Step 7: Trigonometry: Radians, Graphs and Equations
- A radian is the angle that cuts off an arc as long as the radius, so a full turn is \( 2\pi \) radians and \( 180^\circ = \pi \). In radians, arc length is \( s = r\theta \) and sector area is \( A = \tfrac12 r^2\theta \).
- On the unit circle the point at angle \( \theta \) is \( (\cos\theta, \sin\theta) \). That defines sine and cosine for every angle, and the CAST diagram tells you their signs.
- \( y = a\sin(bx + c) + d \) has amplitude \( |a| \), period \( \dfrac{2\pi}{b} \) and centre line \( y = d \). The same idea fits real seasonal data.
- Exam point: your calculator gives only one solution of \( \sin x = k \). Use \( \pi-\alpha \) for sine, \( 2\pi-\alpha \) for cosine and \( \alpha + \pi \) for tangent to find the others, and change the interval first when the angle is \( 2x \) or \( 3x \).
Radians: the circle’s own unit
Trigonometry starts with triangles, but in calculus it is really about cycles: tides, seasons, sound waves and sales that rise and fall every year. To work with cycles we first need a better way to measure angles.
Degrees are a human choice. Someone decided that a full turn is 360. Mathematics prefers a unit that comes from the circle itself. Take a circle of radius \( r \) and walk a distance \( r \) around its edge. The angle you have turned through at the centre is one radian, about \( 57.3^\circ \).
A full circle has circumference \( 2\pi r \), and each radian uses an arc of length \( r \). So a full turn is \( 2\pi \) radians, and half a turn is \( \pi \). Every conversion comes from that one fact.
To change degrees to radians, multiply by \( \tfrac{\pi}{180} \). To change radians to degrees, multiply by \( \tfrac{180}{\pi} \).
Learn the common ones: \( 30^\circ = \tfrac{\pi}{6} \), \( 45^\circ = \tfrac{\pi}{4} \), \( 60^\circ = \tfrac{\pi}{3} \), \( 90^\circ = \tfrac{\pi}{2} \). When an angle has no degree sign, it is in radians, so \( \sin 2 \) means the sine of 2 radians.
Why bother? Because a sector with angle \( \theta \) is the fraction \( \tfrac{\theta}{2\pi} \) of the whole circle. Its arc is that fraction of \( 2\pi r \), and its area is that fraction of \( \pi r^2 \). The \( 2\pi \) cancels and leaves two very clean formulas.
A segment is the region between a chord and its arc. It is the sector minus the triangle, and the triangle has area \( \tfrac12 r^2 \sin\theta \), so the segment has area \( \tfrac12 r^2(\theta-\sin\theta) \).
(a) A sector has radius 5 cm and angle 1.2 radians. Find its arc length, area and perimeter. (b) A chord subtends a right angle at the centre of a circle of radius 4 cm. Find the exact area of the segment it cuts off.
Forgetting the two radii in a perimeter is one of the commonest lost marks in circular measure questions.
Exact values and the unit circle
Exams expect exact values for \( 30^\circ \), \( 45^\circ \) and \( 60^\circ \). You do not need to memorise them. Half of a unit square has sides 1, 1 and \( \sqrt2 \). Half of an equilateral triangle of side 2 has sides 1, \( \sqrt3 \) and 2. Read the ratios off those two triangles and you get this table.
| \( \theta \) | 0 | \( \tfrac{\pi}{6} = 30^\circ \) | \( \tfrac{\pi}{4} = 45^\circ \) | \( \tfrac{\pi}{3} = 60^\circ \) | \( \tfrac{\pi}{2} = 90^\circ \) |
|---|---|---|---|---|---|
| \( \sin\theta \) | 0 | \( \tfrac12 \) | \( \tfrac{\sqrt2}{2} \) | \( \tfrac{\sqrt3}{2} \) | 1 |
| \( \cos\theta \) | 1 | \( \tfrac{\sqrt3}{2} \) | \( \tfrac{\sqrt2}{2} \) | \( \tfrac12 \) | 0 |
| \( \tan\theta \) | 0 | \( \tfrac{\sqrt3}{3} \) | 1 | \( \sqrt3 \) | undefined |
A right-angled triangle only allows angles up to \( 90^\circ \). To go further, put the angle in a circle of radius 1 centred at the origin, the unit circle. Start on the positive \( x \)-axis and turn anticlockwise through \( \theta \) (clockwise if \( \theta \) is negative). You land on a point \( P \).
