 2π radians.

Of course you should not learn these off by heart. Any one of them will do. Practice willyield familiarity.

Figure 3.1.7 may help to illustrate the relationship between degrees and radians. Radianmeasures are written in the interior of the circle, degrees on the outside.


π

7π/4

π/4π/3

0◦

30◦

45◦60◦90◦

120◦

150◦

180◦

240◦315◦

2π/3

5π/6

π/2

0

4π/3

π/6

Figure 3.1.7

Exercise 3.1

1. Sketch the points obtained on the unit circle by the following angles:

(a) 45◦ (b) 210◦ (c) −80◦ (d) 530◦ (e) −390◦ (f) 1800◦.

2. Convert to radians:(a) 45◦ (b) 180◦ (c) 60◦ (d) 270◦ (e) 135◦

(f) 210◦ (g) 240◦ (h) 225◦ (i) 315◦ (j) −30◦

(k) −60◦ (l) −45◦ (m) −75◦ (n) 520◦ (o) −405◦

(p) −315◦ (q) −225◦ (r) −150◦ (s) −135◦ (t) −130◦

(u) 100◦ (v) 5◦ (w) 37◦ (x) 247◦ (y) −543◦

3. Convert to degrees:

(a) 3π/2 rad. (b) π/10 rad. (c) 9π/2 rad. (d) −3π/4 rad.(e) 3 rad. (f) −5π/8 rad. (g) −11π/12 rad. (h) −3π/8 rad.(i) 7π/15 rad. (j) 7 rad. (k) −11π/15 rad. (l) −34π/12 rad.(m) 6.2 rad.

1.2 Trigonometr1c Functions

Let P (x, y) be any point on the unit circle and let A = (1, 0) and O = (0, 0). Let ∠AOP = θand the perpendicular from P to the x-axis meets that axis at K. Then PK = y and OK = x.

...........................

P (x, y)

θO K A

Figure 3.2.1

1.2. TRIGONOMETR1C FUNCTIONS 5

We define:

x = cos θ = cosine θ1x

= sec θ = secant θ (x 6= 0)

y = sin θ = sine θ1y

= csc θ = cosecant θ (y 6= 0)

y

x= tan θ = tangent θ (x 6= 0)

x

y= cot θ = cotangent θ (y 6= 0).

These six functions of θ are called trigonometric functions.

A large number of facts are immediately clear.

(a) Since −1 ≤ x ≤ 1, i.e. |x| ≤ 1, then | cos θ| ≤ 1. Similarly | sin θ| ≤ 1.

(b) It follows directly from the definitions that

tan θ =sin θ

cos θ, cos θ 6= 0 (1.3)

Furthermorecot θ =

cos θ

sin θ, sec θ =

1cos θ

, csc θ =1

sin θ.

(c) Since x2 + y2 = 12, by Pythagoras’ Theorem, then

sin2 θ + cos2 θ = 1. (1.4)

Note that sin2 θ means (sin θ)(sin θ) = (sin θ)2.

This is quite different from sin(θ2).

(d) Dividing (1.4) by sin2 θ (when sin θ 6= 0), we obtain

1 +cos2 θ

sin2 θ=

1sin2 θ

whencecsc2 θ = 1 + cot2 θ (1.5)

Similarly, dividing by cos2 θ, we have

sec2 θ = 1 + tan2 θ. (1.6)

(e) If φ = θ+2π, then the point P ′ on the unit circle determined by φ will be the same as thepoint P determined by θ. In fact φ is the same as θ with one rotation added on. Thensin φ = sin θ (since P ′ = P ), cos φ = cos θ etc. Indeed if n is any integer, then

sin(θ + 2nπ) = sin θ, cos(θ + 2nπ) = cos θ

and the same for the other four trigonometric functions.

(f) Directly from the definitions,

sin 0 = 0, tan 0 = 0, cos 0 = 1.

sinπ/2 = 1, cos π/2 = 0, tan π/2 is undefined

sin π = 0, tanπ = 0, cos π = −1

sin 3π/2 = −1, cos 3π/2 = 0, tan 3π/2 is undefined.


(g) If P is in the first quadrant, sin θ > 0, cos θ > 0, tan θ > 0.

If P is in the second quadrant, sin θ > 0, cos θ < 0, tan θ < 0.

If P is in the third quadrant then sin θ < 0, cos θ < 0, tan θ > 0.

If P is in the fourth quadrant then sin θ < 0, cos θ > 0, tan θ < 0.

It follows that if 0 < θ < π then sin θ > 0 while if π < θ < 2π then sin θ < 0.

Also if −π/2 < θ < π/2, then cos θ > 0, while if π/2 < θ < 3π/2 then cos θ < 0.

Finally, if 0 < θ < π/2 or π < θ < 3π/2 then tan θ > 0, while if π/2 < θ < π or3π/2 < θ < 2π then tan θ < 0. All this information is summarized in Figure 3.2.2.

sin θ > 0cos θ > 0tan θ > 0

sin θ > 0cos θ < 0tan θ < 0

sin θ < 0cos θ < 0tan θ > 0

sin θ < 0cos θ > 0tan θ < 0

III

III IV

Figure 3.2.2

(h) If θ = π/4, then x = y = sin θ = cos θ.

Sincesin2(π/4) + cos2(π/4) = 1

by (1.4), then2 sin2(π/4) = 1,

sosin π/4 = cos π/4 =

1√2, and tan π/4 = 1.

-

6

π/4

1/√

2¾? ...

...

...

...

. P

AO

1/√

2 1............

Figure 3.2.3

(i) If θ =π

3(= 60◦), then ∠OPB = 30◦ = π/6 (see Figure 3.2.4). Hence

OB =12OP =

12.


Also

PB2 = OP 2 −OB2 = 1− 14

=34.

Thus

PB =√

32

,

i.e.

sinπ/3 =√

32

, cosπ/3 =12, tan π/3 =

√3.

1

?

√3/2

-¾

6

...

...

...

...

...P

1/2O A

π/3

.........

Figure 3.2.4

By a similar argument,

sinπ

6=

12, cos

π

6=√

32

, tanπ

6=

1√3.

These results may be set out as follows, in a table of standard angles:

θ 0 π/6 π/4 π/3 π/2 π 3π/2

sin θ 0 1/2 1/√

2√

3/2 1 0 −1

cos θ 1√

3/2 1/√

2 1/2 0 −1 0

tan θ 0 1/√

3 1√

3 undef 0 undef

(j) Let P be the point on the unit circle with coordinates (x, y) and let ∠AOP = θ,i.e. x = cos θ, y = sin θ. Suppose Q is the point with coordinates (x,−y). Evidently∠AOQ = −θ.

