Trigonometry

1Name the sides before you name the ratio

In a right triangle, \sin\theta = \frac{\text{opp}}{\text{hyp}}, \cos\theta = \frac{\text{adj}}{\text{hyp}} and \tan\theta = \frac{\text{opp}}{\text{adj}}. The three reciprocals are \csc\theta = 1/\sin\theta, \sec\theta = 1/\cos\theta and \cot\theta = 1/\tan\theta. Nothing in this chapter is harder than reading which side is opposite the angle you are standing at.

Two rewrite rules fall out of the definitions and are worth writing once: \tan\theta = \sin\theta/\cos\theta and \cot\theta = \cos\theta/\sin\theta. Every later identity is just these two plus Pythagoras, so the first move on a messy expression is always to convert every ratio into \sin and \cos.

Figure. The right angle sits at the bottom-left corner. Standing at θ on the right, the vertical leg is opposite, the horizontal leg is adjacent, and the slant is the hypotenuse. Side lengths are a 3-4-5 triple drawn to scale at aspect 1.6 (Δx = 0.45, Δy = 0.54), so the screen ratios match opp : adj : hyp.

How it works

1. Mark the right angleThe hypotenuse is the side facing it — the longest side, never one of the legs.
2. Stand at θOpposite is the leg you are not touching; adjacent is the leg that forms θ with the hypotenuse.
3. Write the ratioopp/hyp, adj/hyp or opp/adj — and flip for the reciprocal if the question asks for csc, sec or cot.

2The five angles you must know cold

Every numerical trig question in Tier 1 uses one of 0^\circ, 30^\circ, 45^\circ, 60^\circ and 90^\circ. Memorise the table once — \sin 30^\circ = 1/2, \cos 30^\circ = \sqrt{3}/2, \tan 45^\circ = 1, \tan 60^\circ = \sqrt{3} — and the arithmetic afterwards is one multiplication.

A pattern that saves a row when two elevations are 30^\circ and 60^\circ: \tan 60^\circ = 3\,\tan 30^\circ, because \sqrt{3} divided by 1/\sqrt{3} is 3. The two distances then sit in a 1 : 3 ratio without a second square-root evaluation.

Figure. A 30^\circ–60^\circ–90^\circ triangle with sides 1:\sqrt{3}:2 drawn to scale at aspect 1.8. Every Tier-1 standard value for 30^\circ and 60^\circ is a ratio of these three sides; the 45^\circ row is the isosceles twin 1:1:\sqrt{2} held in the concept table beside it.

3Three identities, one Pythagorean fact

Divide a^2 + b^2 = c^2 by c^2 and you get \sin^2\theta + \cos^2\theta = 1. Divide by \cos^2\theta instead and you get 1 + \tan^2\theta = \sec^2\theta; divide by \sin^2\theta and you get 1 + \cot^2\theta = \csc^2\theta. The three formulas in the sheet are one triangle identity written three ways.

The exam form is almost always the rearranged pair \sec^2\theta - \tan^2\theta = 1 and \csc^2\theta - \cot^2\theta = 1. Given one ratio, square it, subtract from 1 (or add 1), and take the square root — keeping the sign that an acute angle demands.

Figure. From the worked ledger with \sin\theta=3/5: \sin^2\theta=9/25 and \cos^2\theta=16/25 add to the unit bar. The other two sheet identities are the same Pythagorean fact divided by \cos^2\theta or \sin^2\theta instead of by the hypotenuse squared — one triangle, three writings.

How it works

1. Square what you knowIf sin θ = 3/5 then sin² θ = 9/25 — do not reach for a second formula yet.
2. Subtract from 1cos² θ = 1 − sin² θ, or sec² θ = 1 + tan² θ, depending on which pair you hold.
3. Root and signFor an acute angle every primary ratio is positive; take the positive square root and stop.

4Angles that add to 90° swap sin for cos

\sin(90^\circ - \theta) = \cos\theta and \tan(90^\circ - \theta) = \cot\theta — and the same swap runs through every co-pair: \cos(90^\circ - \theta) = \sin\theta, \cot(90^\circ - \theta) = \tan\theta, and so on. So \sin 37^\circ = \cos 53^\circ is not a coincidence; it is the definition of complementary.

Simplify questions are built on this. Pair every angle with the one that makes 90^\circ, rewrite, and \sin^2 + \cos^2 = 1 collapses the pair. An expression that looked like four different angles is usually two complementary pairs.

Figure. In a right triangle the two acute angles add to 90^\circ, so each is the complement of the other. The side opposite \theta is adjacent to 90^\circ-\theta, which is why \sin\theta=\cos(90^\circ-\theta) and every co-pair swaps. Simplify questions pair angles that make 90^\circ and rewrite.

How it works

1. Hunt for 90° pairs65° with 25°, 50° with 40°, 53° with 37° — any two that add to 90°.
2. Rewrite the co-functionsin(90° − θ) becomes cos θ; cos(90° − θ) becomes sin θ; tan becomes cot.
3. Collapse with PythagorasEach rewritten pair is sin² + cos² or the same under a different name, and equals 1.

5Elevation up, depression down, same tangent

The angle of elevation is measured up from the horizontal to the line of sight; the angle of depression is measured down from the horizontal. Between an observer and an object on level ground the two angles are equal — they are alternate angles on a transversal — so one right triangle serves both wordings.

Once the triangle is drawn, the height is h = d\,\tan\theta when the ground distance d and the elevation \theta are known. Pick \tan when you hold the adjacent side and want the opposite; pick \sin when you hold the hypotenuse instead.

Figure. Observer at the left end of the base, tower vertical on the right, line of sight as the hypotenuse. Elevation θ sits at the observer. With aspect 1.0 the base Δx = 0.30 and height Δy = 0.52 are in the ratio 1 : √3, matching tan 60° — the drawn geometry is the formula.

How it works

1. Draw the right triangleHorizontal ground, vertical height, line of sight as hypotenuse; mark θ at the observer.
2. Name the known sideA ground distance is adjacent to the elevation angle; a slant range is the hypotenuse.
3. Choose sin, cos or tanh = d tan θ when d is on the ground; rearrange if the question hands you h and asks for d.