You want the height of a tree. You can’t climb it, but you can walk away from its base. You pace out 30 m along flat ground, then use a phone’s inclinometer app to measure the angle from the ground up to the treetop: 35°. That’s all you have, one distance and one angle. Is it enough?

Yes, it is. Once you know the angle and the ground distance, only one right triangle fits, so the height is determined. You just need a way to compute it.

Think back to similar triangles (Chapter 9). Two triangles with the same angles have the same shape, and one is a scaled copy of the other. Scaling multiplies every side by the same factor, so the ratio of any two sides doesn’t change.

So for a right triangle with a 35° angle, the ratio \(\text{height}/\text{ground}\) is the same whether the triangle is drawn on a napkin or stretches across a valley. Measure it once on any 35° right triangle (the napkin will do) and you can use it for every 35° right triangle. People have tabulated these ratios for every angle, and now a calculator does it for you.

The 30 m ground distance is adjacent to the 35° angle, and the height we want is opposite it. The ratio linking opposite and adjacent is tangent:

One refinement: your eye (or phone) isn’t at ground level. If it is 1.5 m up, the triangle starts at that height, and the tree is about \(21.0 + 1.5 = 22.5\) m tall. Real surveying is full of corrections like this.

A 6 m ladder leans against a wall at 70° to the ground. How high up the wall does it reach, and how far is its foot from the wall?

The ladder is the hypotenuse. The height on the wall is opposite the 70° angle, and the distance from the wall is adjacent to it.

Now the reverse problem. A wheelchair ramp rises 1 m over a horizontal run of 12 m. What angle does it make with the ground?

Here we know \(\tan\theta = 1/12\) and want \(\theta\). The function that answers "which angle has this tangent?" is the inverse tangent, written \(\tan^{-1}\) or \(\arctan\):

Three angles come up so often that their exact ratios are worth deriving once.

45°. Take a square with side 1 and cut it along a diagonal. Each half is a right isosceles triangle with legs 1, 1 and hypotenuse \(\sqrt{1^2 + 1^2} = \sqrt2\):

30° and 60°. Take an equilateral triangle with side 2 (all angles 60°) and drop a perpendicular from the top vertex. It bisects the base, giving a right triangle with hypotenuse 2, short leg 1, angles 30° and 60°, and long leg \(\sqrt{2^2 - 1^2} = \sqrt3\):

A pattern to remember: down the sine column the values are \(\tfrac{\sqrt0}{2}, \tfrac{\sqrt1}{2}, \tfrac{\sqrt2}{2}, \tfrac{\sqrt3}{2}, \tfrac{\sqrt4}{2}\), and cosine is the same list read upward. Cosine is the sine of the *co*mplementary angle: \(\cos\theta = \sin(90^\circ - \theta)\). That’s where the "co" in cosine comes from.

Why does a full turn have 360 degrees? The Babylonians liked base 60, and 360 is close to the number of days in a year and divides evenly by many numbers. Those are cultural reasons, not mathematical ones. If you had to pick a unit for angles that came from the geometry of the circle itself, what would you choose?

An angle at the centre of a circle cuts off an arc. Larger angles cut off longer arcs, in proportion. On a circle of radius 1, the arc length could be the measure of the angle. On a circle of radius \(r\) everything scales by \(r\), so the natural measure is

That’s the radian. One radian is the angle whose arc is exactly one radius long. Try it: See one radian.

A full circle has circumference \(2\pi r\), so a full turn is \(2\pi r / r = 2\pi\) radians. Therefore

Convert 135° to radians: \(135 \times \dfrac{\pi}{180} = \dfrac{135\pi}{180} = \dfrac{3\pi}{4} \approx 2.356\). Try it: Convert 135° interactively.

Convert \(\tfrac{5\pi}{6}\) to degrees: \(\dfrac{5\pi}{6} \times \dfrac{180}{\pi} = \dfrac{5 \times 180}{6} = 150^\circ\).

Arc length. A pizza with radius 18 cm is cut into 8 equal slices. How long is the crust edge of one slice? Try it: Arc length of the pizza slice.

Sector area. A sector with angle \(\theta\) is the fraction \(\theta / 2\pi\) of the whole disc, so its area is \(\dfrac{\theta}{2\pi}\cdot \pi r^2 = \tfrac12 r^2\theta\). For the pizza slice: \(\tfrac12 \times 18^2 \times \tfrac{\pi}{4} = 40.5\pi \approx 127.2\ \text{cm}^2\).

