Chapter 10: Trigonometry
10.1 The Six Trigonometric Functions
The three standard functions:
For an acute angle \(\theta\) in a right-angled triangle:
- \(\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}\)
- \(\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}\)
- \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}}\)
The three reciprocal functions:
| Function |
Notation |
Reciprocal of |
| Secant |
\(\sec \theta\) |
\(\frac{1}{\cos \theta}\) |
| Cosecant |
\(\cosec \theta\) |
\(\frac{1}{\sin \theta}\) |
| Cotangent |
\(\cot \theta\) |
\(\frac{1}{\tan \theta}\) |
Tip: A good way to remember which is which is to look at the
third letter:
- \(\sec \theta\) → third letter 'c' → reciprocal of cos
- \(\cosec \theta\) → third letter 's' → reciprocal of sin
- \(\cot \theta\) → third letter 't' → reciprocal of tan
Proof: Relationship between \(\tan \theta\), \(\sin \theta\), and \(\cos \theta\)
From SOHCAHTOA: \(\sin \theta = \frac{O}{H}\) and \(\cos \theta = \frac{A}{H}\).
\[
\frac{\sin \theta}{\cos \theta} = \frac{O/H}{A/H} = \frac{O}{A} = \tan \theta
\]
Thus:
\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\]
Exact values for standard angles (non-calculator required for Paper 1):
| Degrees |
\(0^\circ\) |
\(30^\circ\) |
\(45^\circ\) |
\(60^\circ\) |
\(90^\circ\) |
| Radians |
0 |
\(\frac{\pi}{6}\) |
\(\frac{\pi}{4}\) |
\(\frac{\pi}{3}\) |
\(\frac{\pi}{2}\) |
| \(\sin \theta\) |
0 |
\(\frac{1}{2}\) |
\(\frac{\sqrt{2}}{2}\) |
\(\frac{\sqrt{3}}{2}\) |
1 |
| \(\cos \theta\) |
1 |
\(\frac{\sqrt{3}}{2}\) |
\(\frac{\sqrt{2}}{2}\) |
\(\frac{1}{2}\) |
0 |
| \(\tan \theta\) |
0 |
\(\frac{1}{\sqrt{3}}\) |
1 |
\(\sqrt{3}\) |
undefined |
Worked Example 10.1 (Finding exact values – non-calculator)
Find the exact value of:
- \(\sec 60^\circ\)
- \(\cosec 45^\circ\)
- \(\cot 30^\circ\)
Solution:
(a) \(\sec 60^\circ = \frac{1}{\cos 60^\circ} = \frac{1}{1/2} = 2\)
(b) \(\cosec 45^\circ = \frac{1}{\sin 45^\circ} = \frac{1}{\sqrt{2}/2} = \frac{2}{\sqrt{2}} = \sqrt{2}\)
(c) \(\cot 30^\circ = \frac{1}{\tan 30^\circ} = \frac{1}{1/\sqrt{3}} = \sqrt{3}\)
Worked Example 10.2 (Mauritian context – angle of elevation)
A tourist in Grand Baie looks up at the top of a lighthouse. The angle of elevation is \(30^\circ\), and the distance from the tourist to the base of the lighthouse is 100 m. Find the height of the lighthouse.
Solution:
\[
\tan 30^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{100}
\]
\[
h = 100 \times \tan 30^\circ = 100 \times \frac{1}{\sqrt{3}} = \frac{100}{\sqrt{3}} \text{ m}
\]
Answer: \(\frac{100}{\sqrt{3}}\) m.
10.2 Graphs of Trigonometric Functions
Key features of \(y = \sin x\) and \(y = \cos x\):
- Domain: all real numbers
- Range: \(-1 \le y \le 1\)
- Period: \(360^\circ\) (\(2\pi\) radians)
- Amplitude: \(1\) (maximum distance from principal axis)
Key features of \(y = \tan x\):
- Domain: \(x \neq 90^\circ, 270^\circ, \ldots\) (where \(\cos x = 0\))
- Range: all real numbers
- Period: \(180^\circ\) (\(\pi\) radians)
- Asymptotes: vertical lines at \(x = 90^\circ, 270^\circ, \ldots\)
Transformations: \(y = a\sin(bx) + c\)
| Parameter |
Effect |
| \(a\) |
Amplitude = \(|a|\) (vertical stretch) |
| \(b\) |
Period = \(\frac{360^\circ}{b}\) (horizontal stretch/squash) |
| \(c\) |
Vertical translation (principal axis becomes \(y = c\)) |
Tip for sketching transformations:
- Start with the graph of the original function
- Carry out stretches first, then translations
- Mark the principal axis (\(y = c\)), maximum (\(y = c + a\)), and minimum (\(y = c - a\)) lines
Worked Example 10.3 (Sketching transformed sine graph – non-calculator)
Sketch \(y = 2\sin x + 1\) for \(0^\circ \le x \le 360^\circ\). Label the principal axis, maximum and minimum points.
