Trigonometry

How tall is the Dharahara, a tree in the school ground, or a hill across the river? You cannot climb up with a measuring tape, but you can stand some distance away, measure that distance along the ground, and measure the angle up to the top. Trigonometry (from Greek trigonon, triangle, and metron, measure) is the mathematics that turns such an angle and a distance into a height. It rests on a simple fact: in right-angled triangles with the same angle, the ratios of the sides are always the same. This chapter first explains three ways of measuring angles, then defines the six trigonometric ratios, proves the identities that connect them, finds their exact values for the standard angles, extends them to angles such as 120^\circ and -30^\circ, and finally uses them to find heights and distances.

7.1 Measurement of angles: sexagesimal, centesimal and circular (radian) systems

An angle is formed when a ray turns about its end point from a starting position (the initial side) to a final position (the terminal side). The amount of turning is the size of the angle. Turning anticlockwise gives a positive angle; turning clockwise gives a negative angle. One complete turn brings the ray back to where it started, and a quarter of a complete turn is a right angle.

There are three systems of measuring angles. They differ only in how they divide the right angle.

Sexagesimal (British) system

The right angle is divided into 90 equal parts called degrees (^\circ), each degree into 60 minutes (') and each minute into 60 seconds (''). The name comes from the Latin for sixty.

Centesimal (French) system

The right angle is divided into 100 equal parts called grades (^g), each grade into 100 centesimal minutes (^c) and each of these into 100 centesimal seconds (^{cc}). The name comes from the Latin for hundred.

Circular system and the radian

The circumference of a circle is 2\pi r, so it contains 2\pi arcs of length r. A complete turn is therefore 2\pi radians, a half turn is \pi radians, and a right angle is \frac{\pi}{2} radians:

A radian is the same size for every circle, because doubling the radius doubles the arc as well. That is why scientists prefer it.

Length of an arc

The definition of a radian gives a quick formula: an angle of \theta radians at the centre cuts off an arc \theta times as long as the radius,

A bicycle wheel of radius 35 cm that turns through 2 radians rolls 35\times2=70 cm along the road. A pendulum 50 cm long whose bob moves along an arc of 10 cm swings through \theta=\frac{10}{50}=0.2 radian, about 11.5^\circ.

7.2 Trigonometric ratios of an acute angle

Take a right-angled triangle and fix one of its acute angles, \theta. The three sides are named by their position relative to \theta:

• the hypotenuse h is the longest side, opposite the right angle;
• the perpendicular p is the side opposite \theta;
• the base b is the side next to \theta (other than the hypotenuse).

Why the ratios depend only on the angle

Any two right-angled triangles with the same acute angle \theta are similar (their angles are 90^\circ, \theta and 90^\circ-\theta). In similar triangles corresponding sides are in proportion, so \frac{p}{h}, \frac{b}{h} and \frac{p}{b} come out the same however large the triangle is drawn. The ratios belong to the angle, not to the particular triangle. That is what makes it possible to measure a tall tower with a small triangle drawn in a notebook.

Reciprocal and quotient relations

Straight from the definitions:

(For example, \frac{\sin\theta}{\cos\theta}=\frac{p/h}{b/h}=\frac{p}{b}=\tan\theta.)

7.3 Trigonometric identities and their conversions

An identity is an equation that is true for every value of the angle, like (a+b)^2=a^2+2ab+b^2 in algebra. Three identities follow from Pythagoras' theorem p^2+b^2=h^2.

Conversions: one ratio in terms of another

Every ratio can be written in terms of any one of the others. The easy way is to draw a triangle. To write everything in terms of \sin\theta, take p=\sin\theta and h=1; then b=\sqrt{1-\sin^2\theta}. To write everything in terms of \tan\theta, take p=\tan\theta and b=1; then h=\sqrt{1+\tan^2\theta}.

Ratio	in terms of \sin\theta	in terms of \tan\theta
\sin\theta	\sin\theta	\frac{\tan\theta}{\sqrt{1+\tan^2\theta}}
\cos\theta	\sqrt{1-\sin^2\theta}	\frac{1}{\sqrt{1+\tan^2\theta}}
\tan\theta	\frac{\sin\theta}{\sqrt{1-\sin^2\theta}}	\tan\theta
\mathrm{cosec}\,\theta	\frac{1}{\sin\theta}	\frac{\sqrt{1+\tan^2\theta}}{\tan\theta}
\sec\theta	\frac{1}{\sqrt{1-\sin^2\theta}}	\sqrt{1+\tan^2\theta}
\cot\theta	\frac{\sqrt{1-\sin^2\theta}}{\sin\theta}	\frac{1}{\tan\theta}

Proving identities

To prove an identity, start from one side (usually the more complicated one) and change it, step by step, into the other side. Never move terms across the equals sign as if solving an equation: you are showing that two expressions are the same, not finding \theta. Useful moves are: change everything into \sin and \cos; use \sin^2\theta+\cos^2\theta=1; factorise; and add fractions over a common denominator.

