Questions on trigonometric graphs and equations

Question 1

Write down the exact values of sin 30°, cos 45° and tan 60°.

Accepted Answer

sin 30° = ½, cos 45° = √2/2 and tan 60° = √3. B1 B1 B1 for the three exact values. All three come from two triangles, the halved equilateral and the halved square. On a non-calculator paper a decimal such as 0.866 is an approximation where the exact value √3/2 is asked for.

Question 2

Solve cos x = ½ for 0 ≤ x < 360°, giving exact answers.

Accepted Answer

The principal value is 60°, and cosine's symmetry supplies the partner 360° − 60° = 300°. So x = 60° or x = 300°. B1 for 60°, B1 for 300°. The cosine graph is symmetrical about x = 180° across this interval, so the two roots sit the same distance either side of that line.

Question 3

Write down the period of y = sin x, the period of y = tan x and the period of y = cos 3x, each in degrees.

Accepted Answer

Sine is defined by a point going round a circle, so it repeats once per full turn: its period is 360°. B1 for 360°. Tangent is sin x/cos x, and a half-turn reverses the sign of the top and of the bottom alike, so their quotient is back where it began: the period of tangent is 180°. B1 for 180°. In cos 3x the angle inside runs three times as fast as x, so the cosine completes a cycle when 3x has covered 360°, that is when x has covered 120°. B1 for 120°. The coefficient inside the function divides into the parent period rather than multiplying it, so a bigger coefficient means a shorter period.

Question 4

Given that sin 40° = 0.643 to 3 decimal places, write down the values of sin 140° and sin 220° to the same accuracy.

Accepted Answer

The sine curve is symmetrical about x = 90°, and 140° stands as far beyond 90° as 40° falls short of it, so sin 140° = sin 40° = 0.643. B1 for 0.643. Sine is negative through the third quadrant, and 220° = 180° + 40°, which is the reflection of 40° in the origin, so sin 220° = −sin 40° = −0.643. B1 for −0.643.

Question 5

Solve sin x = −0.5 for 0 ≤ x < 360°.

Accepted Answer

The calculator returns −30°, which is outside the interval. Sine is negative in the third and fourth quadrants, so take the acute angle 30° and place it there: x = 180° + 30° = 210° and x = 360° − 30° = 330°. M1 for the acute angle 30°, A1 for 210°, A1 for 330°. Working from the size of the angle and then choosing quadrants is safer than trying to add 360° to the calculator value, which here gives 330° and hides the other root entirely.

Question 6

Solve 2 sin² x + cos x − 1 = 0 for 0 ≤ x < 360°.

Accepted Answer

Replace sin² x with 1 − cos² x, giving 2 − 2 cos² x + cos x − 1 = 0. That tidies to 2 cos² x − cos x − 1 = 0 and factorises as (2 cos x + 1)(cos x − 1) = 0. From cos x = 1 comes x = 0°, and from cos x = −½ come x = 120° and x = 240°, so the solutions are 0°, 120° and 240°. M1 for using sin² x = 1 − cos² x, A1 for 2 cos² x − cos x − 1 = 0, M1 for factorising, A1 for x = 0°, A1 for 120° and 240°. The identity has to go in first; there is no way to factorise an equation carrying both sin and cos. The interval includes x = 0°, so that solution is in range.

Question 7

Solve sin x cos x = 2 sin x for 0 ≤ x < 360°.

Accepted Answer

Gather everything on one side and factorise: sin x (cos x − 2) = 0. The bracket can never be zero, since cosine never reaches 2, so the only route left is sin x = 0, giving x = 0° or x = 180°. M1 for factorising, A1 for sin x (cos x − 2) = 0, A1 for 0°, A1 for 180°. Dividing both sides by sin x at the start removes the solutions of sin x = 0, which here are the only solutions the equation has.

Question 8

Solve 3 sin x = 2 cos x for 0 ≤ x < 360°, to 1 decimal place.

Accepted Answer

Dividing by cos x is safe here, because cos x = 0 would force sin x = ±1 and leave the left side at ±3 against a right side of 0. So tan x = 2/3, giving x = 33.7° and, one half-turn along, 213.7°. M1 for tan x = 2/3, A1 for 33.7°, A1 for 213.7°. Tangent repeats every 180°, so a tan equation delivers two roots in a 360° window and the second is found by adding 180°, not by subtracting from 360°.

Question 9

Solve tan 3x = 1 for 0 ≤ x < 180°.

Accepted Answer

Let θ = 3x. As x runs over 0 ≤ x < 180°, the angle θ runs over 0 ≤ θ < 540°. M1 for widening the interval to 540°. Tangent equals 1 at θ = 45°, and it repeats every 180°, so inside the widened interval θ = 45°, 225° and 405°; the next one, 585°, is outside. A1 for the three values of θ. Dividing each by 3 gives x = 15°, 75° and 135°. A1 for 15°, A1 for 75° and 135°. Tripling the angle triples the count of solutions, because 3x covers three periods of the tangent while x covers one half-turn. Solving tan x = 1 first and dividing after would give only 15° and 75°.

Question 10

Show that the equation 2 sin² x + 3 cos x = 5 has no solutions.

Accepted Answer

Replace sin² x by 1 − cos² x so that a single ratio is left: 2 − 2 cos² x + 3 cos x = 5, which tidies to 2 cos² x − 3 cos x + 3 = 0. M1 for the identity, A1 for the quadratic. Treat it as a quadratic in cos x and take its discriminant: (−3)² − 4 × 2 × 3 = 9 − 24 = −15. M1 for the discriminant. That is negative, so there is no real number cos x satisfying the quadratic, let alone one between −1 and 1, and the equation has no solutions. A1 for the negative discriminant with the conclusion. A second route: the left-hand side is 2 − 2 cos² x + 3 cos x, whose greatest value is 3.125, at cos x = ¾, so 5 is out of reach.