The nautical mile. The Earth’s radius is about 6,371 km. One minute of arc (\(1/60\) of a degree) along a great circle is

which is (to within rounding) the nautical mile, 1.852 km. Sea and air navigation still use it for this reason. One degree of latitude is about 111 km.

With degrees, \(s = r\theta\) would need a conversion factor: \(s = \tfrac{\pi}{180} r\theta\). Radians remove it. The real payoff comes in calculus. Look at small angles:

Try it: Watch sin θ/θ approach 1.

For small \(\theta\) in radians, \(\sin\theta \approx \theta\). In Chapter 16 this becomes \(\lim_{\theta\to 0} \frac{\sin\theta}{\theta} = 1\), and in Chapter 17 it gives the clean rule "the derivative of \(\sin x\) is \(\cos x\)". In degrees the rule would read "the derivative of \(\sin x\) is \(\tfrac{\pi}{180}\cos x\)", and that factor would follow you through every formula in physics. Radians are the unit in which the circle’s own geometry needs no conversion constant.

The small-angle approximation \(\sin\theta \approx \theta\) is also what makes the simple pendulum formula work (Section 13.8).

Radians are the natural unit whenever something turns and you care how far a point on it travels, because the link is simply \(s = r\theta\), or per second, \(v = r\omega\).

SOH-CAH-TOA needs a right triangle, so it only works for angles between 0° and 90°. But a wheel turns through 120°, 270° and 1,000°, and a robot arm can rotate by \(-45^\circ\). What is \(\sin 150^\circ\)? What is \(\cos(-30^\circ)\)? We need definitions that work for any angle and agree with the triangle definitions where those apply.

Draw a circle of radius 1 centred at the origin, the unit circle. Start at the point \((1, 0)\) and move anticlockwise around the circle through an angle \(\theta\). Call where you end up \(P\).

For an acute angle, drop a perpendicular from \(P\) to the \(x\)-axis. That gives a right triangle with hypotenuse 1, so the adjacent side (the \(x\)-coordinate of \(P\)) is \(\cos\theta\) and the opposite side (the \(y\)-coordinate) is \(\sin\theta\). So for acute angles,

Now turn this around and make it the definition for every angle.

Signs by quadrant. The signs of \(x\) and \(y\) in each quadrant give:

(A traditional mnemonic, anticlockwise from quadrant I: "*A*ll *S*tudents *T*ake *C*alculus". All three are positive in I, Sine in II, Tangent in III, Cosine in IV.)

Bounds. \(P\) is on a circle of radius 1, so \(-1 \le \sin\theta \le 1\) and \(-1 \le \cos\theta \le 1\).

Periodicity. Adding a full turn brings you back to the same point:

Sine and cosine are periodic with period \(2\pi\). Tangent repeats every \(\pi\), because \(P\) and the opposite point \(-P\) give the same slope.

Symmetry. Reflecting in the \(x\)-axis turns angle \(\theta\) into \(-\theta\). That flips \(y\) and leaves \(x\) unchanged:

\(\sin 150^\circ\). Quadrant II, reference angle \(180^\circ - 150^\circ = 30^\circ\), sine is positive in II: \(\sin 150^\circ = +\sin 30^\circ = \tfrac12\). Try it: Evaluate sin 150° step by step.

\(\cos 240^\circ\). Quadrant III, reference angle \(240^\circ - 180^\circ = 60^\circ\), cosine is negative in III: \(\cos 240^\circ = -\cos 60^\circ = -\tfrac12\). Try it: Evaluate cos 240° step by step.

\(\tan\!\left(\tfrac{11\pi}{6}\right)\). \(\tfrac{11\pi}{6} = 330^\circ\): quadrant IV, reference angle 30°, tangent is negative in IV: \(\tan 330^\circ = -\tan 30^\circ = -\tfrac{\sqrt3}{3}\). Try it: Evaluate tan(11π/6) step by step.

\(\sin 765^\circ\). \(765 - 2 \times 360 = 45\), so \(\sin 765^\circ = \sin 45^\circ = \tfrac{\sqrt2}{2}\). Try it: Evaluate sin 765° step by step.

Because the functions repeat, an equation like \(\sin x = \tfrac12\) has infinitely many solutions. Your calculator’s \(\sin^{-1}\) returns just one of them.