.................

−θ

P (x, y)

θO

...........................

A

......

......

.....

Q

Figure 3.2.5


Hence

cos(−θ) = x = cos θ, sin(−θ) = −y = − sin θ, tan(−θ) = −y/x = − tan θ.

In summary,

cos(−θ) = cos θ, sin(−θ) = − sin θ, tan(−θ) = − tan θ. (1.7)

Thus for example, sin(−π/6) = − sin π/6 = −1/2; cos(−π/6) = cos π/6 =√

3/2, etc.

Once the values of any trigonometric function are known from θ = 0 to θ = π/2, all othervalues can be calculated.

Example 1.2.1

Find : (a) sin3π

4(b) tan

11π

6(c) cos

7π

6.

(a) sin 3π/4 is the value of the y coordinate of P when ∠AOP = 3π/4.

....................................

P (x, y)

O

3π/4A

Figure 3.2.6

From Figure 3.2.6 it is clear that sin 3π/4 = sin π/4 = 1/√

2.

(b) It is easy to see from Figure 3.2.7 that

tan = 11π/6 = tan(−π/6) = − tanπ/6 = −1/√

3.

−π/6

P (x, y)

O A11π/6

Figure 3.2.7

(c) From Figure 3.2.8,

cos7π

6= − cos

π

6= −

√3

2.


O

P (x, y)

A7π/6

π/6

Figure 3.2.8

Notice now that we can regard any of the trigonometric functions as either depending onan angle or just on a real number. Thus when we say sin π/4 = 1/

√2, we may have in mind

an angle of 45◦ or the real number π/4 ≈ 0.7854.

We conclude this section with a simple identity.

Example 1.2.2

Prove the identity (cot θ − csc θ)2 =1− cos θ

1 + cos θ.

The way to proceed is to start with one side and show it equals the other. Letting LHSand RHS stand for left-hand side and right-hand side respectively, we have

LHS =(cos θ

sin θ− 1

sin θ

)2

=1

sin2 θ(cos θ − 1)2

=(cos θ − 1)2

1− cos2 θ

=(cos θ − 1)2

(1− cos θ)(1 + cos θ)

= −cos θ − 1cos θ + 1

= RHS.

Exercise 3.2

1. Find, where possible, the exact values of all six trigonometric functions for the followingvalues of θ:(a) 0 (b) 4π/3 (c) 4π/3 (d) 2π/3 (e) π/2(f) π (g) 3π/2 (h) 7π/2 (i) 5π/6 (j) 7π/6(k) −3π/4 (l) −7π/4 (m) −π/3 (n) −π/6 (o) −7π/2(p)−11π/6 (q) 10π/3 (r) 13π/6 (s) 23π/6 (t) 101π/4(u) −13π/3 (v) 3π (w)103π (x) 2780π (y) 53π − π/4

2. Find, where possible, the exact values of all six trigonometric functions for the followingvalues of θ:(a) 30◦ (b) 120◦ (c) 210◦ (d) 135◦ (e) 240◦

(f) 225◦ (g) 315◦ (h) −45◦ (i) −135◦ (j) 420◦

(k) 765◦ (l) −660◦ (m) 1440◦


3. Simplify:

(a) sin(103π

9

)+ sin

(− 103π

9

)(b)

√sin2(π/15 + cos2 π/15

(c) sec2(−π/8)− tan2(−π/8) (d) csc2(π/7)− cot2(π/7)

(e) − tan2 θ + sec2 θ − sin2 θ (f) sin4 θ + 2 sin2 θ cos2 θ + cos4 θ.

4. In what quadrant must θ lie in order that the following be satisfied?

(a) cos θ > 0 and sin θ < 0 (b) cos θ > 0 and tan θ < 0(c) sin θ > 0 and cot θ < 0 (d) csc θ > 0 and sec θ < 0(e) tan θ > 0 and csc θ < 0 (f) sec θ < 0 and csc θ < 0.

5. Prove the following identities:

(a) (1− sin2 θ)(1 + tan2 θ) = 1

(b) (tan θ − sec θ)2 =csc θ − 1csc θ + 1

(c)1− sec θ

tan θ − sin θ= − csc θ

(d)1 + cos2 x

sin2 x= 2 csc2 x− 1.

(e)cot x− tan x

sin x + cos x= csc x− sec x

(f) csc4 x− cot4 x = (1− cot2 x) csc2 x

(g)

√1− cosx

1 + cosx=

1− cos x

| sin x|(h) sin3 x + cos3 x = (1− sinx cosx)(sin x + cos x).

6. Let C1 be the circle passing through the origin with centre (2, 0) and C2 be the circlepassing through the origin with centre (0, 4). The line y = x tan θ cuts C1 again at A andC2 again at B. Let M be the mid-point of AB. Show that M has coordinates

(2 cos2 θ + 4 sin θ cos θ, 2 sin θ cos θ + 4 sin2 θ)

and hence deduce that M lies on the circle x2 + y2 − 2x− 4y = 0.

1.3 Graphs of Trigonometric Functions

In this section we shall discuss the graphs of the elementary trigonometric functions f(θ) =sin θ, f(θ) = cos θ and f(θ) = tan θ as well as some related functions.

(a) f(θ) = sin θ.

Recall that sin θ is the y coordinate of the point P (x, y) on the unit circle (Figure 3.3.1).First note that f(0) = 0. As θ moves from 0 to π/2 , P moves from A to B, y moves from 0 to1 and so f(θ) increases from 0 to 1. Then as θ increases from π/2 to π, P moves from B to C,y moves from 1 back to 0 and thus f(θ) decreases back to 0.

1.3. GRAPHS OF TRIGONOMETRIC FUNCTIONS 11

................C O A

sin θ

6

θ

PB

?

Figure 3.3.1

While θ moves from π to 3π/2, y becomes negative and f(θ) continues to decrease, reaching−1 when θ = 3π/2. Finally, as θ increases to 2π, sin θ increases back to 0 again. After that,the whole process repeats itself, coming back to where it started every time θ increases by 2π.(Recall that f(θ + 2nπ) = sin(θ + 2nπ) = sin θ = f(θ).) The same thing happens when θ isnegative.

Graph of y = sin θ

−π π/2π−π/2

2π 5π/2

y

θ3π/2

−1

1y = sin θ

Figure 3.3.2

Notice that the graph lies entirely between −1 and 1, i.e. range(f) = [−1, 1]. Also each wave(counting a peak and a trough) has length 2π. We say that f is periodic of period 2π.

Once again θ is regarded either as a real number or as an angle measured in radians (andnot degrees).

