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Trigonometric functions and equations

Exact values, graphs and periodicity of sin, cos, tan; tan = sin/cos and sin^2 + cos^2 = 1; solving equations in an interval.

Spec MM4.3-MM4.6Section 1 Part 1: AS-level pure mathematics19 questions33 real past questions
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1Trigonometric functions and equations · spec MM4.6 · Paper 1 style · ★☆☆

How many solutions does \(\sin 2x = \dfrac{1}{2}\) have for \(0 \leq x < 2\pi\)?

  1. A\(2\)
  2. B\(6\)
  3. C\(4\)
  4. D\(8\)
  5. E\(1\)
  6. F\(3\)
Worked solution

\(2x\) runs over \([0, 4\pi)\), which contains two full periods of \(\sin\), each giving two solutions: \(4\) in total.

Full method — how and why
Counting solutions of \(\sin(kx) = c\). Transform the interval for the inside expression and count there.
  1. Let \(u = 2x\). As \(x\) runs over \([0, 2\pi)\), \(u\) runs over \([0, 4\pi)\).
    Multiplying the variable by \(2\) doubles the range of the inside expression - this is the whole trick.
  2. \([0, 4\pi)\) contains exactly two full periods of \(\sin\).
    \(\sin\) has period \(2\pi\), so a \(4\pi\)-long interval holds two complete waves.
  3. In each period, \(\sin u = \frac12\) has two solutions (one on the way up, one on the way down).
    A horizontal line strictly between \(-1\) and \(1\) cuts each wave twice.
  4. Total: \(2 \times 2 = 4\) solutions.
    Each \(u\) gives exactly one \(x\) via \(x = u/2\), so no solutions are lost or duplicated.
Where it goes wrong
  • Answering \(2\) by counting solutions for \(x\) as if the \(2\) were not there.
  • Answering \(8\) by doubling twice.
Takeaway For \(\sin(kx) = c\) on an interval of length \(L\), the inside runs over length \(kL\) - count solutions there.
Answer C · Trigonometric functions and equations (MM4.6) · question TRG-01
2Trigonometric functions and equations · spec MM4.3 · Paper 1 style · ★☆☆

Find the exact value of \(\sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ\).

  1. A\(\dfrac{\sqrt3}{2}\)
  2. B\(\dfrac{3}{4}\)
  3. C\(\dfrac{\sqrt3}{4}\)
  4. D\(1\)
  5. E\(0\)
  6. F\(\dfrac12\)
Worked solution

\(\frac{\sqrt3}{2}\cdot\frac{\sqrt3}{2} + \frac12\cdot\frac12 = \frac34 + \frac14 = 1\).

Full method — how and why
Pure recall of exact values. (It is also \(\sin(60^\circ + 30^\circ) = \sin 90^\circ\), but the addition formula is off-specification.)
  1. Write down the exact values: \(\sin 60^\circ = \frac{\sqrt3}{2}\), \(\cos 30^\circ = \frac{\sqrt3}{2}\), \(\cos 60^\circ = \frac12\), \(\sin 30^\circ = \frac12\).
    These come from the half-equilateral triangle with sides \(1, \sqrt3, 2\) - worth being able to redraw in seconds.
  2. Substitute: \(\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt3}{2} + \dfrac12\cdot\dfrac12 = \dfrac34 + \dfrac14\).
    \((\sqrt3)^2 = 3\), so the first product is \(\frac34\).
  3. Add: \(1\).
    A clean answer - the sign that the exact values were right.
Where it goes wrong
  • Using \(\sin 60^\circ = \frac12\) (that is \(\sin 30^\circ\)), giving \(\frac{\sqrt3}{2}\).
  • Writing \(\left(\frac{\sqrt3}{2}\right)^2 = \frac{\sqrt3}{4}\).
Takeaway \(\sin\) and \(\cos\) of \(30^\circ\) and \(60^\circ\) swap over; square a surd fraction by squaring top and bottom.
Answer D · Trigonometric functions and equations (MM4.3) · question TRG-03
3Trigonometric functions and equations · spec MM4.6 · Paper 1 style · ★☆☆

Find the sum of all solutions of \(\tan x = \sqrt{3}\) in the interval \(0 \leq x \leq 2\pi\).

