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Reciprocal and inverse trigonometric functions questions

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.

8 original questions · 29 marks · the reciprocal and inverse trigonometric functions notes · Trigonometry

Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.

• 3 marks · define · warm-up

  Define sec x, cosec x and cot x, and write down the exact values of sec 60° and cot 45°.

  Worked answer

  sec x = 1/cos x, cosec x = 1/sin x and cot x = 1/tan x = cos x/sin x. Then sec 60° = 1/(½) = 2 and cot 45° = 1/1 = 1. B1 for the three definitions, B1 for sec 60°, B1 for cot 45°. Pair each function with its parent by the third letter, so sec goes with cosine and cosec goes with sine. The mismatch is deliberate and it catches out a fair share of candidates every year.

• 2 marks · warm-up

  Starting from sin² θ + cos² θ = 1, derive the identity sec² θ = 1 + tan² θ.

  Worked answer

  Divide every term by cos² θ, giving sin² θ/cos² θ + 1 = 1/cos² θ, which reads tan² θ + 1 = sec² θ. M1 for the division, A1 for the identity. Papers ask for exactly this one line, and dividing by sin² θ instead produces the cosec twin, 1 + cot² θ = cosec² θ.

• 3 marks

  Given that tan θ = 5/12 and θ is acute, find the exact values of sec θ and cos θ.

  Worked answer

  sec² θ = 1 + tan² θ = 1 + 25/144 = 169/144, so sec θ = 13/12, taking the positive root because θ is acute. Then cos θ is its reciprocal, 12/13. M1 for the identity, A1 for sec θ, A1 for cos θ. No triangle needs drawing and no angle needs finding, since the identity moves straight from one ratio to the other.

• 4 marks

  Given that cosec θ = 5/4 and θ is acute, find the exact values of cot θ and cos θ.

  Worked answer

  cot² θ = cosec² θ − 1 = 25/16 − 1 = 9/16, so cot θ = 3/4, positive because θ is acute. Since cosec θ = 5/4 gives sin θ = 4/5, and cos θ = cot θ × sin θ, we get cos θ = (3/4)(4/5) = 3/5. M1 for the identity, A1 for cot θ, M1 for linking cos to cot and sin, A1 for 3/5. The 3-4-5 triangle was underneath the whole question, and spotting it early is a legitimate check rather than a method.

• 3 marks

  Solve sec x = 2 for 0 ≤ x < 360°.

  Worked answer

  sec x = 2 means cos x = ½, so x = 60° or x = 300°. M1 for flipping to cosine, A1 A1 for the two roots. Every reciprocal equation is solved by turning it back into its parent function first, because the cosine graph is the one whose symmetry you can read. Stopping at 60° loses a mark, since cosine is positive in the fourth quadrant too.

• 4 marks · state

  State the range of values returned by arcsin and by arccos, and evaluate arcsin ½, arccos (−1) and arctan 1 exactly, in radians.

  Worked answer

  arcsin returns values in [−π/2, π/2] and arccos returns values in [0, π]. Then arcsin ½ = π/6, arccos (−1) = π and arctan 1 = π/4. B1 for both ranges, B1 B1 B1 for the three values. The restricted ranges are what make these inverses functions at all. Each has to choose a single angle from infinitely many candidates, and these are the agreed windows.

• 6 marks · stretch

  Solve 2 sec² x − 5 tan x = 1 for 0 ≤ x < 360°, giving your answers to 1 decimal place.

  Worked answer

  Replace sec² x by 1 + tan² x so that the equation involves one function only: 2(1 + tan² x) − 5 tan x = 1, which tidies to 2 tan² x − 5 tan x + 1 = 0. This does not factorise, so use the formula: tan x = (5 ± √17)/4, giving tan x = 0.2192 or tan x = 2.2808. From the first, x = 12.4° and 192.4°; from the second, x = 66.3° and 246.3°. M1 for the identity, A1 for the quadratic, M1 for solving it, A1 for both tan values, M1 for adding 180° to each principal value, A1 for all four roots. Tangent has period 180°, not 360°, so each value of tan x contributes exactly two solutions in the interval. Looking for the second root at 180° − x is the sine habit misapplied, and it gives four wrong answers.

• 4 marks · show · stretch

  Show that arctan (1/2) + arctan (1/3) = π/4.

  Worked answer

  Let A = arctan (1/2) and B = arctan (1/3), so tan A = 1/2 and tan B = 1/3. Then tan (A + B) = (tan A + tan B)/(1 − tan A tan B) = (1/2 + 1/3)/(1 − 1/6) = (5/6)/(5/6) = 1. Now A and B are both positive and both less than π/4, since their tangents are less than 1, so A + B lies strictly between 0 and π/2. The only angle in that interval with tangent 1 is π/4, so A + B = π/4. M1 for the compound tangent formula, M1 for substituting, A1 for the value 1, A1 for the range argument and the conclusion. The last mark is the one most often dropped. tan (A + B) = 1 alone only narrows the answer to π/4 plus any multiple of π, and the bounding argument is what pins it down.

The same practice on paper: the printable workbook for this topic, questions and a worked answer book.

Practise reciprocal and inverse trigonometric functions one question at a time

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