(b) g(θ) = cos θHere the process is almost the same. Recall that cos θ is the x-coordinate of the point

P (x, y) (see Figure 3.3.3). When θ = 0, P lies on A, x = 1 and so g(0) = 1. After that g(θ)decreases to 0 as θ reaches π/2 and further to −1 when θ is π. Then it increases back to 1when θ goes up to 2π.


θ

cos θ-C O A¾

P

B

..........

Figure 3.3.3

Graph of y = cos θ

−π/2 π/2 π 3π/2 2π

y

θ

−1

1y = cos θ

Figure 3.3.4

Notice that the shape is the same as that of the graph of sin θ but it has been moved tothe left by π/2, i.e. sin θ is obtained by moving the graph of cos θ to the right by π/2. Thuscos(θ − π/2) = sin θ. Hence by (1.7)

cos(π/2− θ) = sin θ. (1.8)

Also, if we move the graph of sin θ to the right by π/2 we shall obtain the same graph as thatof cos θ only “turned over”, that is with the sign changed. Hence

sin(θ − π/2) = − cos θ,

i.e.− sin(π/2− θ) = − cos θ

by (1.7). Hence

sin(π/2− θ) = cos θ. (1.9)

Notice that π/2 − θ is the complement of θ. The first two letters of the word “complement”are used in the word “cosine”. So the cosine is the sine of the complement. In the same way

tan(π/2− θ) = cot θ cot(π/2− θ) = tan θ

sec(π/2− θ) = csc θ csc(π/2− θ) = sec θ.


(c) h(θ) = tan θ.In Figure 3.3.5 tan θ = y/x, which is just the gradient of OP .

θ

x-C O A¾

P.................6

B

?

..........y

Figure 3.3.5

Thus as P moves from A and θ increases to π/2, the slope of OP increases from 0 to becomelarger and larger as P approaches π/2, at which point it is undefined. Then as θ increasesto π the slope changes from very large and negative up to zero, and in the next quadrant itbecomes larger and larger (and positive), becoming undefined at 3π/2. Between 3π/2 and 2πthe slope is negative, moving from very large and negative back to zero. The whole processclearly repeats itself every time θ increases by 2π.

Graph of y = tan θ

−π−π/2 π/2

π3π/2

2π5π/2

y

θ

y = tan θ

Figure 3.3.6

In many practical applications wave forms such as those exhibited by sin θ and cos θ arise.However, the amplitude (how high the graph is), the period (how long the waves are) and thevalue at θ = 0 may be changed.

Example 1.3.1

Sketch the graph of y = 2 sin θ.

This is just like sin θ except that the y values are twice as large, so that the graph willvary between y = −2 and y = 2.


y = sin θ

y = 2 sin θ −2

−1

1

2y

π/2π

2πθ

Figure 3.3.7

Example 1.3.2

Sketch the graph of y = −3 cos θ.

This is just like cos θ, except the values will vary from −3 to 3 and the graph has beenturned over.

−π/2 π/2 π 2π

−3

1

3y = −3 cos θ

y = cos θ

θ

Figure 3.3.8

Example 1.3.3

Sketch y = sin 2θ.

This is similar to sin θ, but now when θ increases from 0 to π/2, y increases to sin 2π/4 =sin π/2 = 1. Then as θ moves from π/4 to π/2, y returns to 0. This means everything ishappening twice as fast and we get back to where we started when θ reaches π.


π/2 π 3π/2 2πθ

yy = sin θ

y = sin 2θ

−1

1

Figure 3.3.9

Example 1.3.4

Sketch the graph of y = cos(θ/2).

This is very similar to the previous example, but things are now happening only half asfast as for y = cos θ.

−1

1y

−π/2 π/2 π 2π

y = cos θ

y = cos(θ/2)

Figure 3.3.10

Example 1.3.5

Sketch y = sin(θ + 3π/2).

This is just sin θ shifted to the left by 3π/2.

−1

1y

π/2 π 3π/2 2πθ

y = sin θ

y = sin(θ + 3π/2)

Figure 3.3.11

It is interesting to note that the graph we obtain is the same as that obtained by shiftingsin θ to the right by π/2, i.e.

sin(θ + 3π/2) = sin(θ − π/2).

Also it is evident that the graph is just cos θ “turned over”, i.e.

sin(θ + 3π/2) = − cos θ.


We can combine all three ideas of the last three examples together.

Example 1.3.6

Sketch y = 3 sin(2θ + π/2).

Firstly the amplitude is 3, i.e. y ∈ [−3, 3]. Secondly the graph repeats itself everytime θ increases by π, i.e. the wavelength or period is only π. Finally, when θ = 0,y = 3 sin(π/2) = 3.

−3

3

−π/4 π/4π/2

3π/42π

5π/4

y = 3 sin(2θ + π/2)

Figure 3.3.12

Finally let us sketch the graph of y = csc θ.

First recall that csc θ = 1/ sin θ. Thus when sin θ = 0, csc θ is undefined. When θ is smalland positive, sin θ is also small and positive and so csc θ is large and positive (for example ifsin θ = 1/100, then csc θ = 100). Now as θ increases to 1 at π/2, csc θ decreases to 1. Notethat if θ ∈ (0, π/2), then sin θ ∈ (0, 1) and so csc θ ∈ (1,∞). In the second quadrant the reversehappens. Thus sin θ decreases, so csc θ increases. In the third and fourth quadrant the samesort of thing happens, but the signs are reversed. Figure 3.3.13 shows the graphs of y = sin θand csc θ on the same axes.

−π π 2π−1

1

y = csc θ

y = sin θ

Figure 3.3.13

The graphs of y = sec θ and y = cot θ can be drawn in a similar way.

1.4. ADDITION AND DOUBLE ANGLE FORMULAE 17

Exercise 3.3

1. Sketch the graphs of:(a) y = sin 3θ (b) y = 2 cos θ (c) y = 2 sin θ(d) y = cos 4θ (e) y = 3 cos θ (f) y = − sin 4θ(g) y = 3 cos 2θ (h) y = 1 + sin θ (i) y = −4 sin(θ − π/2)(j) y = 3 cos(2θ − pi) (k) y = 4 cos(θ/2) (l) y = 3sin(π/6 + θ/2)(m) y = 1/(1 + sin θ) (n) y = | sin θ| (o) y = tan 2θ(p) y = − tan θ (q) y = 1 + tanθ (r) y = tan(θ + π/2)(s) y = −2 tan 3θ (t) y = sec θ (u) y = cot θ(v) y = sec 2θ (w) y = sec θ/2 (x) y = csc(θ + π)(y) y = 4 cot(θ − 3π/2) (z) y = −2 sec(2π − θ)

1.4 Addition and Double Angle Formulae

It is useful in many applications to have formulae for the values of such expressions as sin 2θ,cos(θ + φ), etc. It is tempting to think that

sin 2θ = 2 sin θ, cos(θ + φ) = cos θ + cos φ

etc.

