Questions on reciprocal and inverse trigonometric functions

Question 1

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

Accepted 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.

Question 2

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

Accepted 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. Dividing by sin² θ instead produces the cosec twin, 1 + cot² θ = cosec² θ.

Question 3

Write down the value of cot 90°, and explain why 1/tan 90° does not give it.

Accepted Answer

cot is defined as cos/sin, so cot 90° = cos 90°/sin 90° = 0/1 = 0. B1 for 0. The description 1/tan fails at this angle because tan 90° = sin 90°/cos 90° has a zero on the bottom and is undefined, so 1/tan 90° is the reciprocal of nothing. B1 for the reason. The two descriptions agree wherever tan is defined and not zero, and cos/sin is the one that also covers the odd multiples of 90°, which is exactly where the graph of cot crosses the axis.

Question 4

Sketch y = sec x for −π ≤ x ≤ π, giving the equations of the asymptotes and the coordinates of the turning points.

Accepted Answer

sec x = 1/cos x, so the curve is undefined wherever cos x = 0, and its vertical asymptotes in this interval are x = −π/2 and x = π/2. B1 for both asymptotes. Between them lies a branch shaped like a U, coming down from the left asymptote to a lowest point and rising to the right one; outside them lie two pieces below the axis, each shaped like an upside-down U with its highest point at the far end of the interval. B1 for the three branches, correctly placed above and below the axis. The turning points sit where cos x = ±1, since that is where 1/cos x is smallest in size: a minimum at (0, 1) and maxima at (−π, −1) and (π, −1). B1 for the three turning points. No part of the curve lies strictly between y = −1 and y = 1, because |cos x| ≤ 1 forces |sec x| ≥ 1.

Question 5

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

Accepted 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.

Question 6

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

Accepted 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.

Question 7

Solve sec x = 2 for 0 ≤ x < 360°.

Accepted 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. Cosine is positive in the fourth quadrant as well, so 300° is a second solution in range.

Question 8

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

Accepted 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.

Question 9

Solve 2 sec² x + 3 sec x − 2 = 0 for 0 ≤ x < 360°, explaining why one of the two values of sec x gives no solutions.

Accepted Answer

Only one ratio appears, so treat the equation as a quadratic in sec x. It factorises as (2 sec x − 1)(sec x + 2) = 0, giving sec x = ½ or sec x = −2. M1 for factorising as a quadratic in sec x, A1 for both values. Since |cos x| ≤ 1, the reciprocal satisfies |sec x| ≥ 1, so sec x = ½ is impossible and that bracket contributes nothing. B1 for rejecting sec x = ½ with the reason. From sec x = −2 comes cos x = −½, and cosine is negative through the second and third quadrants, so x = 120° and x = 240°. A1 for both roots.

Question 10

Prove that 1/(1 + cos θ) + 1/(1 − cos θ) ≡ 2 cosec² θ.

Accepted Answer

Work down the left-hand side. Over the common denominator (1 + cos θ)(1 − cos θ) the sum becomes [(1 − cos θ) + (1 + cos θ)]/[(1 + cos θ)(1 − cos θ)]. M1 for the common denominator. The numerator collapses to 2, since the cosines cancel, and the denominator is a difference of two squares, 1 − cos² θ. A1 for 2/(1 − cos² θ). By sin² θ + cos² θ = 1 that denominator is sin² θ. M1 for the Pythagorean identity. So the expression is 2/sin² θ = 2 cosec² θ, as required. A1 for the conclusion. Both fractions on the left are undefined when cos θ = ±1, and so is the right-hand side, so the identity holds wherever either side means anything.

Question 11

Explain why arcsin (sin 200°) is not 200°, and find its value.

Accepted Answer

arcsin returns values in [−90°, 90°] and nothing outside that, so 200° cannot be an output of arcsin whatever the input. B1 for the range of arcsin. Evaluate the inside first: 200° = 180° + 20°, and sine is negative through the third quadrant, so sin 200° = −sin 20°. M1 for sin 200° = −sin 20°. The angle in [−90°, 90°] whose sine is −sin 20° is −20°, so arcsin (sin 200°) = −20°. A1 for −20°.

Question 12

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

Accepted 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 two wrong answers, 167.6° and 113.7°.

Question 13

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

Accepted 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. 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.

Question 14

The function f is defined by f(x) = arccos x, with domain −1 ≤ x ≤ 1. Sketch y = f(x), stating the coordinates of its endpoints and its range. Solve arccos (2x − 1) = π/3, and explain why the equation has no other solution.

Accepted Answer

arccos is the inverse of the cosine restricted to [0, π], so its graph is that piece of the cosine curve reflected in the line y = x. It falls steadily from (−1, π) on the left to (1, 0) on the right, steeply near both ends and least steeply through (0, π/2). B1 for a decreasing curve with those two endpoints. Its range is 0 ≤ f(x) ≤ π. B1 for the range. To solve, take the cosine of both sides: 2x − 1 = cos (π/3) = ½. M1 for taking the cosine of both sides, A1 for 2x − 1 = ½. Then 2x = 3/2, so x = ¾, which is admissible because 2 × ¾ − 1 = ½ lies in [−1, 1], the domain of arccos. A1 for x = ¾. There is no second solution because arccos decreases across its whole domain and so takes each value of its range exactly once: one input gives π/3, and no other can. B1 for the uniqueness argument. The right-hand side π/3 had to lie in [0, π] for a solution to exist at all; arccos (2x − 1) = 4 would have none.

Question 15

The function g is defined by g(x) = 4 − 2 cosec x, for 0 < x < 2π with x ≠ π. Find the set of values that g cannot take, and solve g(x) = 8, giving your answers in terms of π.

Accepted Answer

Since |sin x| ≤ 1, its reciprocal satisfies |cosec x| ≥ 1, so either cosec x ≥ 1 or cosec x ≤ −1 and nothing in between. B1 for the range of cosec. Follow each branch through. Where cosec x ≥ 1, multiplying by −2 turns the inequality round to give −2 cosec x ≤ −2, and adding 4 gives g(x) ≤ 2. Where cosec x ≤ −1, the same steps give −2 cosec x ≥ 2 and g(x) ≥ 6. M1 for tracking both branches through the multiplication by −2 and the addition of 4. So the values g cannot take are 2 < g(x) < 6. A1 for the excluded set. Now 4 − 2 cosec x = 8 gives cosec x = −2, so sin x = −½. M1 for sin x = −½. Sine is negative through the third and fourth quadrants, so in 0 < x < 2π the solutions are x = π + π/6 = 7π/6 and x = 2π − π/6 = 11π/6. A1 for both. The value 8 is above 6, so it is one that g does reach; a value such as 4 sits inside the excluded band and would give no solution at all.