Solve \(\sin x = \tfrac12\) for \(0 \le x < 2\pi\): Try it: Find every solution of sin x = ½.

The general solution is \(x = \tfrac{\pi}{6} + 2\pi k\) or \(x = \tfrac{5\pi}{6} + 2\pi k\) for any integer \(k\).

Plot the \(y\)-coordinate of the moving point against the angle it has turned through, and you get the sine wave. Plot the \(x\)-coordinate and you get the cosine wave, which is the same shape shifted left by a quarter turn: \(\cos x = \sin(x + \tfrac{\pi}{2})\).

Key features:

Real oscillations don’t swing exactly between −1 and 1 with period \(2\pi\). You get the general sinusoid by applying the function transformations from Chapter 11:

Try it: Adjust amplitude, period and phase.

In a city at about 40° N, the longest day (around 21 June, day 172) has about 15 hours of daylight and the shortest (around 21 December, day 355) about 9 hours. Model daylight hours \(H\) against day of the year \(t\) with a sinusoid. Try it: Fit the daylight model.

So \(H(t) = 3\sin\!\big(\tfrac{2\pi}{365}(t - 80.75)\big) + 12\). Check day 172: the argument is \(\tfrac{2\pi}{365} \times 91.25 = \tfrac{\pi}{2}\), so \(H = 3 + 12 = 15\). Correct.

Anything that repeats on a schedule can be modelled, at least roughly, by a sinusoid, and the four parameters answer practical questions.

In a physics problem you get \(\sin\theta = 0.6\) and need \(\cos\theta\). A graphics routine needs the sine of a sum of two rotations. A signal-processing derivation needs \(\sin^2 x\) written without the square. These all call for identities, equations that hold for every angle. They let you rewrite one trig expression as another.

The point \(P = (\cos\theta, \sin\theta)\) lies on the unit circle \(x^2 + y^2 = 1\) (Chapter 10). Substitute:

That’s Pythagoras' theorem for a right triangle with hypotenuse 1, and the unit-circle definition makes it hold for every angle, not just acute ones. Dividing through by \(\cos^2\theta\) or \(\sin^2\theta\) gives two variants:

Worked example. \(\sin\theta = 0.6\) and \(\theta\) is in quadrant II. Find \(\cos\theta\) and \(\tan\theta\).

Is \(\sin(\alpha + \beta) = \sin\alpha + \sin\beta\)? Try \(\alpha = \beta = 90^\circ\): the left side is \(\sin 180^\circ = 0\), the right side is \(1 + 1 = 2\). No. So what is it?

The construction in the figure:

Now read off \(P\)'s coordinates two ways:

The picture assumes acute angles with an acute sum, but the formulas hold for all angles. (The cleanest general proof uses rotation matrices, Chapter 27, or complex numbers, Chapter 15.) Replacing \(\beta\) with \(-\beta\) and using even/odd symmetry gives the difference formulas.

Worked example: an exact value for \(\sin 75^\circ\).

Put \(\beta = \alpha\) in the sum formulas:

Using \(\sin^2\alpha + \cos^2\alpha = 1\) to remove one or the other gives two more forms of \(\cos 2\alpha\):

Rearranged, these are the power-reduction formulas that calculus uses all the time (Chapter 19):

Note what they say: the square of a wave at frequency \(f\) is a constant plus a wave at frequency \(2f\). That fact explains why the power of an AC signal pulses at twice the mains frequency.

A surveyor on one bank of a river marks two points \(A\) and \(B\), 500 m apart. A tree \(C\) stands on the far bank. From \(A\), the angle between \(B\) and the tree is 52°. From \(B\), the angle between \(A\) and the tree is 71°. How far is the tree from each point, and how wide is the river?

No right angle appears anywhere, so SOH-CAH-TOA doesn’t apply directly. We need tools for any triangle.

Drop the altitude \(h\) from \(C\) to side \(AB\). It splits the triangle into two right triangles, and in each one SOH gives:

Dropping an altitude from a different vertex brings \(c\) and \(C\) in the same way. The result is the law of sines:

Each side is proportional to the sine of the angle opposite it. (The common ratio has a meaning: it’s the diameter of the triangle’s circumscribed circle.)

The river, solved. \(C = 180^\circ - 52^\circ - 71^\circ = 57^\circ\), and \(c = AB = 500\). Try it: Solve the river triangle.

This is triangulation. From one carefully measured baseline and many angle measurements, surveyors mapped whole countries. The 18th–19th-century Great Trigonometrical Survey of India measured the height of Everest this way.