7.4 Values of trigonometric ratios of 0°, 30°, 45°, 60° and 90°

For most angles the ratios are long decimals found with a calculator. For five special angles we can find them exactly from simple triangles.

45°

Cut a square of side 1 along a diagonal. Each half is a right triangle with two angles of 45^\circ, legs 1 and 1, and hypotenuse \sqrt{1^2+1^2}=\sqrt{2}.

30° and 60°

Take an equilateral triangle of side 2 and drop a perpendicular from the top vertex. It bisects the base and the top angle, giving a right triangle with angles 30^\circ, 60^\circ, 90^\circ, hypotenuse 2, short side 1, and height \sqrt{2^2-1^2}=\sqrt{3}.

For the 30^\circ angle at the top, p=1, b=\sqrt{3}, h=2; for the 60^\circ angle at the base, p=\sqrt{3}, b=1, h=2:

0° and 90°

Imagine a right triangle with hypotenuse 1 and let its angle \theta shrink towards 0^\circ: the perpendicular shrinks to 0 and the base grows to 1. So \sin0^\circ=0, \cos0^\circ=1 and \tan0^\circ=0. Now let \theta grow towards 90^\circ: the perpendicular grows to 1 and the base shrinks to 0. So \sin90^\circ=1 and \cos90^\circ=0, while \tan90^\circ=\frac{1}{0} is not defined. (This is the limit idea of chapter 4 at work.)

7.5 Trigonometric ratios of (90° ± θ), (180° ± θ) and (−θ)

Ratios of any angle

A right triangle can only hold acute angles. To give 120^\circ or -30^\circ a sine, we use coordinates. Let the terminal side of an angle \theta (measured from the positive x-axis) pass through a point P(x,y), and let r=OP=\sqrt{x^2+y^2}. Define

and the other three as their reciprocals. For an acute angle, y is the perpendicular, x the base and r the hypotenuse, so these agree with section 7.2. For larger angles, x or y may be negative, and the ratios take the signs of x and y.

How the allied angles arise

Let \theta be acute and P(x,y) a point on its terminal side, so \sin\theta=\frac{y}{r} and \cos\theta=\frac{x}{r}. Each allied angle comes from moving P by a reflection or a rotation that keeps its distance r from O (you will meet these transformations again in chapter 9).

Reading the ratios from the new coordinates (with the same r) gives every rule at once:

Angle	New point	sin	cos	tan
-\theta	(x,-y)	-\sin\theta	\cos\theta	-\tan\theta
90^\circ-\theta	(y,x)	\cos\theta	\sin\theta	\cot\theta
90^\circ+\theta	(-y,x)	\cos\theta	-\sin\theta	-\cot\theta
180^\circ-\theta	(-x,y)	\sin\theta	-\cos\theta	-\tan\theta
180^\circ+\theta	(-x,-y)	-\sin\theta	-\cos\theta	\tan\theta

For example, for 90^\circ+\theta the new point is (-y,x), so \sin(90^\circ+\theta)=\frac{x}{r}=\cos\theta and \cos(90^\circ+\theta)=\frac{-y}{r}=-\sin\theta. For 90^\circ-\theta, the two acute angles of one right triangle are \theta and 90^\circ-\theta, so the perpendicular of one is the base of the other: the sine of one is the cosine of the other.

7.6 Height and distance with one angle of elevation or depression

Angles of elevation and depression

When you look at an object, the straight line from your eye to the object is the line of sight. The angle is always measured from the horizontal line through your eye.

The method

1. Draw a neat figure: the object, the observer, the horizontal line, the line of sight, the right angle, and the given angle.
2. Name the unknown side (usually h for height or x for distance).
3. Choose the ratio that connects the known side and the unknown side with the given angle: \tan when both are the legs (height and horizontal distance), \sin or \cos when the hypotenuse (a ladder, a string, a line of sight) is involved.
4. Substitute the standard value and solve. Use \sqrt{3}\approx1.732 and \sqrt{2}\approx1.414 for a decimal answer, and write the unit.