Solution:
- Amplitude = \(2\) (range: \(-1 \le y \le 3\))
- Principal axis: \(y = 1\)
- Period: \(360^\circ\)
- Maximum points: \(y = 3\) at \(x = 90^\circ\)
- Minimum points: \(y = -1\) at \(x = 270^\circ\)
- Passes through \((0, 1)\), \((180^\circ, 1)\), \((360^\circ, 1)\)
Sketch: sine wave oscillating between \(-1\) and \(3\), centred on \(y = 1\).
Worked Example 10.4 (Finding period – non-calculator)
Find the period of:
- \(y = \cos 3x\)
- \(y = \sin \frac{x}{2}\)
Solution:
(a) Period = \(\frac{360^\circ}{3} = 120^\circ\)
(b) Period = \(\frac{360^\circ}{1/2} = 720^\circ\)
Worked Example 10.5 (Mauritian context – tidal wave)
The height of the tide at a beach in Mauritius is modelled by:
\[
h(t) = 3\sin(30t) + 4
\]
where \(t\) is the number of hours after midnight. Find:
- The amplitude
- The principal axis
- The period in hours
- The maximum height of the tide
Solution:
(a) Amplitude = \(3\) m
(b) Principal axis: \(h = 4\) m
(c) Period = \(\frac{360^\circ}{30} = 12\) hours
(d) Maximum height = \(4 + 3 = 7\) m
10.4 Trigonometric Identities
Three Pythagorean identities:
\(\sin^2 \theta + \cos^2 \theta = 1\)
\(\sec^2 \theta = 1 + \tan^2 \theta\)
\(\cosec^2 \theta = 1 + \cot^2 \theta\)
Proof: Derivation of \(\sec^2 \theta = 1 + \tan^2 \theta\)
Start with \(\sin^2 \theta + \cos^2 \theta = 1\).
Divide both sides by \(\cos^2 \theta\):
\[
\frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}
\]
\[
\tan^2 \theta + 1 = \sec^2 \theta
\]
Proof: Derivation of \(\cosec^2 \theta = 1 + \cot^2 \theta\)
Start with \(\sin^2 \theta + \cos^2 \theta = 1\).
Divide both sides by \(\sin^2 \theta\):
\[
1 + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta}
\]
\[
1 + \cot^2 \theta = \cosec^2 \theta
\]
Reciprocal identities:
\(\sec \theta = \frac{1}{\cos \theta}, \quad \cosec \theta = \frac{1}{\sin \theta}, \quad \cot \theta = \frac{1}{\tan \theta}\)
Worked Example 10.7 (Using Pythagorean identity – non-calculator)
Given that \(\sin \theta = \frac{3}{5}\) and \(0^\circ < \theta < 90^\circ\), find \(\cos \theta\) and \(\tan \theta\).
Solution:
\[
\sin^2 \theta + \cos^2 \theta = 1
\]
\[
\left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1 \Rightarrow \frac{9}{25} + \cos^2 \theta = 1
\]
\[
\cos^2 \theta = \frac{16}{25} \Rightarrow \cos \theta = \frac{4}{5} \quad (\text{positive, since } 0^\circ < \theta < 90^\circ)
\]
\[
\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{4/5} = \frac{3}{4}
\]
Answer: \(\cos \theta = \frac{4}{5}\), \(\tan \theta = \frac{3}{4}\).
Worked Example 10.8 (Using \(\sec^2\) identity – calculator required)
Solve \(\sec^2 \theta = 4 \tan \theta + 2\) for \(0^\circ \le \theta \le 360^\circ\).