The law of sines needs at least one angle–opposite side pair. If you know two sides and the angle between them (or all three sides), you need the law of cosines instead.

Derivation. Use the triangle in the figure. Put \(A\) at the origin and \(B\) at \((c, 0)\). \(C\) is at distance \(b\) from \(A\) at angle \(A\), so \(C = (b\cos A,\ b\sin A)\). Then \(a\) is the distance from \(B\) to \(C\) (distance formula, Chapter 10):

Worked example: three sides, find an angle. A triangle has sides 5, 7 and 8. Find the angle opposite the side of length 7. Try it: Solve the 5-7-8 triangle.

Worked example: navigation. A ship sails 20 km on a bearing of 040°, then turns and sails 30 km on a bearing of 110°. How far is it from port? Try it: Solve the navigation triangle.

Bearings are measured clockwise from north. The ship’s heading changes by \(110^\circ - 40^\circ = 70^\circ\), so the interior angle of the triangle at the turning point is \(180^\circ - 70^\circ = 110^\circ\).

The obtuse angle makes the distance longer than \(\sqrt{20^2 + 30^2} \approx 36.06\) km, as the sign rule predicts. (In Section 13.9 we solve the same problem with components and atan2, which also gives the bearing back to port.)

Try it: See both SSA solutions.

A radar screen reports a contact "5 km away at 30°". A robot vacuum knows its wheel has turned through some angle at some distance from its axle. Spirals, flower petals and orbits are awkward to describe as \(y = f(x)\). All of these are naturally given by how far and which direction, not by how far across and how far up.

Cartesian to polar: \((3, 2)\).

Cartesian to polar: \((-3, 2)\). \(r = \sqrt{13}\) again. \(\tan^{-1}(2/(-3)) = \tan^{-1}(-0.667) \approx -33.69^\circ\), which points into quadrant IV, the wrong quadrant. The point is in quadrant II, so add 180°: \(\theta \approx 146.31^\circ\). (atan2 does this correction for you.)

Polar to Cartesian: \(r = 4,\ \theta = \tfrac{5\pi}{6}\). \(x = 4\cos\tfrac{5\pi}{6} = 4 \times (-\tfrac{\sqrt3}{2}) = -2\sqrt3 \approx -3.464\), \(y = 4\sin\tfrac{5\pi}{6} = 4 \times \tfrac12 = 2\).

Curves. \(r = 3\) is a circle (compare \(x^2 + y^2 = 9\)). \(r = \theta\) is a spiral. \(r = 1 + \cos\theta\) is a heart-shaped cardioid, which is also the pickup pattern of a cardioid microphone.

Radar and sonar; robot arm kinematics (joint angles and link lengths); orbital mechanics; antenna and microphone directional patterns; and complex numbers, where \(r\) and \(\theta\) become modulus and argument (Chapter 15). Two concrete cases:

Pull a mass on a spring down and let go. It bounces up and down with a regular rhythm. A pendulum, a guitar string, the air pressure at your eardrum and the voltage in a wall socket all do something similar. What function describes this motion, and why that one?

If you studied Mechanics, you met simple harmonic motion (SHM): motion in which the restoring force is proportional to the displacement and directed back towards equilibrium. For a spring with stiffness \(k\) and mass \(m\), that’s Hooke’s law with Newton’s second law:

where \(\ddot{x}\) (read "x double-dot") is the acceleration, the second derivative of position with respect to time (Chapter 17). The solution is a sinusoid:

Why does a sinusoid appear? Think of the unit circle again. A point moving round a circle at a steady angular speed \(\omega\) has \(x\)-coordinate \(\cos(\omega t)\), and its acceleration always points to the centre, so the \(x\)-component of the acceleration is proportional to \(-x\). SHM is the shadow of uniform circular motion. A cylinder on a turntable lit from the side casts a shadow that moves in exactly simple harmonic motion. Try it: Watch SHM as a shadow.

Worked example. A 0.5 kg mass hangs on a spring with \(k = 200\) N/m. It is pulled 4 cm below equilibrium and released. Try it: Run the spring example.