The signs follow from the signs of \( x \) and \( y \) in each quadrant. In the first quadrant all three ratios are positive. In the second only sine is, in the third only tangent, and in the fourth only cosine. Reading anticlockwise from the fourth quadrant gives C, A, S, T, so this is called the CAST diagram.
To find any value, find the basic angle (the acute angle between the arm and the \( x \)-axis), take its exact value, then attach the sign from CAST.
Find the exact values of (a) \( \sin 240^\circ \) (b) \( \cos\tfrac{5\pi}{4} \) (c) \( \tan 330^\circ \).
Sketching the four quadrants takes five seconds and stops almost every sign error.
Graphs and transformations
Let \( \theta \) run from 0 to \( 2\pi \) and watch the point \( P \). Its height, \( \sin\theta \), starts at 0, rises to 1 at \( \tfrac{\pi}{2} \), falls to 0 at \( \pi \), reaches \( -1 \) at \( \tfrac{3\pi}{2} \) and returns to 0 at \( 2\pi \). Then \( P \) goes round again and the pattern repeats. Plot height against angle and you get the sine wave (Figure 1).
Sine and cosine both have period \( 2\pi \): the graph repeats every \( 2\pi \). Tangent has period \( \pi \), because half a turn sends \( (x, y) \) to \( (-x, -y) \) and \( \tfrac{-y}{-x} = \tfrac{y}{x} \). It has vertical asymptotes wherever \( \cos x = 0 \), at \( x = \pm\tfrac{\pi}{2}, \pm\tfrac{3\pi}{2}, \ldots \), because there it would divide by zero.
The general sine wave \( y = a\sin(bx + c) + d \) is the basic wave after the transformations of Step 3. Each letter has one job.
| Letter | Effect on \( y = \sin x \) | What you read off |
|---|---|---|
| \( a \) | vertical stretch, factor \( a \) | amplitude \( |a| \) |
| \( b > 0 \) | horizontal stretch, factor \( \tfrac1b \) | period \( \tfrac{2\pi}{b} \) |
| \( c \) | horizontal shift: \( \sin(bx + c) = \sin\big(b(x + \tfrac{c}{b})\big) \) | moves the wave \( \tfrac{c}{b} \) to the left |
| \( d \) | vertical shift | centre line \( y = d \); max \( d + |a| \), min \( d-|a| \) |
Why is the period \( \tfrac{2\pi}{b} \) and not \( 2\pi b \)? The wave finishes one cycle when its input has gone from 0 to \( 2\pi \). In \( \sin bx \) the input is \( bx \), which reaches \( 2\pi \) when \( x = \tfrac{2\pi}{b} \). A bigger \( b \) makes the input race round faster, so the cycle ends sooner.
For \( y = 2\sin\left(3x-\tfrac{\pi}{2}\right) + 1 \), state the amplitude, period and range, and find the first maximum point with \( x > 0 \).
Figure 2 confirms it: three full cycles between 0 and \( 2\pi \), each peaking at height 3.
An economics example: seasonal sales. A product sells 55 thousand units in its best month, August, and 25 thousand in its worst, February. Fit \( S = a\sin(bt + c) + d \), with \( t = 1 \) for January. The centre line is the midpoint, \( d = \tfrac{55 + 25}{2} = 40 \). The amplitude is half the gap, \( a = \tfrac{55-25}{2} = 15 \). The cycle lasts 12 months, so \( \tfrac{2\pi}{b} = 12 \) and \( b = \tfrac{\pi}{6} \). The peak at \( t = 8 \) needs an input of \( \tfrac{\pi}{2} \): \( \tfrac{8\pi}{6} + c = \tfrac{\pi}{2} \), so \( c = -\tfrac{5\pi}{6} \). The model is \( S = 15\sin\left(\tfrac{\pi}{6}t-\tfrac{5\pi}{6}\right) + 40 \) (Figure 3). Economists use exactly this kind of seasonal component when they publish seasonally adjusted figures: subtract it, and the months can be compared fairly.