These are totally WRONG.

The situation is more complicated. For example if we take θ = π/4, then 2θ = π/2. Hence

sin 2θ = sin π/2 = 1, but 2 sin θ = 2 sin π/2 =2√2.

Further, taking θ = φ = 0,

cos(θ + φ) = cos 0 = 1, but cos θ + cos φ = cos 0 + cos 0 = 2.

Writing f(θ) = sin θ or cos θ or tan θ it is NOT TRUE that

f(θ + φ) = f(θ) + f(φ) or that f(kθ) = kf(θ). (1.10)

In fact, the only functions for which the relations (1.10) hold are straight lines through theorigin. THEY DO NOT HOLD FOR ANY TRIGONOMETRIC FUNCTIONS.

Now let P be any point on the unit circle such that ∠AOP = θ, let Q be a point on the unitcircle such that ∠AOQ = θ +φ, and let R be a point on the unit circle such that ∠AOR = −φ,where as usual, A is the point (1,0) (see Figure 3.4.1).

Notice then that∠AOQ = ∠ROP = θ + φ.

Hencechord AQ = chord RP. (1.11)

Furthermore,P = (cos θ, sin θ), Q = (cos(θ + φ), sin(θ + φ)),

andR = (cos(−φ), sin(−φ)) = (cos φ,− sinφ).

Now by (1.11),AQ = RP.


−φθ

φ

Y

?

θ + φ

O

R

A

P

Q

Figure 3.4.1

Using the distance formula, we have

AQ2 = [cos θ + φ)− 1]2 + [sin(θ + φ)]2

= cos2(θ + φ)− 2 cos(θ + φ) + 1 + sin2(θ + φ)= 2− 2 cos(θ + φ)

and

RP 2 = [cos θ − cosφ]2 + [sin θ + sin φ]2

= cos2 θ − 2 cos θ cosφ + cos2 φ + sin2 θ + 2 sin θ sin φ + sin2 φ

= 2 + 2 sin θ sin φ− 2 cos θ cosφ.

Since AQ2 = RP 2, we obtain

cos(θ + φ) = cos θ cos φ− sin θ sin φ. (1.12)

Also, since cos(−φ) = cos φ and sin(−φ) = − sin φ, changing φ to −φ in (1.12) yields

cos(θ − φ) = cos θ cos φ + sin θ sin φ. (1.13)

Equations (1.12) and (1.13) may be summarized in the crucial addition formula below.

Addition Formula for Cosine

cos(θ ± φ) = cos θ cos φ∓ sin θ sin φ (1.14)

Now let us consider sin(θ + φ). By (1.8) and (1.9),

sin(π/2− x) = cos x, and cos(π/2− x) = sin x.

Thus

sin(θ + φ) = cos(π/2− (θ + φ))= cos((π/2− θ)− φ)= cos(π/2− θ) cos φ + sin(π/2− θ) sin φ (by (1.14))= sin θ cosφ + cos θ sin φ (by (1.8) and (1.9)).

Similarly,sin(θ − φ) = sin θ cos φ− cos θ sin φ,

and we obtain the fundamental formula below.


Addition Formula for Sine

sin(θ ± φ) = sin θ cos φ± cos θ sin φ (1.15)

Finally let us consider tan(θ + φ). By (1.3) we have

tan(θ + φ) =sin(θ + φ)cos(θ + φ)

=sin θ cosφ + cos θ sin φ

cos θ cos φ− sin θ sin φ

(by (1.14) and (1.15)). Now dividing the numerator and denominator by cos θ cosφ, we obtain

tan(θ + φ) =tan θ + tan φ

1− tan θ tan φ

A similar formula may be derived for tan(θ − φ).

Addition Formula for Tangent

tan(θ ± φ) =tan θ ± tan φ

1∓ tan θ tan φ(1.16)

Example 1.4.1

Evaluate sinπ

12and cos

7π

12.

Nowπ

12=

π

3− π

4, while

7π

12=

π

3+

π

4.

Thus (by 1.15)

sin π/12 = sin(π/3− π/4)= sin π/3 cos π/4− cos π/3 sin π/4

=√

32

1√2− 1

21√2

=√

3√

24

−√

24

=√

6−√24

.

Similarly

cos7π

12= cos(π/3 + π/4)

= cos π/3 cos π/4− sin π/3 sin π/4

=12

1√2−√

32

1√2

=1−√32√

2=√

2−√64

.

The following formulae are also very useful. Setting θ = φ in (1.14), we have

cos 2θ = cos2 θ − sin2 θ.


Now bearing in mind that cos2 θ + sin2 θ = 1, it follows that

cos2 θ = (1− sin2 θ)− sin2 θ = 1− 2 sin2 θ,

and alsocos 2θ = cos2 θ − (1− cos2 θ) = 2 cos2 θ − 1.

Double Angle Formulae for Cosine

In summary we obtain the double-angle formulae:

cos 2θ =

cos2 θ − sin2 θ

1− 2 sinθ

2 cos2 θ − 1(1.17)

Double Angle Formula for Sine

Also setting φ = θ in (1.15), we obtain

sin 2θ = 2 sin θ cos θ (1.18)

while from (1.16) we have a similar formula for tan 2θ.

Double Angle Formula for Tangent

tan 2θ =2 tan θ

1− tanθ(1.19)

It is obviously a nearly impossible task to memorize all of these formulae – and there aremore to come! Some however are essential. In particular (1.3), (1.4), (1.14) and (1.15) shouldbe learned. It should be known how to derive all the others. If a large number of practiceexercises are done, many of the others will stick in ones memory without conscious effort.

Example 1.4.2

Prove that 2 sin2 x = tanx sin 2x.

Now

tan x sin 2x = 2 tan x sin x cosx by (1.18)

= 2sin x

cos xsin x cosx by (1.3)

= 2 sin2 x.

Example 1.4.3

Prove that (sinx + cos x)2 = 1 + sin 2x.

Multiplying out the left-hand side, we have

(sinx + cosx)2 = sin2 x + 2 sin x cosx + cos2 x

= 1 + 2 sin x cosx (by (1.4))= 1 + sin 2x, by (1.18).


Here is an incorrect “proof” of the above.

(sin x + cos x)2 = 1 + sin 2x,

sosin2 x + 2 sin x cosx + cos2 x = 1 + 2 sin x cosx

or1 + 2 sin x cos x = 1 + 2 sin x cos x,

i.e.0 = 0.