  1. A\(\dfrac{2\pi}{3}\)
  2. B\(\dfrac{4\pi}{3}\)
  3. C\(\pi\)
  4. D\(\dfrac{5\pi}{3}\)
  5. E\(2\pi\)
  6. F\(\dfrac{7\pi}{3}\)
Worked solution

\(x = \frac\pi3\) and \(x = \frac\pi3 + \pi = \frac{4\pi}{3}\). Sum \(\frac{5\pi}{3}\).

Full method — how and why
\(\tan\) has period \(\pi\), so solutions come one per period. Find the first, add \(\pi\), then check the interval.
  1. The principal solution: \(\tan x = \sqrt3\) at \(x = \dfrac{\pi}{3}\).
    From the exact values: \(\tan 60^\circ = \sqrt3\).
  2. \(\tan\) repeats every \(\pi\), so the next solution is \(\dfrac{\pi}{3} + \pi = \dfrac{4\pi}{3}\).
    Unlike \(\sin\) and \(\cos\), \(\tan\) has period \(\pi\), not \(2\pi\).
  3. Check the interval \([0, 2\pi]\): \(\frac{\pi}{3}\) and \(\frac{4\pi}{3}\) are in it; the next, \(\frac{7\pi}{3}\), is not.
    \(\frac{7\pi}{3} > 2\pi\), so stop there.
  4. Sum: \(\dfrac{\pi}{3} + \dfrac{4\pi}{3} = \dfrac{5\pi}{3}\).
    Common denominator \(3\).
Where it goes wrong
  • Using period \(2\pi\) and finding only \(\frac\pi3\).
  • Including \(\frac{7\pi}{3}\) and answering \(\frac{7\pi}{3}\) or more - which is offered.
Takeaway \(\tan\) repeats every \(\pi\): solutions are \(x_0 + k\pi\), so there are usually two in a \(2\pi\) interval.
Answer D · Trigonometric functions and equations (MM4.6) · question TRG-04
4Trigonometric functions and equations · spec MM4.5 · Paper 1 style · ★☆☆

Given that \(\sin\theta = \dfrac{3}{5}\) and \(\theta\) is obtuse, find \(\tan\theta\).

  1. A\(-\dfrac{3}{4}\)
  2. B\(\dfrac{4}{3}\)
  3. C\(-\dfrac{4}{3}\)
  4. D\(-\dfrac{3}{5}\)
  5. E\(\dfrac{3}{4}\)
  6. F\(-\dfrac{4}{5}\)
Worked solution

\(\cos\theta = -\frac45\) (negative in the second quadrant), so \(\tan\theta = \frac{3/5}{-4/5} = -\frac34\).

Full method — how and why
From one ratio to another, with a quadrant condition fixing the signs. Use a right-angled triangle plus CAST.
  1. From \(\sin\theta = \frac35\), the reference triangle has opposite \(3\) and hypotenuse \(5\), so the third side is \(4\) (a 3-4-5 triangle).
    Pythagoras gives the missing side; the \(3\)-\(4\)-\(5\) triple appears constantly.
  2. \(\theta\) is obtuse, so it lies in the second quadrant where \(\sin\) is positive but \(\cos\) and \(\tan\) are negative.
    CAST: only sine is positive in the second quadrant.
  3. So \(\cos\theta = -\frac45\) and \(\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{-4/5} = -\dfrac34\).
    The fives cancel; the sign comes from the quadrant.
Where it goes wrong
  • Ignoring 'obtuse' and answering \(\frac34\).
  • Inverting the ratio and answering \(-\frac43\).
Takeaway Build the reference triangle from the given ratio, then let the quadrant decide the signs.
Answer A · Trigonometric functions and equations (MM4.5) · question TRG-06
5Trigonometric functions and equations · spec MM4.6 · Paper 1 style · ★☆☆

Solve \(\sin^2 x = \dfrac{3}{4}\) for \(0 \leq x < 2\pi\). How many solutions are there?