Question 11

On one set of axes, sketch y = sin x and y = cos x for 0 ≤ x ≤ 360°. State how many solutions the equation sin x = cos x has in that interval, and find them exactly.

Accepted Answer

Both curves run between −1 and 1 across the interval, cosine starting at 1 and falling, sine starting at 0 and rising, with cosine a quarter of a turn ahead throughout. B1 for the two curves, correctly placed relative to one another. They meet twice: once in the first quadrant, where both are positive, and once in the third, where both are negative. B1 for two solutions. To find them, note that cos x = 0 would force sin x = ±1 and cannot give a crossing, so dividing by cos x is safe and gives tan x = 1. The first solution is x = 45°, and tangent repeats every 180°, so the second is x = 45° + 180° = 225°. A1 for 45°, A1 for 225°.

Question 12

The graph shows the curve y = f(x) for 0 ≤ x ≤ 360°, where f(x) = a sin bx and a and b are positive constants. Use the graph to write down the amplitude and the period of the curve, and hence state the values of a and b. Use the graph to find all the solutions of f(x) = 1.5 in the interval 0 ≤ x ≤ 360°.

Accepted Answer

The curve reaches 3 at its highest and −3 at its lowest, so the amplitude is 3 and a = 3. B1 for the amplitude. One complete cycle is finished at x = 180°, where the curve has returned to the axis going upwards for the first time, so the period is 180°. B1 for the period. A curve y = a sin bx has period 360°/b, and 360/b = 180 gives b = 2, so f(x) = 3 sin 2x. B1 for a = 3 and b = 2 together in the equation. Reading across at the height y = 1.5, the curve is at that height at four places in the interval, and each falls on a gridline: x = 15° and x = 75° on the first arch, then x = 195° and x = 255° on the second. M1 for two correct values read off, A1 for all four with no extras. The solutions come in pairs about each maximum, 45° and 225°, spaced 30° either side.

Question 13

Solve cos (x − 40°) = ½ for 0 ≤ x < 360°.

Accepted Answer

Let θ = x − 40°, so θ runs over −40° ≤ θ < 320° while x covers its interval. cos θ = ½ at θ = 60° and 300°, and also at −60°, which the widened interval excludes (−60° < −40°). Translating back: x = 100° or 340°. M1 for the substitution θ = x − 40°, B1 for the widened interval, A1 for θ = 60° and 300°, A1 for 100°, A1 for 340°.

Question 14

Sketch y = tan x for −180° ≤ x ≤ 180°, giving the equations of the asymptotes in that interval. Show that, for every real number k, the equation tan x = k has exactly two solutions in 0 ≤ x < 360°, and state the relation between them.

Accepted Answer

The graph comes in three pieces, each rising from left to right and crossing the axis at −180°, 0° and 180°, with vertical asymptotes at x = −90° and x = 90°, where cos x is zero and the quotient sin x/cos x is undefined. B1 for the shape with the three crossings, B1 for the two asymptotes. Now take the interval 0 ≤ x < 360°. It carries three pieces of the curve: on 0° ≤ x < 90° the graph rises from 0 through every positive value; on 90° < x < 270° it rises through every real value once; and on 270° < x < 360° it rises through every negative value, reaching 0 only at 360°, which the interval excludes. M1 for describing the three branches and the values each one covers. A positive k is met once on the first branch and once on the middle one; a negative k once on the middle branch and once on the last; and k = 0 at x = 0° and 180°. So in every case there are exactly two solutions. A1 for the count with that argument. The two always differ by 180°, which is the period of the tangent. B1 for the relation.

Question 15

Solve 2 cos (2x − 30°) = √3 for 0 ≤ x < 360°, giving exact answers.

Accepted Answer

Divide by 2 first: cos (2x − 30°) = √3/2. Let θ = 2x − 30°. At x = 0 the angle θ is −30°, and as x rises towards 360° the angle θ rises towards 690° without reaching it, so θ runs over −30° ≤ θ < 690°. M1 for the substitution, B1 for the widened interval. Cosine equals √3/2 at 30°; it is symmetrical about 0°, which supplies −30°, and it repeats every 360°, which supplies 330° and 390° and then 690°. Of these, 690° is excluded by the open end, leaving θ = −30°, 30°, 330° and 390°. M1 for generating the values of θ, A1 for exactly those four. Then x = (θ + 30°)/2, so x = 0°, 30°, 180° and 210°. A1 for 0° and 30°, A1 for 180° and 210°.

Question 16

Prove that sin⁴ x − cos⁴ x ≡ 1 − 2 cos² x. Hence solve sin⁴ x − cos⁴ x = −½ for 0 ≤ x < 360°, giving exact answers.

Accepted Answer

The left-hand side is a difference of two squares: sin⁴ x − cos⁴ x = (sin² x − cos² x)(sin² x + cos² x). M1 for the factorisation. The second bracket is 1, and replacing sin² x in the first by 1 − cos² x leaves (1 − cos² x) − cos² x = 1 − 2 cos² x, as required. A1 for the identity, worked down one side only. For the equation, 1 − 2 cos² x = −½ gives cos² x = ¾, so cos x = √3/2 or cos x = −√3/2, both signs of the square root being available. M1 for both signs. From cos x = √3/2 come x = 30° and 330°, and from cos x = −√3/2 come x = 150° and 210°. A1 for 30° and 330°, A1 for 150° and 210°. The fourth powers hide the two signs, and taking only the positive square root halves the solution set.