1The idea in one paragraph

Turn each of sin, cos and tan upside down and you get three new ratios: cosec θ = 1/sin θ, sec θ = 1/cos θ and cot θ = 1/tan θ. They are new names, not new information, and they make some expressions much shorter. Dividing cos²θ + sin²θ = 1 by cos²θ or by sin²θ gives two more Pythagorean identities, 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ, and those turn many equations into quadratics. The second half of the subtopic goes the other way: from a ratio back to an angle. Sin, cos and tan take every value infinitely often, so they have no inverse until you cut each one down to a piece where it takes every value once. The inverses of those pieces are arcsin, arccos and arctan, and their domains and ranges come straight from how the cut was made.

2Three new ratios

The three reciprocal trigonometric ratios are defined by

sec θ = 1/cos θ, cosec θ = 1/sin θ, cot θ = 1/tan θ = cos θ/sin θ.

The formula booklet prints sec and cosec with the new identities. Learn cot too: it is 1/tan θ, and also cos θ/sin θ, which is the form that still works when tan θ is undefined. In a right-angled triangle each one is its partner turned upside down, as Figure 1 shows. The third letter pairs them: sec goes with cos, cosec with sin, cot with tan. Pairing sec with sin, because both start with s, is the classic slip.

Two facts follow from the definitions and save marks.

Each reciprocal has the sign of its partner. A number and its reciprocal are positive together and negative together. So sec θ is negative wherever cos θ is (the second and third quadrants), cosec θ wherever sin θ is (the third and fourth), and cot θ wherever tan θ is (the second and fourth).

Each reciprocal is undefined where its partner is 0. sec θ is undefined at θ = π/2, 3π/2, …, cosec θ and cot θ at θ = 0, π, 2π, … And because |sin θ| ≤ 1 and |cos θ| ≤ 1, turning them upside down gives |sec θ| ≥ 1 and |cosec θ| ≥ 1: sec and cosec never take a value strictly between −1 and 1. That rejects roots in equations later. cot θ, like tan θ, takes every real value.

Worked example 1 (Paper 1). Find the exact values of sec(π/3), cosec(π/4), cot(π/6), sec(5π/6) and cot(π/2).

Your calculator has no sec, cosec or cot keys: type 1 ÷ cos(…), never cos⁻¹, which is a different function (section 5).

Figure 2 draws the three new graphs over one turn, each with its partner dashed. Where the partner crosses zero, the reciprocal shoots off to a vertical asymptote; where the partner reaches 1 or −1, the two graphs touch, because 1 and −1 are their own reciprocals. sec and cosec have period 2π and cot has period π, like their partners.

3Two more Pythagorean identities

Start from the one you know, cos²θ + sin²θ = 1, and divide every term by cos²θ (allowed wherever cos θ ≠ 0).

Divide the same identity by sin²θ instead (wherever sin θ ≠ 0).

1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ · both in the formula booklet. Each is cos²θ + sin²θ = 1 divided through.

If you forget which way round the second one goes, rebuild it in two lines; a "show that" question may ask for exactly that.

Use 1: from one ratio to another without finding the angle.

Worked example 2 (Paper 1). Given that tan θ = −3/4 and π/2 < θ < π, find the exact values of sec θ, cos θ, cosec θ and cot θ.

The square root gives two signs and the quadrant decides; the written reason is the mark.

Use 2: proving identities. An identity (≡) holds for every θ where both sides are defined. Start from the more complicated side and turn it into the other; rewriting everything in sin and cos is the reliable first move.

Worked example 3 (Paper 1). Show that sec θ − cos θ ≡ sin θ tan θ.

Use 3: equations that become quadratics. If an equation mixes tan²x with sec x, or cot²x with cosec x, replace the square with the identity so only one ratio is left.

Worked example 4 (Paper 1). Solve 2cot²x = 5cosec x − 4 for 0 ≤ x ≤ 2π.

Rejecting cosec x = 1/2 with a reason is worth a mark; dropping it silently may not be.

Worked example 5 (Paper 2). Solve sec²x + tan x = 3 for 0 ≤ x ≤ π.

The calculator's arctan 2 = 1.107… is in the first quadrant; the solution in [0, π] with a negative tangent is π minus it.

4Why sin, cos and tan need cutting down before they can be undone

An inverse function exists only for a one-to-one function, where each output comes from exactly one input. Sine fails badly. Figure 3 draws y = sin x with the line y = 0.5: they meet again and again, at π/6, 5π/6, π/6 − 2π and so on. "The angle whose sine is 0.5" has infinitely many answers.