Note that in this case what was required has NOT been proved. What has been proved is thatif (sin x + cos x)2 = 1 + sin 2x, then 0 = 0. This is quite useless. Everybody knows that 0 = 0anyway.

To emphasize the point consider the following “proof” that 2 = 1.

2 = 1.

So0.2 = 0.1

or0 = 0.

Exercise 3.4

1. (a) Prove that cos 3θ = 4 cos3 θ − 3 cos θ. [Hint : cos 3θ = cos(2θ + θ).]

(b) Derive a similar formula for sin 3θ.

(c) Use (a) and (b) to deduce an expression for tan 3θ in terms of tan θ.

2. By using the addition formulae, evaluate exactly:

(a) sin 105◦ (b) cos 75◦ (c) tan 15◦.

3. (a) Show that sinθ

2= 2 sin

θ

2cos

θ

2.

(b) Deduce formulae for cos θ and tan θ in terms of θ/2.

(c) Evaluate cos π/8, sin π/8 and tan π/8.

4. Suppose a = sin 18◦. Express the following in terms of a:

(a) sin(−18◦) (b) cos2 18◦ (c) cos 108◦ (d) tan 162◦ (e) cos 288◦.

5. Verify that x = 18◦ satisfies the equation sin 2x = cos 3x. Using the double angle formulaand the formula for cos 3x given above in question 1, obtain an equation for sin 18◦, andhence find sin 18◦.

6. Prove the (very easy) identity 2 cos 2x = 1 + cos 2x. Hence sketch the curve y = 2 cos2 x.

7. Generalise formulae (1.14) and (1.15) to obtain formulae for sin(x+y+z) and cos(x+y+z).

8. Obtain formulae for cot(x + y) in terms of cot x and cot y, and sec(x + y) in terms ofsecants and tangents.

9. Obtain formulae for cot 2x and sec 2x in terms of cot x and sec x respectively.


10. Prove the following identities:

(a) sin(π − θ) = sin θ (b) cos(π + θ) = − cos θ

(c) sin 4θ = 4 sin θ cos θ cos 2θ (d) tan θ =sin 2θ

1 + cos 2θ

(e) tan θ + cot θ = sec θ csc θ (f)1 + cot θ

tan θ=

sin θ + cos θ

sin θ − cos θ

(g) (tan θ + sec θ)2 =1 + sin θ

1− sin θ(h) 8 cos4 θ = 3 + 4 cos 2θ + cos 4θ.

(i) (sin θ − cos θ)2 = 1− sin 2θ (j) cot θ − tan θ = 2 cot 2θ

(k) tan θ + cot θ = 2 csc 2θ (l) sin(x + y) sin(x− y) = sin2 x− sin2 y

(m) cos(x + y) cos(x− y) = cos2 x− sin2 y

11. Use the identities for cos(A−B) and cos(A + B) to prove that

1 + cos(A−B) cos(A + B) = cos2 A + cos2 B.

Hence show that 1 + cos 2x = 2 cos2 x and

3+cos(x−y) cos(x+y)+cos(y−z) cos(y+z)+cos(z−x) cos(z+x) = 2(cos2 x+cos2 y+cos2 z).

12. Suppose θ + φ = π/4.

(a) Show that tan φ =1− tan θ

1 + tan θ. (b) Hence prove that (1 + tan θ)(1 + tan φ) = 2.

(c)Deduce that tan π/8 =√

2− 1.

13. Show that:

(a) tan 2A =2 tan A

1− tan2 A(b) tan 3A =

3 tan A− tan3 A

1− 3 tan2 A.

From (a) deduce that tan θ =2t

1− t2, where t = tan θ/2. Derive similar expressions for

cos θ and sin θ in terms of t .

1.5 Applications to Triangles etc.

The treatment so far may seem very far removed from secondary school trigonometry. Thisdifference is, in fact, very slight. We shall then apply the theory to the solution of triangles aswell as to the derivation of a number of other simple facts.

θO

KC

y

xL B

Figure 3.5.1

1.5. APPLICATIONS TO TRIANGLES ETC. 23

Suppose then we have a triangle OBC in which ∠OBC is a right angle. Fix coordinates sothat the origin is O and the positive x-axis lies along OB. Suppose OC = r and ∠BOC = θ.With centre O and radius 1 draw another circle. This will cut OC (or OC produced) in K.Let L be the foot of the perpendicular from K to OB (see Figure 3.5.1).

Then by definition

sin θ = KL

=KL

OK(since OK = 1)

=CB

OCby similar triangles.

In the same way cos θ = OL = OL/OK = OB/OC and tan θ = CB/OB.

O B

C

θ

Figure 3.5.2

Deleting all the excess information and concentrating only on 4OBC, (Figure 3.5.2), we see

sin θ =CB

OP=

“opposite”“hypoteneuse”

cos θ =OB

OP=

“adjacent”“hypoteneuse”

tan θ =CB

OB=

“opposite”“adjacent”

.

These are just secondary school definitions. Notice, however, that because of our definitionsin Section 3.2, it is now clear why, if θ is obtuse, we should have cos θ < 0 and tan θ < 0. Alsoby the very definition of triangles this method will not work unless θ ∈ (0, π), whereas ourmore advanced definition from Section 3.2 gives a meaning to the trigonometric functions forall θ ∈ R.

Example 1.5.1

Let ABC be a triangle with B a right angle, a = 2, and b = 3. (Here a is the side opposite∠A etc.) Find c and ∠C.


a = 2C B

A

b = 3

Figure 3.5.3

By Pythagoras’ TheoremAC2 = AB2 + BC2,

so 9 = AB2 + 4 andAB =

√5.

Also BC/AC = cos C. Thus cos C = 2/3 and hence C = 48.19◦ (using a calculator).

(a) Area of a Triangle, Sine Rule.Suppose ABC is any triangle with ∠BAC = θ. Let D be the foot of the perpendicular from

B to AC (Figure 3.5.4).

.......................¾ -=

>

A

B

CD

b

c

θ

Figure 3.5.4

Nowsin θ =

BD

BA,

soBD = BA sin θ = c sin θ.

Hencearea ABC =

12

base × height =12

AC.BD =12

bc sin θ =12

bc sin A.

Area of a Triangle

Thus

area of triangle =12

bc sin A (1.20)

Similarly

area =12

ab sin C =12

ac sin B.

Applications of this formula are well-known from secondary school. It follows that

ab sin C = ac sin B = bc sin A.


Sine Rule

Dividing by abc, we obtain the sine rule:

sin A

a=

sin B

b=

sin C

c(1.21)

(b) Cosine RuleLet ABC be a triangle. Arrange axes so that A = (0, 0), B = (c, 0) and C = (x, y). Let the

foot of the perpendicular from C to AB be K (see Figure 3.5.5).