  1. A\(3\)
  2. B\(6\)
  3. C\(2\)
  4. D\(8\)
  5. E\(4\)
  6. F\(1\)
Worked solution

\(\sin x = \pm\frac{\sqrt3}{2}\): \(x = \frac\pi3, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\). Four solutions.

Full method — how and why
A squared trig equation. Taking the square root produces two values, each with its own solutions.
  1. Take square roots: \(\sin x = \pm\dfrac{\sqrt3}{2}\).
    \(X^2 = k\) has two roots - forgetting the negative one halves the answer.
  2. \(\sin x = \frac{\sqrt3}{2}\) gives \(x = \frac{\pi}{3}\) and \(\frac{2\pi}{3}\).
    Second-quadrant partner: \(\pi - \frac\pi3\).
  3. \(\sin x = -\frac{\sqrt3}{2}\) gives \(x = \frac{4\pi}{3}\) and \(\frac{5\pi}{3}\).
    Third and fourth quadrants, by symmetry below the axis.
  4. Total: \(4\) solutions in \([0, 2\pi)\).
    Count them; symmetry says there should be an even number here.
Where it goes wrong
  • Taking only the positive root and answering \(2\).
  • Reading \(\sin^2 x\) as \(\sin(x^2)\).
Takeaway \(\sin^2 x = c\) gives \(\sin x = \pm\sqrt c\), and each sign contributes its own solutions.
Answer E · Trigonometric functions and equations (MM4.6) · question TRG-07
6Trigonometric functions and equations · spec MM4.6 · Paper 1 style · ★☆☆

Find the smallest positive solution of \(\cos\left(x - \dfrac{\pi}{6}\right) = -\dfrac{1}{2}\).

  1. A\(\pi\)
  2. B\(\dfrac{5\pi}{6}\)
  3. C\(\dfrac{7\pi}{6}\)
  4. D\(\dfrac{\pi}{2}\)
  5. E\(\dfrac{3\pi}{2}\)
  6. F\(\dfrac{2\pi}{3}\)
Worked solution

\(x - \frac\pi6 = \frac{2\pi}{3}\) gives \(x = \frac{5\pi}{6}\) (the other solution \(x - \frac\pi6 = \frac{4\pi}{3}\) gives \(\frac{3\pi}{2}\), larger).

Full method — how and why
An equation with a shifted argument. Solve for the bracket first, then unshift - and be careful about which solution is smallest.
  1. Let \(u = x - \frac\pi6\) and solve \(\cos u = -\frac12\).
    Substituting keeps the shift out of the way until the end.
  2. \(\cos u = -\frac12\) at \(u = \frac{2\pi}{3}\) and \(u = \frac{4\pi}{3}\) (second and third quadrants).
    \(\cos\) is negative in quadrants 2 and 3; the reference angle is \(\frac\pi3\).
  3. Convert back: \(x = u + \frac\pi6\), giving \(x = \frac{2\pi}{3} + \frac{\pi}{6} = \frac{5\pi}{6}\) and \(x = \frac{4\pi}{3} + \frac{\pi}{6} = \frac{3\pi}{2}\).
    Common denominator \(6\): \(\frac{4\pi}{6} + \frac{\pi}{6} = \frac{5\pi}{6}\).
  4. The smallest positive one is \(\dfrac{5\pi}{6}\).
    Compare the candidates; \(\frac{5\pi}{6} < \frac{3\pi}{2}\).
Where it goes wrong
  • Subtracting the shift instead of adding when converting back.
  • Using the reference angle \(\frac\pi3\) directly, which gives \(\frac\pi2\) - an offered option.
Takeaway Solve for the whole bracket, then undo the shift; adjust the interval for the bracket if the question restricts \(x\).
Answer B · Trigonometric functions and equations (MM4.6) · question TRG-08
7Trigonometric functions and equations · spec MM4.5 · Paper 1 style · ★☆☆

Simplify \(\dfrac{1 - \cos^2\theta}{\cos\theta \tan\theta}\) for values of \(\theta\) where the expression is defined.

