Maths › Trigonometry › Reciprocal and inverse trigonometric functions
Reciprocal and inverse trigonometric functions
Six more functions arrive and none of them is genuinely new. Three are reciprocals of sine, cosine and tangent. Three run those functions backwards. The reciprocals buy you two fresh identities from one old one, and the inverses only exist at all because somebody restricted a domain first.
Builds on Radians, arcs and small angles and Functions, inverses and the modulus.
IN THIS TOPIC
- Work with sec, cosec and cot, including their graphs, asymptotes and ranges.
- Derive and use sec² = 1 + tan² and cosec² = 1 + cot².
- Solve equations that are quadratics in a reciprocal ratio.
- Use arcsin, arccos and arctan with their restricted domains and ranges.
COMMON MISCONCEPTION
arcsin x means 1 over sin x.
The reciprocal three
Three new names and no new mathematics. sec x = 1/cos x, cosec x = 1/sin x, and cot x = cos x/sin x. Each one inherits its parent's period and blows up wherever the parent hits zero, so every one of the three comes in branches pinned between vertical asymptotes. The branches do not all look alike. sec and cosec have period 2π and each branch is a U or an upside-down U, turning at a height of 1 or −1 where the parent wave peaks. cot has period π, and each of its branches falls steadily from +∞ to −∞ across the gap between consecutive asymptotes, crossing zero halfway. Nothing turns, so cot has no maximum or minimum anywhere.
Two facts about these graphs get asked and almost never revised. Cosine never exceeds 1 in size, so sec x never drops below 1 in size, and neither sec x nor cosec x ever takes a value strictly between −1 and 1. And define cot as cos/sin, not as 1/tan, because cot (π/2) = 0 while 1/tan (π/2) invites you to divide by something undefined.
Two identities from one
Divide sin²x + cos²x = 1 through by cos²x and one new identity drops out. Divide the same line by sin²x instead and you get the other.
Both belong on the must-learn list, and each derivation is a single line. Rehearse them, because papers set the derivation itself as a question and not only the result.
WORKED EXAMPLE
An exact value through the identity
Given that tan θ = 3/4 with θ acute, find sec θ and cos θ exactly.
sec2 θ = 1 + 9/16 = 25/16, so sec θ = 5/4, positive because θ is acute.
cos θ is its reciprocal, 4/5.
No triangle drawn, no angle found. The identity moved straight between the ratios, and that is faster than anything involving a sketch.
GUIDED PRACTICE
A quadratic in cosec
Solve 2cot2 x + cosec x = 1 for 0 ≤ x < 2π, before opening the working.
Show the working
Trade cot² for cosec² − 1. That gives 2cosec2 x + cosec x − 3 = 0, which factorises as (2cosec x + 3)(cosec x − 1) = 0.
cosec x = 1 gives sin x = 1, so x = π/2. cosec x = −3/2 gives sin x = −2/3, so x = π + 0.730 and 2π − 0.730.
Solutions: x = π/2, 3.87, 5.55 (radians, 3 significant figures).
The identity converted two unknown ratios into one. After that it was ordinary factorising, in radians throughout because the interval was given in radians.
Running trig backwards
arcsin, arccos and arctan, also written sin−1 and so on, answer the question “which angle has this sine”. Sine is many-one, so an inverse only exists once the domain has been restricted. arcsin and arccos accept inputs in [−1, 1] and nothing outside it, since no angle has a sine of 2. arctan will accept any real number. Their outputs are pinned down too, to [−π/2, π/2] for arcsin, [0, π] for arccos and (−π/2, π/2) for arctan.
The notation is what breeds the confusion. sin−1 x is an inverse function, not a reciprocal. The reciprocal already has a name of its own, and that name is cosec.
INDEPENDENT PRACTICE
Exact inverses, no calculator
Write down the exact values of arcsin (½), arccos (½) and arctan (−1).
Show the working
arcsin (½) = π/6, the angle in [−π/2, π/2] whose sine is ½.
arccos (½) = π/3, from the range [0, π].
arctan (−1) = −π/4. Not 3π/4, because arctan's range stops at π/2 and the function has to pick the branch it owns.
Each answer is the exact-value table read backwards and then filtered through the stated range. When two angles compete, the range does the choosing.
ASSESSMENT FOCUS
- Rewrite sec, cosec and cot in terms of sin, cos and tan the moment an expression stops looking familiar.
- Derive the two squared identities by dividing sin² + cos² = 1. Examiners set that derivation on its own.
- Quadratics in sec or cosec factorise like any other quadratic. Convert the odd ratio out first so only one remains.
- An inverse answer outside the standard range is wrong even when its trig ratio checks out. Quote the range when you reject an angle.
- sin−1 means arcsin and 1/sin means cosec. Muddling those two is the oldest trap in the topic.
CHECK YOURSELF
Given that cosec θ = 3 with θ acute, find cot θ and cos θ exactly.
Show a hint
cosec² = 1 + cot², then cot = cos/sin.
Show the answer
cot2 θ = cosec2 θ − 1 = 8, so cot θ = 2√2, positive for acute θ.
sin θ = 1/3, and cos θ = cot θ × sin θ = 2√2/3.
Check it against Pythagoras: sin² + cos² = 1/9 + 8/9 = 1, as it has to.
sec, cosec and cot are reciprocals that blow up where their parents vanish.
Divide the Pythagorean identity for two new ones; restrict before inverting, and the range picks the angle.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the reciprocal and inverse trigonometric functions questions page.
CHECK YOUR PROGRESS
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- Work with sec, cosec and cot, including their graphs, asymptotes and ranges.
- Derive and use sec² = 1 + tan² and cosec² = 1 + cot².
- Solve equations that are quadratics in a reciprocal ratio.
- Use arcsin, arccos and arctan with their restricted domains and ranges.
Open the full revision checklist to see every objective in the course in one place.