NowCK

CA= sin A and

AK

CA= cos A,

i.e.CK = CA sinA and AK = CA cos A.

.......................¾ -

a

B(c, 0)A

C(x, y)

Kc

b

Figure 3.5.5

Soy = b sin A and x = b cos A.

By the distance formulaBC2 = (c− x)2 + y2,

i.e.a2 = (c− b cosA)2 + (b sinA)2.

Hence using (1.4),

a2 = b2 cos2 A− 2bc cosA + c2 + b2 sin2 A = b2 + c2 − 2bc cosA.

Cosine Rule

This is the cosine rule:a2 = b2 + c2 − 2bc cos A (1.22)

Of course, the results

b2 = a2 + c2 − 2ac cos B and c2 = a2 + b2 − 2ab cos C

follow in exactly the same way.


(c) Arc Length, Segment AreaSuppose we have a circle centre O, radius r. Let P and Q be points on the circle. We wish

to find the length of arc PQ. Let ∠POQ = θ (see Figure 3.5.6).

O P

Q

θ

Figure 3.5.6

If the angle at the centre were 2π, the arc length would be the circumference, i.e. 2πr. Hence,

if the angle at the centre is θ, the arc length isθ

2π× 2πr = rθ.

Arc Length

We have shown

arc length PQ = rθ (1.23)

Notice that this formula works only when θ is measured in radians.

Example 1.5.2

Let C be a circle with centre O, radius 3. Find the arc length subtended by an angle of60◦ at the centre (Figure 3.5.7).

O

360◦

Figure 3.5.7

Now 60◦ = π/3 radians, so the arc length is rθ = 3π

3= π.

Next suppose C is a circle centre O, radius r. Let P and Q be points on the circumference.Suppose we wish to find the area of the segment POQ (see Figure 3.5.8).


O

r

θ

Figure 3.5.8

If the angle at the centre were 2π, the area described would be πr2. So if the angle at thecentre is θ (in radians) then the area described is

A =12πr2.

Segment Area

A =12r2θ (1.24)

Again this formula works only for θ measured in radians. The simplicity of formulae (1.23)and (1.24) gives another reason why it is advantageous to work in radians. The reader shouldwork out what the corresponding formulae are when θ is measured in degrees.

Exercise 3.5

1. Suppose 0 < θ < π/2 and sin θ = 3/4. Draw a right-angled triangle one of whose anglesis θ. Find(a) cos θ (b) tan θ (c) cot θ(d) sin 2θ (e) sin(π/2− θ) (f) cos(π/2 + θ)(g) cos(3π/2− θ) (h) sin(π/3 + θ) (i) sec(π + θ)(j) csc(3π/2 + θ) (k) tan(53π + θ) (l) csc(55π/2− θ)

2. Let C be a circle of radius 4, centre O. Points P and Q lie on C and ∠POQ = 45◦.Calculate the area of the shaded sector shown.

O

Q

P45◦

3. Solve the following triangles ABC, with the standard notation, i.e. find the unknownsides and angles. (sin or cos of the angle is enough). Find also the area of the triangles.

(a) a = 5, b = 6, c = 7 (b) a = 3, b = 3, C = 30◦

(c) a = 3, B = 45◦, C = 30◦ (d) a = 3, b = 4, C = 60◦

(e) a = 3, b = 5, c = 7 (f) cos C = 11/14, b = 7, a = 3(g) a = 3, c = 5, B = 120◦ (h) B = 90◦, b = 3, sin A = 2/3.


4. The triangle ABC is defined by: a = 1 ; b =√

3; A = 30◦. Find sinB. Deduce that thereare 2 triangles which satisfy these conditions, and find the 3 sides and 3 angles of each.

5. Angle A is acute with tan A = 1/2. Angle B is obtuse with tan B = −1/3. Find withoutusing a calculator the value of cos(A−B).

6. The triangle ABC is defined by: B = 90◦; sin A = 1/3; b = 3. Find the perpendiculardistance from B to AC.

7. A parallelogram has sides of 4 and 7 and an included angle of 60◦. Find the lengths ofthe diagonals and the area of the parallelogram.

8. Show that the points P = (a cos θ, b sin θ) and Q = (−a sin θ, b cos θ) lie on the ellipse

x2

a2+

y2

b2= 1.

Prove that, as θ varies, the values of OP 2+OQ2 (where O is the origin) remains constant.

9. (a) By using the cosine rule with the usual notation show that

2ac(1− cos B) = (b + c− a)(b− c + a),

and obtain a similar expressions for

2ac(1 + cos B).

(b) Using formula (1.20), deduce that

Area(4ABC) =√

s(s− a)(s− b)(s− c),

where s = (a + b + c)/2.

(c) Find the area of the triangle whose sides are 4, 5 and 7.

1.6 The Solution of Trigonometric Equations

In this section we shall investigate how to solve various trigonometric equations. Usually thesecan be reduced to one of the forms cosθ = a, sin θ = a, or tan θ = a. We wish to determine θ.This is not always easy. In general there may be many solutions and sometimes no solution.For example the very simple equation

cos θ = 1 (1.25)

is clearly satisfiedby θ = 0; but it is equally satisfied by θ = ±2π,±4π, . . . The solution is thusθ = 2nπ, n ∈ Z, or if you prefer the solution set is {2nπ | n ∈ Z}.

By way of contrast, the equation cos θ = 2 has no solutions, since | cos θ| ≤ 1.

The Equation cos θ = a

We shall now suppose that |a| ≤ 1, so that the equation has solutions.

We wish to determine for what values of θ we have cos θ = a. By tables (or calculator,previous knowledge or whatever) it will be possible to find a value θ1 of θ such that 0 ≤ θ1 ≤ πand cos θ1 = a (see Figure 3.6.1).

1.6. THE SOLUTION OF TRIGONOMETRIC EQUATIONS 29

−π/2 π/2 π 3π/2 2π

θ3 θ1 θ4 θ2

θ3 = −θ1 θ2 = θ1 + 2π

y = ay = cos θ

−1

1

Figure 3.6.1

This means θ1 is one solution. Then clearly if θ2 = θ1 + 2π, we have

cos θ2 = cos(θ1 + 2π) = cos θ1 = a.

So θ2 is another solution, and indeed θ1 + 2nπ will be a solution for any integer n. Howeverthis is not all. If θ3 = −θ1, then

cos θ3 = cos(−θ1) = cos θ1 = a,

so that θ3 is also a solution and so is θ3 + 2nπ for each integer n. Hence all solutions are

±θ1 + 2nπ, n ∈ Z.