  1. A\(1\)
  2. B\(\sin\theta\)
  3. C\(\tan\theta\)
  4. D\(\sin^2\theta\)
  5. E\(\cos\theta\)
  6. F\(\dfrac{\sin\theta}{\cos\theta}\)
Worked solution

\(\cos\theta\tan\theta = \sin\theta\), and \(1 - \cos^2\theta = \sin^2\theta\). So the expression is \(\sin\theta\).

Full method — how and why
Simplification using the two permitted identities. Work on the denominator and numerator separately.
  1. Denominator: \(\cos\theta\tan\theta = \cos\theta \times \dfrac{\sin\theta}{\cos\theta} = \sin\theta\).
    \(\tan\theta = \frac{\sin\theta}{\cos\theta}\), and the \(\cos\theta\) cancels (legitimate where the expression is defined).
  2. Numerator: \(1 - \cos^2\theta = \sin^2\theta\).
    Rearranged Pythagorean identity.
  3. Divide: \(\dfrac{\sin^2\theta}{\sin\theta} = \sin\theta\).
    One power of \(\sin\theta\) cancels.
Where it goes wrong
  • Cancelling to \(\sin^2\theta\) by forgetting to simplify the denominator.
  • Writing \(1 - \cos^2\theta = \cos^2\theta - 1\) (sign flip), giving \(-\sin\theta\).
Takeaway Convert \(\tan\) into \(\frac{\sin}{\cos}\) and use \(1 - \cos^2 = \sin^2\); then cancel.
Answer B · Trigonometric functions and equations (MM4.5) · question TRG-09
8Trigonometric functions and equations · spec MM4.4 · Paper 1 style · ★☆☆

The function \(f(x) = 3\sin(2x) - 1\) is defined for all real \(x\). Find the difference between its maximum and minimum values.

  1. A\(1\)
  2. B\(2\)
  3. C\(4\)
  4. D\(3\)
  5. E\(6\)
  6. F\(7\)
Worked solution

Range is \([-4, 2]\); the difference is \(6\).

Full method — how and why
Range of \(a\sin(bx) + c\). Only \(a\) and \(c\) matter - the \(b\) affects the period, not the values taken.
  1. \(\sin(2x)\) takes every value in \([-1, 1]\).
    Compressing horizontally changes how fast the wave oscillates, not how high it goes.
  2. Multiply by \(3\): the range becomes \([-3, 3]\).
    Amplitude \(|a| = 3\).
  3. Subtract \(1\): the range becomes \([-4, 2]\).
    The whole wave shifts down by \(1\).
  4. Difference between maximum and minimum: \(2 - (-4) = 6\).
    Equivalently \(2 \times\) amplitude \(= 2 \times 3 = 6\) - the vertical shift cancels out of a difference.
Where it goes wrong
  • Answering \(3\) (the amplitude) instead of the full range \(2a\).
  • Letting the \(-1\) affect the difference, giving \(7\) or \(5\).
Takeaway Max minus min for \(a\sin(\cdot) + c\) is always \(2|a|\); vertical shifts cancel.
Answer E · Trigonometric functions and equations (MM4.4) · question TRG-11
9Trigonometric functions and equations · spec MM4.6 · Paper 1 style · ★☆☆

Find the number of solutions of \(\tan^2 x = 3\) for \(0 \leq x < 2\pi\).

  1. A\(1\)
  2. B\(4\)
  3. C\(6\)
  4. D\(3\)
  5. E\(8\)
  6. F\(2\)
Worked solution

\(\tan x = \pm\sqrt3\): \(x = \frac\pi3, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\). Four solutions.