Two identities and solving equations
Two facts come straight from the unit circle. Since \( P = (\cos\theta, \sin\theta) \), the gradient of \( OP \) gives \( \tan\theta = \dfrac{\sin\theta}{\cos\theta} \). And since \( P \) is 1 unit from the origin, Pythagoras gives \( \cos^2\theta + \sin^2\theta = 1 \) in every quadrant.
The sign \( \equiv \) marks an identity: true for every angle where both sides are defined. Note that \( \sin^2\theta \) means \( (\sin\theta)^2 \).
Given \( \cos\theta = -\tfrac{5}{13} \) and \( 180^\circ < \theta < 270^\circ \), find the exact values of \( \sin\theta \) and \( \tan\theta \).
The identity gives the size; the quadrant gives the sign.
Now equations. \( \sin x = 0.5 \) has infinitely many solutions, because the wave keeps crossing the line \( y = 0.5 \). Exams always give an interval, and you must find every solution in it. Your calculator gives one, the principal value \( \alpha \). The symmetry of each graph gives the rest.
| Equation | Solutions in one cycle | Then |
|---|---|---|
| \( \sin x = k \) | \( \alpha \) and \( 180^\circ-\alpha \) | add or subtract \( 360^\circ \) |
| \( \cos x = k \) | \( \alpha \) and \( 360^\circ-\alpha \) | add or subtract \( 360^\circ \) |
| \( \tan x = k \) | \( \alpha \) | add or subtract \( 180^\circ \) |
In radians, replace \( 180^\circ \) by \( \pi \) and \( 360^\circ \) by \( 2\pi \).
Solve \( 2\sin 2x = 1 \) for \( 0^\circ \le x \le 360^\circ \).
Four answers, as Figure 4 shows. Halving only 30° and 150° would lose half the marks.
When an equation mixes \( \sin^2 x \) and \( \cos x \), use \( \sin^2 x = 1-\cos^2 x \) to leave one ratio, then solve a quadratic as in Step 2.
Solve \( 2\sin^2 x + 3\cos x = 0 \) for \( 0^\circ \le x \le 360^\circ \).
Always check each root of the quadratic against the range of the ratio before solving it.
Inverse trigonometric functions
To undo sine we want \( \sin^{-1} \), “the angle whose sine is”. But sine is many-one: \( \tfrac{\pi}{6} \), \( \tfrac{5\pi}{6} \) and \( \tfrac{13\pi}{6} \) all have sine \( \tfrac12 \). As Step 3 showed, only one-one functions have inverses, so we restrict the domain to a piece where the function is one-one and still takes every value. The inverse then returns one principal value.
| Function | Domain | Principal range |
|---|---|---|
| \( y = \sin^{-1} x \) | \( -1 \le x \le 1 \) | \( -\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2} \) |
| \( y = \cos^{-1} x \) | \( -1 \le x \le 1 \) | \( 0 \le y \le \pi \) |
| \( y = \tan^{-1} x \) | all real \( x \) | \( -\tfrac{\pi}{2} < y < \tfrac{\pi}{2} \) |
Find exactly (a) \( \cos^{-1}\left(-\tfrac{\sqrt3}{2}\right) \) (b) \( \tan^{-1}(-1) \) (c) \( \sin^{-1}\left(\sin\tfrac{2\pi}{3}\right) \).
\( \sin^{-1} \) and \( \sin \) only cancel when the angle is already in the principal range.
The \( -1 \) in \( \sin^{-1} x \) means “inverse function”, exactly as in \( f^{-1} \). It is not \( \dfrac{1}{\sin x} \). Many books write \( \arcsin x \) to avoid the confusion. IB students also need the sine rule, the cosine rule and the area formula \( \tfrac12 ab\sin C \) for non-right-angled triangles (IB SL 3.2); they follow from the same unit-circle coordinates.