This can be visualised in another way using the unit circle. We need to find all values of θsuch that the x coordinate of the corresponding point on the circumference is a. Suppose θ1 issuch a value with corresponding point P .

....................................

θ1

0

P

a

Q

Figure 3.6.2

Then clearly θ1 +2nπ, n ∈ Z will generate the same point on the circumference, and hence thesame x-coordinate.

Furthermore, at Q (see Figure 3.6.2) the x-coordinate is still a and the angle is −θ1. Anymultiple of 2π added on will amount to a number of full rotations added on, taking us back towhere we started, at −θ1 + 2nπ, n ∈ Z and all solutions are given by

±θ1 + 2nπ, n ∈ Z.


Example 1.6.1

Solve cos θ =12.

Now cos π/3 = 1/2. So all solutions are θ = ±π/3 + 2nπ, n ∈ Z.

Example 1.6.2

Solve the equation cos θ = −√

32

.

We firstly note that cos 5π/6 = −√3/2.

Hence all solutions are θ = 5π/6 + 2nπ, n ∈ Z.

Example 1.6.3

Find all values of θ in degrees between 0◦ and 360◦ for which cos 3θ = 1.

Now noting cos 0 = 1 all solutions are given by

3θ = 0 + 2nπ,

i.e.θ =

2nπ

3, n ∈ Z.

Changing to degrees,θ = n120◦, n = 0,±1,±2, . . .

We want only values between 0◦ and 360◦, so θ = 0◦, 120◦, 240◦ or 360◦, taking n = 0, 1, 2, 3.

The Equation sin θ = a

Again we suppose |a| ≤ 1, so that the equation does have solutions. We wish to determine forwhat values of θ we have sin θ = a. Suppose one solution θ1 between −π/2 and π/2 is known,so that sin θ1 = a.

−π/2 π/2 π 3π/2 2π

θ1 θ3 θ2

θ3 = π − θ1 θ2 = θ1 + 2π

y = ay = sin θ

−1

1

Figure 3.6.3

Clearly (see Figure 3.6.3), if θ2 = θ1 + 2π, then θ2 is a solution, since

sin θ2 = sin(θ1 + 2π) = sin θ1 = a.

Indeed, θ1 + 2nπ is a solution for any integer n, since

sin(θ1 + 2nπ) = sin θ1 = a.


Notice however that θ3 = π − θ1 is also a solution. For

sin θ3 = sin(π − θ1)= sin π cos θ1 − cosπ sin θ1

= 0− (−1) sin θ1 = sin θ1 = a.

Hence (π − θ1) + 2nπ is also a solution for any n ∈ Z, and the general solution is

θ1 + 2nπ or (π − θ1) + 2nπ, n ∈ Z.

Alternatively, using the unit circle, we wish to find all values of θ such that the y-coordinateof the corresponding point is a (Figure 3.6.4). If θ1 is such a value with corresponding point Pthen clearly so are the points θ1 + 2nπ, n ∈ Z.

θ1

0

aP................................Q

Figure 3.6.4

However, it is also easy to see that the point Q, on the circumference, corresponding to thevalue π − θ1, also has the same y-coordinate and so will any number of full rotations added tothis. This the complete solution is

θ1 + 2nπ, n ∈ Z or π − θ1 + 2nπ, n ∈ Z.

Example 1.6.4

Solve sin θ =12.

We know thatsin

π

6=

12.

Hence the general solution is

θ =π

6+ 2nπ or

(π − π

6

)+ 2nπ, n ∈ Z,

i.e.θ =

π

6+ 2nπ or

5π

6+ 2nπ, n ∈ Z.

Example 1.6.5

Solve sin θ = −√

32

.

We knowsin(−π/3) = −

√32.


Henceθ = −π

3+ 2nπ or π −

(− π

3

)+ 2nπ, n ∈ Z

i.e.θ = −π

3+ 2nπ or

4π

3+ 2nπ, n ∈ Z.

Example 1.6.6

Find all values of θ between 0◦ and 90◦ for which sin 10θ = 0.

Now sin 0◦ = 0, so the general solution is given by

10θ = 0 + 2nπ or (π − 0) + 2nπ, n ∈ Z,

i.e.10θ = nπ = n180◦

Thusθ = 18n◦.

We want solutions between 0◦ and 90◦, hence

θ = 0◦, 18◦, 36◦, 54◦, 72◦ or 90◦.

The Equation tan θ = a

−π π 3π/2θ1 θ2

θ2 = θ1 + π

y = a

Figure 3.6.5

This is the simplest case. If θ1 is a solution between −π/2 and π/2, then θ2 = θ1 + π is also asolution, and indeed the general solution is

θ = θ1 + nπ, n ∈ Z,

sincetan(θ1 + nπ) =

tan θ1 + tanπ

1− tan θ1 tanπ=

tan θ1 + 01− 0

= tan θ1 = a.

Example 1.6.7

Solve the equation tan θ = 1.

We note that tan π/4 = 1. So θ = π/4 + nπ, n ∈ Z.


Example 1.6.8

Solve the equation cot 2θ = −√3.

This is equivalent to tan 2θ = −1/√

3.

Now we knowtan(−π/6) = −1/

√3.

Hence2θ = −π

6+ nπ, n ∈ Z

soθ = − π

12+

nπ

2, n ∈ Z.

Now that we have these preliminaries out of the way, we can tackle some harder examples.

Example 1.6.9

Solve sin x + sin 3x + sin 5x = 0.

We have, using (??),2 sin 3x cos 2x + sin 3x = 0,

i.e.sin 3x(2 cos 2x + 1) = 0.

Thussin 3x = 0 or cos 2x = −1

2,

so3x = nπ or 2x = ±2π

3+ 2nπ.

Hencex =

nπ

3or ± π

3+ nπ, n ∈ Z.

Example 1.6.10

Solve the equation cos 6θ + cos 4θ + cos 2θ + 1 = 0, for θ between 0◦ and 360◦.

From (26.8) and (28.8),2 cos 5θ cos θ + 2 cos2 θ = 0,

i.e.cos θ(cos 5θ + cos θ) = 0.

Hencecos θ 2 cos 3θ cos 2θ = 0.

Thuscos θ = 0 or cos 3θ = 0 or cos 2θ = 0.

Soθ = ±π

2+ 2nπ or 3θ = ±π

2+ 2nπ or 2θ = ±π

2+ 2nπ,

where n ∈ Z.

Setting n = 0, 1, 2 and 3 and remembering that we wish to have solutions only between0 and 2π, we obtain

θ =π

2,

3π

2,

π

6,

5π

6,

7π

6,

11π

6,

π

4,

3π

4,

5π

4or

7π

4,

which are easily put into degrees.