Full method — how and why
A squared tangent equation. Two values of \(\tan x\), each giving solutions once per \(\pi\).
  1. Take square roots: \(\tan x = \pm\sqrt3\).
    Both signs are needed.
  2. \(\tan x = \sqrt3\) gives \(x = \frac{\pi}{3}\) and \(\frac{\pi}{3} + \pi = \frac{4\pi}{3}\).
    Period \(\pi\) for \(\tan\).
  3. \(\tan x = -\sqrt3\) gives \(x = \frac{2\pi}{3}\) and \(\frac{2\pi}{3} + \pi = \frac{5\pi}{3}\).
    \(\tan\) is negative in quadrants 2 and 4; the reference angle is still \(\frac\pi3\).
  4. Four solutions in \([0, 2\pi)\).
    Two values \(\times\) two periods.
Where it goes wrong
  • Taking only the positive root and answering \(2\).
  • Using period \(2\pi\) for \(\tan\).
Takeaway \(\tan^2 x = k\) gives \(\tan x = \pm\sqrt k\), and each has one solution per interval of length \(\pi\).
Answer B · Trigonometric functions and equations (MM4.6) · question TRG-12
10Trigonometric functions and equations · spec MM4.5 · Paper 1 style · ★☆☆

For all \(\theta\) where the expressions are defined, \((\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2\) equals

  1. A\(2\sin^2\theta\)
  2. B\(2 + 2\sin\theta\cos\theta\)
  3. C\(0\)
  4. D\(2\)
  5. E\(2\cos^2\theta\)
  6. F\(1\)
Worked solution

Expanding: \((1 + 2\sin\theta\cos\theta) + (1 - 2\sin\theta\cos\theta) = 2\).

Full method — how and why
Expanding two brackets whose cross terms cancel. The Pythagorean identity finishes it.
  1. Expand the first: \((\sin\theta + \cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta\).
    Standard \((a+b)^2\) expansion.
  2. Expand the second: \((\sin\theta - \cos\theta)^2 = \sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta\).
    Only the middle sign changes.
  3. Add: the \(\pm2\sin\theta\cos\theta\) terms cancel, leaving \(2(\sin^2\theta + \cos^2\theta)\).
    This cancellation is the point of the question.
  4. Apply the identity: \(2 \times 1 = 2\).
    \(\sin^2 + \cos^2 = 1\) for every \(\theta\), so the answer is a constant.
Where it goes wrong
  • Keeping a cross term and choosing \(2 + 2\sin\theta\cos\theta\) - an offered option.
  • Answering \(1\) by forgetting the factor \(2\).
Takeaway \((a+b)^2 + (a-b)^2 = 2(a^2+b^2)\); with \(\sin\) and \(\cos\) that is always \(2\).
Answer D · Trigonometric functions and equations (MM4.5) · question TRG-13
11Trigonometric functions and equations · spec MM4.6 · Paper 1 style · ★☆☆

Given that \(\sin x + \cos x = 0\) and \(0 \leq x < 2\pi\), find the sum of all possible values of \(x\).

  1. A\(\pi\)
  2. B\(\dfrac{5\pi}{2}\)
  3. C\(\dfrac{7\pi}{4}\)
  4. D\(\dfrac{3\pi}{4}\)
  5. E\(\dfrac{3\pi}{2}\)
  6. F\(2\pi\)
Worked solution

\(\tan x = -1\): \(x = \frac{3\pi}{4}\) or \(\frac{7\pi}{4}\). Sum \(= \frac{10\pi}{4} = \frac{5\pi}{2}\).