Practise: 6 questions
Try each one on paper first, then check the worked answer.
EasyQ1. Convert \( 225^\circ \) to radians exactly, and \( \tfrac{7\pi}{6} \) to degrees. Show worked answer
\( \frac{7\pi}{6} \times \frac{180}{\pi} = 7 \times 30 = 210^\circ \).
EasyQ2. A sector of radius 9 cm has area 54 cm². Find its angle and its arc length. Show worked answer
Arc: \( s = 9 \times \frac43 = 12 \) cm.
MediumQ3. Find the exact values of \( \cos\tfrac{4\pi}{3} \), \( \tan\tfrac{3\pi}{4} \) and \( \sin\left(-\tfrac{\pi}{3}\right) \). Show worked answer
\( \tfrac{3\pi}{4} \): second quadrant, basic angle \( \tfrac{\pi}{4} \), tangent negative, so \( -1 \).
\( -\tfrac{\pi}{3} \): fourth quadrant, sine negative, so \( -\tfrac{\sqrt3}{2} \).
MediumQ4. For \( y = 3\cos 2x-1 \), state the amplitude, period and range, and find the values of \( x \) in \( 0 \le x \le 2\pi \) where \( y \) is greatest. Show worked answer
Range: \( -1-3 \le y \le -1 + 3 \), that is \( -4 \le y \le 2 \).
Greatest when \( \cos 2x = 1 \): \( 2x = 0, 2\pi, 4\pi \), so \( x = 0, \pi, 2\pi \).
MediumQ5. Solve \( 2\cos^2 x + \cos x-1 = 0 \) for \( 0 \le x \le 2\pi \). Show worked answer
\( \cos x = \tfrac12 \): \( x = \tfrac{\pi}{3} \) or \( 2\pi-\tfrac{\pi}{3} = \tfrac{5\pi}{3} \).
\( \cos x = -1 \): \( x = \pi \).
So \( x = \tfrac{\pi}{3}, \pi, \tfrac{5\pi}{3} \).
ExamQ6. (a) Show that the equation \( 3\sin x\tan x = 8 \) can be written as \( 3\cos^2 x + 8\cos x-3 = 0 \). [3] (b) Hence solve \( 3\sin x\tan x = 8 \) for \( 0^\circ \le x \le 360^\circ \), giving answers to 1 decimal place. [3] Show worked answer
Use \( \sin^2 x = 1-\cos^2 x \): \( 3-3\cos^2 x = 8\cos x \) [M1].
Rearrange: \( 3\cos^2 x + 8\cos x-3 = 0 \) [A1].
(b) \( (3\cos x-1)(\cos x + 3) = 0 \) [M1].
\( \cos x = -3 \) is impossible, so \( \cos x = \tfrac13 \) [A1].
\( x = 70.5^\circ \) or \( 360^\circ-70.5^\circ = 289.5^\circ \) [A1].
Total: 6 marks.
Check yourself
1. What is the period of \( y = 5\sin 4x + 2 \)?
2. What is the arc length of a sector with radius 10 cm and angle \( 36^\circ \)?
3. What is \( \sin^{-1}\left(\sin\tfrac{5\pi}{6}\right) \)?
Your score: 0 out of 3
The short version
- A full turn is \( 2\pi \) radians. In radians, arc length is \( r\theta \), sector area is \( \tfrac12 r^2\theta \), and a segment is \( \tfrac12 r^2(\theta-\sin\theta) \).
- The point at angle \( \theta \) on the unit circle is \( (\cos\theta, \sin\theta) \). Exact values come from two triangles; signs come from CAST.
- \( y = a\sin(bx + c) + d \) has amplitude \( |a| \), period \( \tfrac{2\pi}{b} \) and centre line \( y = d \), which is how seasonal data is modelled.
- \( \tan\theta = \tfrac{\sin\theta}{\cos\theta} \) and \( \sin^2\theta + \cos^2\theta = 1 \). Solve equations by finding the principal value, then every other solution in the interval; factorise rather than divide.
- Inverse trigonometric functions return one principal value from a restricted range.
- The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.