Maths › Trigonometry › Reciprocal and inverse trigonometric functions

Reciprocal and inverse trigonometric functions

Six more functions arrive, none genuinely new: three are reciprocals of sine, cosine and tangent, and three run them backwards. The reciprocals bring two fresh identities from one old one, and the inverses only exist because somebody restricted a domain first.

Year 12-13EDEXCEL 9MA0 5.4, 5.5

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 and undefined points.
  • Use sec² = 1 + tan² and cosec² = 1 + cot² in equations and identities.
  • Use arcsin, arccos and arctan with their restricted domains and ranges.

WHAT YOU PROBABLY THINK

arcsin x means 1 over sin x.

The reciprocal three

Define sec x = 1/cos x, cosec x = 1/sin x and cot x = 1/tan x = cos x/sin x. Each inherits its parent's period and blows up where its parent vanishes, so the graphs are families of U-shaped branches pinned between vertical asymptotes.

The graph of sec x above the cosine it is built from: where cosine touches 1 the secant touches 1, and where cosine crosses zero the secant blows up against vertical asymptotes at pi over 2 and 3 pi over 2sec xcos xsec touches where cos peaks, and explodes where cos vanishes
FIG. 1sec x built from cos x: touching at the peaks, undefined at the zeros. The other two reciprocal graphs follow the same recipe from sin and tan.

Dividing sin²x + cos²x = 1 through by cos²x, and then instead by sin²x, produces the two new must-learn identities,

sec2 θ = 1 + tan2 θNOT IN THE BOOKLET — LEARN IT
cosec2 θ = 1 + cot2 θNOT IN THE BOOKLET — LEARN IT

and both derivations are one line each, worth rehearsing because papers ask for them.

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 was drawn and no angle found; the identity moved straight between the ratios, which is exactly what it is for.

YOUR TURN

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: 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, and the algebra lesson's factorising habit did the rest, in radians throughout.

Running trig backwards

The inverse functions arcsin, arccos and arctan, also written sin−1 and so on, answer “which angle has this sine”. But sine is many-one, so the functions lesson demands a restricted domain first: arcsin returns angles in [−π/2, π/2], arccos in [0, π], arctan in (−π/2, π/2). And the notation sets up the opening lie: sin−1 x is an inverse function, never a reciprocal. The reciprocal already has its own name, cosec.

The restricted sine curve from minus pi over 2 to pi over 2 and its inverse arcsin, drawn at equal scales as reflections in the dashed line y equals xarcsin xsin x, restrictedy = xrestrict first, then reflect
FIG. 2arcsin as the restricted sine reflected in y = x. The restriction to [−π/2, π/2] is what makes the reflection a function.

TRY IT UNSEEN

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 must pick the branch it owns.

Each answer is the exact-value table read backwards, filtered through the stated range; the range does the choosing when two angles compete.

THE EXAM BIT

  • Write sec, cosec and cot in terms of sin, cos and tan before manipulating anything unfamiliar.
  • Derive the two squared identities from sin² + cos² = 1 by dividing; the derivation is itself a stock question.
  • Quadratics in cosec or sec factorise like any quadratic; convert cot² or tan² first so only one ratio remains.
  • Inverse-function answers must land in the standard ranges; state the range when justifying a rejected angle.
  • sin−1 means arcsin, and 1/sin means cosec; muddling the two is this topic's oldest trap.

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.

A quick identity check: sin² + cos² = 1/9 + 8/9 = 1, as it must.

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.

CHECK YOUR PROGRESS

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  • Work with sec, cosec and cot, including their graphs and undefined points.
  • Use sec² = 1 + tan² and cosec² = 1 + cot² in equations and identities.
  • Use arcsin, arccos and arctan with their restricted domains and ranges.

No animated video for this topic yet; these notes stand alone.