Full method — how and why
Convert a sum of \(\sin\) and \(\cos\) into a single \(\tan\) equation by dividing.
  1. Rearrange: \(\sin x = -\cos x\).
    Move one term across.
  2. Divide by \(\cos x\): \(\tan x = -1\). (Check \(\cos x = 0\) separately: then \(\sin x = 0\) too, impossible.)
    Dividing by \(\cos x\) is safe only after ruling out \(\cos x = 0\) - and here that case gives a contradiction.
  3. Solve on \([0, 2\pi)\): \(\tan x = -1\) at \(x = \frac{3\pi}{4}\) and \(x = \frac{3\pi}{4} + \pi = \frac{7\pi}{4}\).
    Reference angle \(\frac\pi4\), negative tangent in quadrants 2 and 4.
  4. Sum: \(\frac{3\pi}{4} + \frac{7\pi}{4} = \frac{10\pi}{4} = \frac{5\pi}{2}\).
    Add over the common denominator and simplify.
Where it goes wrong
  • Finding only one solution and answering \(\frac{3\pi}{4}\) - an offered option.
  • Forgetting to justify dividing by \(\cos x\) (here it is fine, but the habit matters).
Takeaway \(a\sin x + b\cos x = 0\) becomes \(\tan x = -\frac ba\); check the \(\cos x = 0\) case before dividing.
Answer B · Trigonometric functions and equations (MM4.6) · question TRG-16
12Trigonometric functions and equations · spec MM4.3 · Paper 1 style · ★☆☆

Find the value of \(\tan 30^\circ \tan 60^\circ + \sin^2 45^\circ\).

  1. A\(\dfrac{\sqrt2}{2}\)
  2. B\(1\)
  3. C\(\dfrac{3}{2}\)
  4. D\(2\)
  5. E\(1 + \sqrt3\)
  6. F\(\dfrac{1}{2}\)
Worked solution

\(\tan30^\circ\tan60^\circ = \frac{1}{\sqrt3}\cdot\sqrt3 = 1\), \(\sin^2 45^\circ = \frac12\). Total \(\frac32\).

Full method — how and why
Exact-value recall plus one squared value. The first product is designed to collapse to \(1\).
  1. \(\tan 30^\circ = \dfrac{1}{\sqrt3}\) and \(\tan 60^\circ = \sqrt3\), so their product is \(1\).
    \(30^\circ\) and \(60^\circ\) are complementary, and \(\tan\theta\tan(90^\circ - \theta) = 1\) always.
  2. \(\sin 45^\circ = \dfrac{\sqrt2}{2}\), so \(\sin^2 45^\circ = \dfrac{2}{4} = \dfrac12\).
    Square top and bottom: \((\sqrt2)^2 = 2\) and \(2^2 = 4\).
  3. Add: \(1 + \frac12 = \frac32\).
    Straightforward once both pieces are exact.
Where it goes wrong
  • Writing \(\sin^2 45^\circ = \frac{\sqrt2}{2}\) (forgetting to square), giving \(1 + \frac{\sqrt2}{2}\).
  • Using \(\tan 30^\circ = \sqrt3\), which makes the product \(3\).
Takeaway \(\tan\theta \tan(90^\circ - \theta) = 1\), and \(\sin^2 45^\circ = \cos^2 45^\circ = \frac12\).
Answer C · Trigonometric functions and equations (MM4.3) · question TRG-18
13Trigonometric functions and equations · spec MM4.6 · Paper 2 style · ★☆☆

Consider the equation \(\sin x = k\) where \(k\) is a real constant. Which of the following statements is/are true?

  1. IFor every \(k\) with \(-1 < k < 1\) the equation has exactly two solutions in \(0 \leq x < 2\pi\).
  2. IIFor \(k = 1\) the equation has exactly one solution in \(0 \leq x < 2\pi\).
  3. IIIFor every real \(k\) the equation has at least one solution.
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I and II are read from the graph of \(\sin\) over one period. III is false: \(\sin x = 2\) has no solutions.