Chapter summary

• Right angle =90^\circ=100^g=\left(\frac{\pi}{2}\right)^c; 1^\circ=60', 1'=60''; 1^g=100^c, 1^c=100^{cc}; \pi^c=180^\circ=200^g; \frac{D}{90}=\frac{G}{100}=\frac{2R}{\pi}; arc l=r\theta.
• \sin\theta=\frac{p}{h}, \cos\theta=\frac{b}{h}, \tan\theta=\frac{p}{b}; cosec, sec and cot are their reciprocals; \tan\theta=\frac{\sin\theta}{\cos\theta}.
• \sin^2\theta+\cos^2\theta=1, 1+\tan^2\theta=\sec^2\theta, 1+\cot^2\theta=\mathrm{cosec}^2\theta; prove identities from one side, one change per line.
• \sin0^\circ, \sin30^\circ, \sin45^\circ, \sin60^\circ, \sin90^\circ=0, \frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{\sqrt{3}}{2}, 1; cosines in reverse order; \tan45^\circ=1, \tan60^\circ=\sqrt{3}, \tan90^\circ not defined.
• Signs by quadrant: All, Sin, Tan, Cos.
• \sin(-\theta)=-\sin\theta, \cos(-\theta)=\cos\theta; with 90^\circ\pm\theta the ratio changes to its partner, with 180^\circ\pm\theta it keeps its name; the sign comes from the quadrant.
• Elevation and depression are measured from the horizontal; the depression from A to B equals the elevation from B to A. Use \tan for height and distance, \sin or \cos with a hypotenuse, and add the observer's eye height.

Practice questions

Group A — Objective (1 mark each)

1. One right angle in the centesimal system is [1]a90^gb100^gc60^gd180^g
2. \pi^c is equal to [1]a90^\circb180^\circc360^\circd57.3^\circ
3. 45^\circ expressed in grades is [1]a40^gb45^gc50^gd60^g
4. In a right-angled triangle, \sin\theta is [1]a\frac{b}{h}b\frac{p}{h}c\frac{p}{b}d\frac{h}{p}
5. If \tan\theta=\frac{3}{4}, then \cot\theta is [1]a\frac{4}{3}b\frac{3}{5}c\frac{4}{5}d\frac{5}{3}
6. \sin^2\theta+\cos^2\theta is equal to [1]a0b1c2d\tan^2\theta
7. \sec^2\theta-\tan^2\theta is equal to [1]a0b-1c1d\cot^2\theta
8. The value of \sin30^\circ+\cos60^\circ is [1]a\frac{1}{2}b1c\sqrt{3}d0
9. \tan45^\circ is equal to [1]a0b\frac{1}{\sqrt{3}}c1d\sqrt{3}
10. \sin(90^\circ-\theta) is equal to [1]a\sin\thetab-\cos\thetac\cos\thetad-\sin\theta
11. \cos(180^\circ-\theta) is equal to [1]a\cos\thetab-\cos\thetac\sin\thetad-\sin\theta
12. When the shadow of a pole is equal to its height, the angle of elevation of the sun is [1]a30^\circb45^\circc60^\circd90^\circ

Answers to Group A

1. (b) — the centesimal system divides the right angle into 100 grades.

2. (b) — half a turn is \pi radians and 180^\circ.

3. (c) — 45\times\frac{10}{9}=50^g.

4. (b) — perpendicular over hypotenuse.

5. (a) — \cot\theta=\frac{1}{\tan\theta}.

6. (b) — the fundamental identity.

7. (c) — from 1+\tan^2\theta=\sec^2\theta.

8. (b) — \frac{1}{2}+\frac{1}{2}=1.

9. (c) — in the half-square triangle the legs are equal.

10. (c) — complementary angles: sine of one is cosine of the other.

11. (b) — 180^\circ-\theta is in quadrant II, where cosine is negative.

12. (b) — \tan\theta=1, so \theta=45^\circ.

Group B — Short answer (2 marks each)

1. Convert 67^\circ30' into grades. [2]
2. Convert \left(\frac{3\pi}{5}\right)^c into degrees and into grades. [2]
3. Two angles of a triangle are 60^\circ and 50^g. Find the third angle in radians. [2]
4. If \sin\theta=\frac{5}{13} and \theta is acute, find \cos\theta and \tan\theta. [2]
5. Prove that (1-\sin^2\theta)\sec^2\theta=1. [2]
6. Prove that \tan\theta+\cot\theta=\sec\theta\,\mathrm{cosec}\,\theta. [2]
7. Find the value of \tan^260^\circ+2\tan^245^\circ. [2]
8. Find the values of \sin120^\circ and \cos135^\circ. [2]
9. Find the acute angle \theta if 2\cos\theta=1. [2]
10. A ladder 10 m long makes an angle of 60^\circ with the ground. How high up the wall does it reach? [2]

Answers to Group B

1. 30'=0.5^\circ, so the angle is 67.5^\circ, and 67.5\times\frac{10}{9}=75^g.

2. \frac{3\times180^\circ}{5}=108^\circ and \frac{3\times200^g}{5}=120^g.

3. 50^g=45^\circ, so the third angle is 180^\circ-60^\circ-45^\circ=75^\circ=75\times\frac{\pi}{180}=\frac{5\pi}{12} radians.

4. p=5, h=13, b=\sqrt{169-25}=12. So \cos\theta=\frac{12}{13} and \tan\theta=\frac{5}{12}.