Exercise 3.6

1. Find all solutions to:

(a) sin x = 0 (b) cos x = 0 (c) cos 4x = 0(d) cos x = 1 (e) tan 3x = 0 (f) tan x = −1(g) sec x = 1 (h) cos x = −1 (i) csc x = 1(j) sec x = −1 (k) sin x = 1/2 (l) cos x = −1/2(m) csc x = 2 (n) sec x = −2 (o) sin x > 0(p) sec x ≥ 0 (q) sin 2x = 1 (r) cos 2x = −1(s) tan x = −√3 (t) cot x =

√3 (u) 2 sin2 x = 1

(v) 2 sin2 3x = 1 (w) sin x = −√3/2 (x) cos2 x = 1(y) 3 tan2 x = 1 (z) tan2 x = 1.

2. Find all solutions x, lying in the interval [0, 2π] to the following:

(a) cos x = 0 (b) cos 3x = 1 (c) sin 2x = 1(d) cos x = −1/2 (e) sin2 x = 1/2 (f) sin x < 0(g) cos 6x = 0 (h) sin 2x < 0 (i) 2 sin2 3x = 1(j) tan(x/2) = 1 (k) tan2 x = 1 (l) sec 2x = 2(m) tan x ≤ 0 (n) tan(x/2) ≤ 0 (o) cot 2x = 1(p) 4 sin2 x = 1 (q) tan2 x = 3

3. Suppose 0 < α < π/2 and cot α = 4/3.

(a) By drawing a suitable triangle, or otherwise, find cosα and sin α.

(b) Now find in terms of α all solutions to the following equations:

(i) sin x = 3/5 (ii) cos x = −4/5 (iii) cos x = −3/5 (iv) tan 2x = 4/3.

4. Suppose 0 < α < π/2 and sin α = 5/13.

(a) Draw an appropriate triangle and find cos α and cos 2α.

(b) Now find in terms of α all solutions to the following equations:

(i) cot x = 5/12 (ii) sin x = −5/13 (iii) sin 3x = 12/13 (iv) sec 2x = 13/12.

5. By writing the expression in the form C sin(x+α) or C cos(x+α) find the general solutionof the following:

(a) cos x− sin x = 1/√

2 (b) cosx− sin x = 1 (c) cos x−√3 sin x = 1

(d)√

3 cos x− sin x = 1.

6. (a) Solve the equation sin x = −1/2.

(b) Sketch the graph of y = 1/2 + sin x, showing points of intersection with the axes.

(c) From your graph or otherwise solve the inequality 0 < 1/2 + sin x.

7. (a) Solve tan x = 1 (for all x).

(b) Sketch the graph of y = −1 + tan x

(c) Solve the inequality tan x ≥ 1.

8. Find the general solutions of the following:


(a) cos3 x = cos x (b) sin3 x = sin x(c) sin 2x = cos x (d) sin2 x = cos2 x(e) sin(x + π/6) = 0 (f) 3 cos2 x + sin2 x + 3 cos x = 3(g) sin x = sin 3x (h) tan4 x− 4 tan2 x + 3 = 0(i) 4 sin2 x cosx = cos x (j) sin2 x = 1− cos x(k) cos 3x = 3 cos x (l) 2 sin3 x− 3 sin2 x + sin x = 0(m) 4 cos 2x = 4 sin x + 5 (n) sin 3x = sin x(o) 2 sin2 x− 5 sinx = −2 (p) cos2 x = − sin2 x− 2 sin x(q) 2 cos3 x− 2 cos2 x− cosx = −1 (r) x2(1/2− cosx) = cosx− 1/2(s) 2 cos 2x > 1 (t) sin x = 1/2 and tan x < 0(u) cos x = 1/2 and sin x = −√3/2.

9. Find all solutions x lying in the interval [−π, π] to:

(a) 2x sin x = x (b) 2 sin 2x = 1(c) sin x + cos x = 0 (d) sin x + cos x = 1(e) sin x = cos 2x (f) cos x = 2 sin2 x− 1(g) sin 2x sin x = cos x (h) tan2 2x = 3(i) cos 3x = −2 cos x (j) 2 sin 2x sin x + 3 cos 2x = 0(k) cos 2x− 3 cos x + 2 = 0 (l) 2 sin2 x + 3 cos x ≤ 3(m) tan x +

√3 cotx = 1 +

√3 (n) |2 tan x− 1| = 1

(o) |2 tan x− 1| < 1 (p) 4 sin3 x− 4 sin2 x− 3 sin x = −3

10. (a) Show that if x ≥ 0, then −x ≤ x sin x ≤ x.

(b) Find where the curve y = x sin x meets the straight lines y = 0, y = −x and y = x.

(c) Sketch the graphs of y = x sin x, y = 0, y = −x and y = x on the same diagram.

11. The triangle ABC is defined by: B = 90◦, a = 12, c = 5. Find:

(a) b (b) sin A (c) cos C (d) cot A.

Find in terms of A all solutions to the equation tanx = −12/5.

12. The triangle ABC is defined by: B = 90◦, sin A = 3/4, b = 4. Find:

(a) a (b) c (c) csc C (d) cos A.

Now find in terms of A all solutions to the equations

(e) sin x = −3/4 (f) sin 2x = 3/4.

13. The triangle ABC is defined by : B = 90◦, cos A = 3/4, b = 4. Find:

(a) a (b) c (c) sin(π −A) (d) cot A (e) cos(π −A) (f) tan A.

Find in terms of A all solutions to:

(g) cos x = −3/4 (h) sin 2x = sin 7A.

14. A certain student believes that sin 2x = 2 sin x for any x. Give an example to prove himwrong, and find all values of x for which he is right. Give a sketch of the curves y = sin 2xand y = 2 sin x on the same diagram.

15. (a) Solve the equation sin(1/x) = 0.

(b) Sketch the curve y = sin(1/x).

16. By considering sin 3θ = 1/2, and expressing the left-hand side in terms of sin θ, show thatthe equation 8x3 − 6x + 1 = 0 has roots sin π/18, sin 5π/18 and − sin 7π/18.

17. Prove that sin 5A − sin A + sin 2A = sin 2A(2 cos 3A + 1). Hence solve the equationsin 5θ + sin 2θ = sin θ for values of θ in the interval 0◦ < θ < 180◦.

18. (a) Factorise x3 − 4x2 + 6x− 4


(b) Show that if4− tan θ = 5 sin θ cos θ

then tan θ is a root of x3 − 4x2 + 6x− 4 = 0.

(c) Hence find the values of tan θ for which

4− tan θ = 5 sin θ cos θ.

Trigonometry - Mathematics · Trigonometry 1.1 Angular Measure Consider the circle in the plane of...

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