Full method — how and why
Read the claims off the graph of \(y = \sin x\) over one period, then remember the range.
  1. I: for \(-1True.
    Sine rises from \(0\) to \(1\), falls to \(-1\) and returns - a strictly-between value is hit once on the way down and once elsewhere in \([0, 2\pi)\).
  2. II: \(k=1\) is reached only at the single peak \(x = \frac{\pi}{2}\). True.
    The maximum is attained once per period.
  3. III: for \(|k| > 1\) there is no solution at all - \(\sin x = 2\) is impossible. False.
    The range of sine is \([-1,1]\).
  4. So I and II.
    Match to the list.
Where it goes wrong
  • Accepting III by forgetting the range restriction.
  • Rejecting II by counting \(x = \frac{\pi}{2}\) and \(x = \frac{5\pi}{2}\) - the second is outside \(0 \leq x < 2\pi\).
  • Rejecting I because \(k=0\) feels special: \(\sin x = 0\) gives \(x = 0\) and \(x=\pi\) in the interval - still exactly two.
Takeaway Per period, \(\sin x = k\) has two solutions for \(|k|<1\), one for \(|k|=1\) and none for \(|k|>1\). Sketch the line across the curve.
Answer E · Trigonometric functions and equations (MM4.6) · question X2-07
14Trigonometric functions and equations · spec MM4.5 · Paper 1 style · ★★☆

Solve \(2\cos^2 x + 3\sin x - 3 = 0\) for \(0^\circ \leq x \leq 360^\circ\). How many solutions are there?

  1. A\(2\)
  2. B\(0\)
  3. C\(4\)
  4. D\(5\)
  5. E\(3\)
  6. F\(1\)
Worked solution

\(2(1 - \sin^2 x) + 3\sin x - 3 = 0 \Rightarrow 2\sin^2 x - 3\sin x + 1 = 0 \Rightarrow (2\sin x - 1)(\sin x - 1) = 0\). \(\sin x = \frac12\): \(30^\circ, 150^\circ\); \(\sin x = 1\): \(90^\circ\). Three solutions.

Full method — how and why
A mixed \(\sin\)/\(\cos\) equation. Use \(\cos^2 = 1 - \sin^2\) to make it a quadratic in \(\sin x\), then count solutions carefully.
  1. Replace \(\cos^2 x\) with \(1 - \sin^2 x\): \(2(1 - \sin^2 x) + 3\sin x - 3 = 0\).
    The identity \(\sin^2+\cos^2=1\) is the only way to get a single trig function - always convert the squared one.
  2. Tidy: \(-2\sin^2 x + 3\sin x - 1 = 0\), or \(2\sin^2 x - 3\sin x + 1 = 0\).
    Multiply through by \(-1\) so the leading coefficient is positive.
  3. Factorise: \((2\sin x - 1)(\sin x - 1) = 0\), giving \(\sin x = \frac12\) or \(\sin x = 1\).
    Treat \(\sin x\) as a single unknown.
  4. Count solutions in \(0^\circ \leq x \leq 360^\circ\): \(\sin x = \frac12\) gives \(30^\circ\) and \(150^\circ\); \(\sin x = 1\) gives only \(90^\circ\).
    A maximum value like \(\sin x = 1\) is reached just once per period, not twice - this is what makes the total odd.
  5. Total: \(3\).
    Count, do not assume 'two roots means four solutions'.
Where it goes wrong
  • Assuming each value of \(\sin x\) gives two solutions and answering \(4\).
  • Sign slips when substituting the identity, producing an unfactorisable quadratic.
Takeaway Convert to one trig function, factorise, then count solutions per value - endpoints \(\pm1\) give only one each.
Answer E · Trigonometric functions and equations (MM4.5) · question TRG-02
15Trigonometric functions and equations · spec MM4.4 · Paper 2 style · ★★☆

Which of the following statements about the graph of \(y = \cos x\) is/are true?

  1. IIt is symmetric about the \(y\)-axis.
  2. IIIt is a translation of the graph of \(y = \sin x\) by \(\dfrac{\pi}{2}\) in the negative \(x\)-direction.
  3. IIIIt has period \(\pi\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

\(\cos(-x) = \cos x\) so I is true. \(\sin\left(x + \frac\pi2\right) = \cos x\), so II is true. The period of \(\cos\) is \(2\pi\), so III is false.