The pendulum. For a pendulum of length \(L\), the restoring force is proportional to \(\sin\theta\), not \(\theta\), so the motion isn’t exactly SHM. For small swings, though, \(\sin\theta \approx \theta\) (in radians, Section 13.2), and the motion becomes SHM with \(\omega = \sqrt{g/L}\), giving period \(T = 2\pi\sqrt{L/g}\). A 1 m pendulum has \(T = 2\pi\sqrt{1/9.81} \approx 2.006\) s, which is why a "seconds pendulum" (one second per swing) is about a metre long.

A wave is an oscillation that travels. A sinusoidal wave moving along a string is

where \(k = 2\pi/\lambda\) is the wavenumber (\(\lambda\), lambda, is the wavelength) and the wave travels at speed \(v = \omega/k = f\lambda\). Sound in air at 343 m/s with frequency 440 Hz (concert A) has wavelength \(343/440 \approx 0.78\) m.

A pure sine wave sounds like a tuning fork: clean and rather dull. Real instruments produce a fundamental frequency plus harmonics at whole-number multiples of it, each with its own amplitude. The mix of harmonics is what makes a violin sound different from a clarinet at the same pitch. Try it: Spectrum of three sines.

The problem. A game sprite at \((x, y)\) relative to its pivot has to rotate by angle \(\theta\). Where does it end up?

Derivation. Write the point in polar form: \(x = r\cos\varphi\), \(y = r\sin\varphi\), with \(r\) its distance from the origin and \(\varphi\) its current angle. Rotating keeps \(r\) the same and adds \(\theta\) to the angle, so the new point is

Worked example. Rotate \((2, 0)\) by 30°: \(x' = 2\cos 30^\circ - 0 = \sqrt3 \approx 1.732\), \(y' = 2\sin 30^\circ + 0 = 1\). The distance is still \(\sqrt{3 + 1} = 2\). Correct.

Worked example. Rotate \((3, 1)\) by 90°: \(\cos 90^\circ = 0\), \(\sin 90^\circ = 1\), so \(x' = -1\), \(y' = 3\). Rotating by 90° sends \((x, y) \mapsto (-y, x)\), a trick worth remembering for perpendicular vectors.

Notice 0.9999999999999999 and -0.9999999999999998. Neither \(\pi/6\) nor \(\pi/2\) is exactly representable as a float, so math.cos(math.radians(90)) is about \(6 \times 10^{-17}\), not 0. Compare floating-point trig results with a tolerance (math.isclose), never with ==.

The problem. An enemy at \((e_x, e_y)\) has to turn to face the player at \((p_x, p_y)\). The direction vector is \((d_x, d_y) = (p_x - e_x, p_y - e_y)\), and the facing angle \(\theta\) satisfies \(\tan\theta = d_y/d_x\). So is the answer atan(dy/dx)? Try it: Watch atan aim the wrong way.

No, for two reasons:

atan2(y, x) takes the two components separately. It uses their signs to pick the correct quadrant and returns an angle in \((-\pi, \pi]\) (−180° to 180°). It handles \(x = 0\) without trouble. Note the argument order: y first, matching \(\tan^{-1}(y/x)\).

The navigation problem by components. Let east be \(x\) and north be \(y\). A bearing \(\beta\) (clockwise from north) points in the direction \((\sin\beta, \cos\beta)\). Note that this is the reverse of the usual math convention.

This matches the law-of-cosines answer and also gives the direction. Swapping the arguments to atan2 converts between the math convention (anticlockwise from east) and the compass convention (clockwise from north). Swapping them by accident is a classic bug.

Try it: See the degrees bug.

A digital audio signal is a list of samples, numbers taken at a fixed sample rate (44,100 per second for CD audio). A pure tone of frequency \(f\) Hz and amplitude \(A\) is

where \(f_s\) is the sample rate and \(n/f_s\) is the time of sample \(n\) in seconds. Here is a complete, standard-library-only synthesiser that writes a one-second chord (A3, its octave, and a fifth above that) to a WAV file:

The amplitudes add up to 0.875, so the sum never goes past full scale and doesn’t clip.

Fourier analysis runs synthesis in reverse: given the summed waveform, recover which frequencies are in it and how strong each one is. NumPy’s FFT does this in \(O(n\log n)\) time. (That’s the logarithm from Chapter 12 again.) Try it: See the spectrum live; Find the three sines.

Try it: Run this in your browser.

The FFT recovers exactly the three sines that went in. In Chapter 15 you’ll see why it uses complex numbers: Euler’s formula \(e^{i\theta} = \cos\theta + i\sin\theta\) packs a sine and a cosine into one exponential, which ties this chapter to the previous one.