5. LHS =\cos^2\theta\times\frac{1}{\cos^2\theta}=1= RHS.

6. LHS =\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta\cos\theta}=\sec\theta\,\mathrm{cosec}\,\theta= RHS.

7. (\sqrt{3})^2+2(1)^2=3+2=5.

8. \sin120^\circ=\sin(180^\circ-60^\circ)=\sin60^\circ=\frac{\sqrt{3}}{2}; \cos135^\circ=\cos(180^\circ-45^\circ)=-\cos45^\circ=-\frac{1}{\sqrt{2}}.

9. \cos\theta=\frac{1}{2}=\cos60^\circ, so \theta=60^\circ.

10. \sin60^\circ=\frac{h}{10}, so h=10\times\frac{\sqrt{3}}{2}=5\sqrt{3}\approx8.66 m.

Group C — Long answer (4 marks each)

1. (i) Define one radian. (ii) Show that \pi^c=180^\circ. (iii) Express 1^c in degrees, minutes and seconds (to the nearest second). (iv) The angles of a triangle are in the ratio 2:3:5. Find them in degrees, in grades and in radians. [4]
2. If \tan\theta=\frac{8}{15}, find the other trigonometric ratios of the acute angle \theta and verify that \sin^2\theta+\cos^2\theta=1 and \sec^2\theta-\tan^2\theta=1. [4]
3. Prove that (i) \frac{1}{1+\sin\theta}+\frac{1}{1-\sin\theta}=2\sec^2\theta (ii) (\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1. [4]
4. (i) Evaluate 4\sin^230^\circ+3\tan^230^\circ-8\sin45^\circ\cos45^\circ. (ii) Evaluate \sin150^\circ+\cos120^\circ-\tan135^\circ. (iii) Show that \sin(90^\circ+\theta)\cos(180^\circ-\theta)+\cos(90^\circ+\theta)\sin(180^\circ-\theta)=-1. [4]
5. From the top of a cliff 45 m high, the angle of depression of a boat is 30^\circ. (i) Draw a labelled figure. (ii) Find the distance of the boat from the foot of the cliff. (iii) Find the length of the line of sight from the top of the cliff to the boat. [4]

Answers to Group C

1. (i) One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. (ii) The circumference 2\pi r contains 2\pi arcs of length r, so a full turn is 2\pi radians; a full turn is also 360^\circ, so 2\pi^c=360^\circ and \pi^c=180^\circ. (iii) 1^c=\frac{180^\circ}{\pi}\approx57.2958^\circ; 0.2958^\circ\times60=17.75' and 0.75'\times60=45'', so 1^c\approx57^\circ17'45''. (iv) The parts are 2+3+5=10, so the angles are \frac{2}{10}\times180^\circ=36^\circ, 54^\circ and 90^\circ; in grades 40^g, 60^g, 100^g; in radians \frac{\pi}{5}, \frac{3\pi}{10}, \frac{\pi}{2}.

2. p=8, b=15, h=\sqrt{64+225}=17. So \sin\theta=\frac{8}{17}, \cos\theta=\frac{15}{17}, \mathrm{cosec}\,\theta=\frac{17}{8}, \sec\theta=\frac{17}{15}, \cot\theta=\frac{15}{8}. Check: \sin^2\theta+\cos^2\theta=\frac{64}{289}+\frac{225}{289}=\frac{289}{289}=1 ✓; \sec^2\theta-\tan^2\theta=\frac{289}{225}-\frac{64}{225}=\frac{225}{225}=1 ✓.

3. (i) LHS =\frac{(1-\sin\theta)+(1+\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}=\frac{2}{1-\sin^2\theta}=\frac{2}{\cos^2\theta}=2\sec^2\theta= RHS. (ii) LHS =\sec^2\theta-\tan^2\theta (difference of two squares) =1= RHS, using 1+\tan^2\theta=\sec^2\theta.

4. (i) 4\left(\frac{1}{2}\right)^2+3\left(\frac{1}{\sqrt{3}}\right)^2-8\cdot\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}=1+1-4=-2. (ii) \frac{1}{2}+\left(-\frac{1}{2}\right)-(-1)=1. (iii) LHS =\cos\theta(-\cos\theta)+(-\sin\theta)(\sin\theta)=-(\cos^2\theta+\sin^2\theta)=-1= RHS.

5. (i) A vertical line CF of 45 m for the cliff, a horizontal line through C, the boat D on the ground level with F, and the line of sight CD; the 30^\circ angle between the horizontal and CD at C, and an equal 30^\circ at D (alternate angles). (ii) \tan30^\circ=\frac{45}{FD}, so FD=45\sqrt{3}\approx77.94 m. (iii) \sin30^\circ=\frac{45}{CD}, so CD=90 m. Check: 45^2+77.94^2\approx2025+6075=8100=90^2 ✓.