Solution:
Use \(\sec^2 \theta = 1 + \tan^2 \theta\):
\[
1 + \tan^2 \theta = 4\tan \theta + 2
\]
\[
\tan^2 \theta - 4\tan \theta - 1 = 0
\]
Let \(t = \tan \theta\):
\[
t^2 - 4t - 1 = 0 \Rightarrow t = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}
\]
So \(\tan \theta = 2 + \sqrt{5}\) or \(\tan \theta = 2 - \sqrt{5}\).
Using a calculator (Paper 2) or exact values (Paper 1 would require a sketch):
\[
\theta = \tan^{-1}(2 + \sqrt{5}) \approx 76.7^\circ, \quad \theta = \tan^{-1}(2 - \sqrt{5}) \approx -11.3^\circ
\]
For \(0^\circ \le \theta \le 360^\circ\), the solutions are:
From \(76.7^\circ\): add \(180^\circ\) \(\rightarrow\) \(76.7^\circ, 256.7^\circ\)
From \(-11.3^\circ\): add \(180^\circ\) \(\rightarrow\) \(168.7^\circ, 348.7^\circ\)
Answer: \(76.7^\circ, 168.7^\circ, 256.7^\circ, 348.7^\circ\) (to 1 d.p.)
10.5 Solving Trigonometric Equations
Steps to solve trigonometric equations:
- Use identities to simplify if necessary
- Find the principal value using inverse trig functions or exact values
- Use the CAST diagram or graphs to find all solutions in the given interval
Secondary values:
| Equation |
Secondary value |
Period for all solutions |
| \(\sin x = k\) |
\(180^\circ - \text{PV}\) |
Add \(360^\circ\) |
| \(\cos x = k\) |
\(-\text{PV}\) (or \(360^\circ - \text{PV}\)) |
Add \(360^\circ\) |
| \(\tan x = k\) |
\(\text{PV} + 180^\circ\) |
Add \(180^\circ\) |
Worked Example 10.9 (Solving basic trig equation – non-calculator)
Solve \(\sin x = \frac{1}{2}\) for \(0^\circ \le x \le 360^\circ\).
Solution:
Principal value: \(x = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ\)
Secondary value: \(180^\circ - 30^\circ = 150^\circ\)
Answer: \(x = 30^\circ, 150^\circ\)
Worked Example 10.10 (Solving with transformations – non-calculator)
Solve \(\cos(2x + 30^\circ) = \frac{1}{2}\) for \(0^\circ \le x \le 180^\circ\).
Solution:
Let \(\theta = 2x + 30^\circ\). The range becomes: \(30^\circ \le \theta \le 390^\circ\).
Solve \(\cos \theta = \frac{1}{2}\):
Principal: \(\theta = 60^\circ\)
Secondary: \(\theta = 360^\circ - 60^\circ = 300^\circ\)
In the range \(30^\circ \le \theta \le 390^\circ\), solutions are: \(60^\circ, 300^\circ\)
Convert back to \(x\):
\(2x + 30^\circ = 60^\circ \Rightarrow 2x = 30^\circ \Rightarrow x = 15^\circ\)
\(2x + 30^\circ = 300^\circ \Rightarrow 2x = 270^\circ \Rightarrow x = 135^\circ\)
Answer: \(x = 15^\circ, 135^\circ\)
Worked Example 10.11 (Quadratic trig equation)
Solve \(2\sin^2 x - \sin x - 1 = 0\) for \(0^\circ \le x \le 360^\circ\).
Solution:
Let \(s = \sin x\):
\(2s^2 - s - 1 = 0 \Rightarrow (2s + 1)(s - 1) = 0\)
So \(\sin x = 1\) or \(\sin x = -\frac{1}{2}\).
For \(\sin x = 1\): \(x = 90^\circ\)
For \(\sin x = -\frac{1}{2}\):
Principal: \(\sin^{-1}(-\frac{1}{2}) = -30^\circ\) (or \(210^\circ\))
Solutions in range: \(210^\circ, 330^\circ\)
Answer: \(x = 90^\circ, 210^\circ, 330^\circ\)
10.6 Proving Trigonometric Identities
Strategy for proving identities:
- Start with the more complicated side
- Use known identities to simplify step by step
- Work towards the other side
- Do not work on both sides simultaneously
Worked Example 10.12 (Proving an identity – non-calculator)
Prove that \(\frac{1}{\cos x} - \cos x = \sin x \tan x\).