The fix is to restrict the domain to a piece of the graph that takes every value between −1 and 1 exactly once, and includes the angles we use most. For sine, the agreed piece is −π/2 ≤ x ≤ π/2, drawn in teal. On it, y = 0.5 is met once, at π/6. For cosine the agreed piece is 0 ≤ x ≤ π (cos is even, so a piece centred on 0 would repeat values). For tangent it is −π/2 < x < π/2, one full branch between asymptotes. These are conventions, so every calculator gives the same answer.

5arcsin, arccos and arctan

The inverse sine function, written arcsin x or sin⁻¹x, is the inverse of sin x on −π/2 ≤ x ≤ π/2. It answers: which angle in [−π/2, π/2] has sine equal to x? Figure 4 draws it as the reflection of that piece of sine in y = x. The reflection swaps every coordinate pair, so the domain and range swap too.

The same idea gives the other two. Figure 5 sets all three side by side.

arcsin and arccos accept only −1 ≤ x ≤ 1. arcsin and arctan return angles in the first or fourth quadrant; arccos returns angles in the first or second.

In Figure 5, arcsin rises from (−1, −π/2) to (1, π/2) with end points, not arrows; arccos falls from (−1, π) through (0, π/2) to (1, 0); arctan rises through the origin, flattening towards the horizontal asymptotes y = π/2 and y = −π/2 without reaching them; they are the images of tan's vertical asymptotes after reflection in y = x. arcsin and arctan are odd functions, so arcsin(−x) = −arcsin x and arctan(−x) = −arctan x. arccos is not: arccos(−x) = π − arccos x.

The notation trap. sin⁻¹x means arcsin x, the inverse function. It does not mean 1/sin x, which is cosec x. The −1 here is the same −1 as in f⁻¹(x), not a power. (sin²x does mean (sin x)²; only the −1 is special.)

Worked example 6 (Paper 1). Find the exact values of arcsin(−1/2), arccos(−√3/2), arctan(−1) and arccos 0.

Other angles share each ratio, but only the one in the range is the value of the function.

6Composites: undoing, and not quite undoing

arcsin undoes sin only on its restricted piece, so the two orders of composition differ.

sin(arcsin x) = x for every x in −1 ≤ x ≤ 1. arcsin x is some angle whose sine is x; take its sine and you get x back. The same holds for cos(arccos x) and, for every real x, tan(arctan x).

arcsin(sin x) = x only when −π/2 ≤ x ≤ π/2. Outside that, arcsin still has to return an angle in its own range. Take x = 5π/6: sin(5π/6) = 1/2, and arcsin(1/2) = π/6, not 5π/6. Likewise arctan(tan(3π/4)) = arctan(−1) = −π/4, and arccos(cos(−π/3)) = arccos(1/2) = π/3.

A ratio of an inverse function: draw the angle. To find cos(arcsin(3/5)), call the angle α = arcsin(3/5). Then sin α = 3/5 and α is in [−π/2, π/2]; since 3/5 > 0, α is acute. Draw a right-angled triangle with opposite 3 and hypotenuse 5, as in Figure 6(a). Pythagoras gives the adjacent side 4, so cos(arcsin(3/5)) = 4/5 and tan(arcsin(3/5)) = 3/4.

When the number inside is negative, the triangle alone can mislead you about the sign, so use the unit circle and the range.

Worked example 7 (Paper 1). Find the exact values of tan(arccos(−1/3)) and sec(arccos(−1/3)).

The same reasoning gives cos(arcsin x) = √(1 − x²) for −1 ≤ x ≤ 1, with the positive root because cosine is never negative on arcsin's range.

7Transformed inverse functions and inverses of restricted functions

Two skills from Topic 2 combine with the table in section 5.

Domain and range of a transformed inverse. The input of arcsin or arccos must lie in [−1, 1]; solve that for x to get the domain. The output starts in the standard range; apply the outside transformations to get the new range.

Worked example 8 (Paper 1). f(x) = 2arcsin(x − 1). Find the largest possible domain of f and the range of f.

For g(x) = 3arctan(2x), the domain is ℝ (arctan takes anything) and the range is −3π/2 < g(x) < 3π/2, with strict inequalities because the asymptotes are never reached.

The inverse of a restricted trigonometric function. f(x) = 3sin 2x on −π/4 ≤ x ≤ π/4 is one-to-one (2x runs over [−π/2, π/2], exactly sine's piece), so it has an inverse.

The step 2y = arcsin(x/3) is valid only because 2y already lies in arcsin's range; on any other piece of the graph it would give the wrong angle.