Solution:
Start with LHS:
\[
\text{LHS} = \frac{1}{\cos x} - \cos x
\]
Write with common denominator:
\[
= \frac{1 - \cos^2 x}{\cos x}
\]
Use \(\sin^2 x + \cos^2 x = 1\) \(\rightarrow\) \(1 - \cos^2 x = \sin^2 x\):
\[
= \frac{\sin^2 x}{\cos x}
\]
\[
= \sin x \times \frac{\sin x}{\cos x} = \sin x \tan x = \text{RHS}
\]
Worked Example 10.13 (Mauritian context – proving identity)
Prove that \(\frac{1}{1 - \sin x} + \frac{1}{1 + \sin x} = 2\sec^2 x\).
Solution:
Start with LHS:
\[
\frac{1}{1 - \sin x} + \frac{1}{1 + \sin x} = \frac{(1 + \sin x) + (1 - \sin x)}{(1 - \sin x)(1 + \sin x)}
\]
\[
= \frac{2}{1 - \sin^2 x}
\]
Use \(1 - \sin^2 x = \cos^2 x\):
\[
= \frac{2}{\cos^2 x} = 2\sec^2 x = \text{RHS}
\]
Chapter 10 Summary
| Concept |
Key points |
| Six functions |
\(\sin\), \(\cos\), \(\tan\), \(\sec\), \(\cosec\), \(\cot\) |
| Reciprocals |
\(\sec = \frac{1}{\cos}\), \(\cosec = \frac{1}{\sin}\), \(\cot = \frac{1}{\tan}\) |
| Graphs |
\(y = a\sin(bx) + c\): amplitude \(|a|\), period \(\frac{360^\circ}{b}\) |
| Principal axis |
\(y = c\) for sin/cos |
| Identities |
\(\sin^2 + \cos^2 = 1\), \(\sec^2 = 1 + \tan^2\), \(\cosec^2 = 1 + \cot^2\) |
| Solving |
Use principal value and secondary values |
| Proving |
Start with one side, simplify to the other |
Exercises – Chapter 10
Easy (Drill – Non-Calculator)
- Write down the exact values of:
(a) \(\sin 60^\circ\)
(b) \(\cos 45^\circ\)
(c) \(\tan 30^\circ\)
(d) \(\sec 45^\circ\)
- State the amplitude and period of:
(a) \(y = 3\sin 2x\)
(b) \(y = 4\cos \frac{x}{2}\)
- Solve for \(0^\circ \le x \le 360^\circ\):
(a) \(\sin x = \frac{\sqrt{3}}{2}\)
(b) \(\cos x = -\frac{1}{2}\)
Medium (Examination Style)
- Sketch \(y = 2\cos x - 1\) for \(0^\circ \le x \le 360^\circ\). Label the principal axis, maximum and minimum points.
- Solve \(\tan x = -\sqrt{3}\) for \(0^\circ \le x \le 360^\circ\).
- Mauritian context: The number of tourists visiting a beach in Mauritius can be modelled by:
\[
N(t) = 200\sin(30t) + 500
\]
where \(t\) is the number of hours after 6 am.
(a) Find the amplitude and principal axis.
(b) Find the maximum number of tourists.
(c) Find the period in hours.
- Prove that \(\frac{\sin x}{1 - \cos x} + \frac{\sin x}{1 + \cos x} = 2\cosec x\).
Hard (Challenge for A*)
- Solve \(\cos 2x = \sin x\) for \(0^\circ \le x \le 360^\circ\).
Hint: use \(\cos 2x = 1 - 2\sin^2 x\)
- Prove that \(\frac{1 - \tan x}{1 + \tan x} = \frac{\cos x - \sin x}{\cos x + \sin x}\).
- Mauritian context: A Ferris wheel at a funfair in Grand Baie has a height \(h\) (in metres) above the ground given by:
\[
h(t) = 15\sin(15t) + 18
\]
where \(t\) is the number of minutes after the ride starts.
(a) What is the maximum height reached?
(b) What is the minimum height?
(c) How long does it take for one complete revolution?
(d) Find the first two times when the height is exactly 25.5 m.