Full method — how and why
Three properties of the cosine graph. Each is a standard fact; the false one is about the period.
  1. I: \(\cos(-x) = \cos x\), so the graph is unchanged by reflection in the \(y\)-axis. True.
    Cosine is an even function - its graph is symmetric about the \(y\)-axis, unlike sine.
  2. II: \(\sin\left(x + \frac\pi2\right) = \cos x\), and replacing \(x\) by \(x + \frac\pi2\) shifts the graph \(\frac\pi2\) to the left. True.
    Check a point: \(\sin\) at \(x = \frac\pi2\) equals \(1\), and \(\cos\) reaches \(1\) at \(x = 0\) - shifted left by \(\frac\pi2\) ✓.
  3. III: the period of \(\cos\) is \(2\pi\), not \(\pi\). False.
    \(\cos(x + \pi) = -\cos x\), so a shift of \(\pi\) flips the graph rather than repeating it.
  4. I and II only.
    Match to the option list.
Where it goes wrong
  • Thinking the period is \(\pi\) because \(\cos\) returns to zero every \(\pi\) - the whole shape must repeat.
  • Getting the direction of the shift in II backwards.
Takeaway \(\cos\) is even with period \(2\pi\); \(\cos x = \sin(x + \frac\pi2)\) is a shift to the left.
Answer E · Trigonometric functions and equations (MM4.4) · question TRG-05
16Trigonometric functions and equations · spec MM4.6 · Paper 1 style · ★★☆

Solve \(12\cos^2 x + 6\sin x - 10 = 2\) for \(0^\circ < x < 360^\circ\). Which of the following is the complete set of solutions?

  1. A\(0^\circ, 30^\circ, 150^\circ, 180^\circ\)
  2. B\(30^\circ, 150^\circ, 210^\circ, 330^\circ\)
  3. C\(30^\circ, 150^\circ, 270^\circ\)
  4. D\(30^\circ, 150^\circ\)
  5. E\(210^\circ, 330^\circ\)
  6. F\(30^\circ, 150^\circ, 180^\circ\)
Worked solution

\(12(1 - \sin^2x) + 6\sin x - 12 = 0 \Rightarrow 12\sin^2 x - 6\sin x = 0 \Rightarrow 6\sin x(2\sin x - 1) = 0\). \(\sin x = 0\) gives \(180^\circ\) (the endpoints \(0^\circ\) and \(360^\circ\) are excluded); \(\sin x = \frac12\) gives \(30^\circ\) and \(150^\circ\).

Full method — how and why
Another quadratic in \(\sin x\), but this one has no constant term, so factorising gives \(\sin x = 0\) as well - and the interval is open.
  1. Replace \(\cos^2 x\): \(12(1 - \sin^2 x) + 6\sin x - 10 = 2\).
    Convert to a single trig function.
  2. Tidy: \(12 - 12\sin^2x + 6\sin x - 12 = 0 \Rightarrow -12\sin^2 x + 6\sin x = 0 \Rightarrow 6\sin x(2\sin x - 1) = 0\).
    The constants cancel completely, leaving a factorisable expression with no constant term - do not divide by \(\sin x\), or you lose solutions.
  3. \(\sin x = 0\) gives \(x = 180^\circ\) within the open interval (\(0^\circ\) and \(360^\circ\) are excluded).
    The interval is \(0^\circ < x < 360^\circ\), strict at both ends - so the endpoints are out but \(180^\circ\) is in.
  4. \(\sin x = \frac12\) gives \(x = 30^\circ\) and \(150^\circ\).
    First and second quadrants.
  5. Complete set: \(30^\circ, 150^\circ, 180^\circ\).
    Three solutions - the \(180^\circ\) is what distinguishes the correct option.
Where it goes wrong
  • Dividing by \(\sin x\) and losing \(180^\circ\), giving just \(30^\circ, 150^\circ\) - an offered option.
  • Including \(0^\circ\) and \(360^\circ\), which the strict inequalities exclude.
Takeaway Never divide by a trig function: factorise, so the zero solutions survive; then read the interval's endpoints carefully.
Answer F · Trigonometric functions and equations (MM4.6) · question TRG-14
17Trigonometric functions and equations · spec MM4.4 · Paper 1 style · ★★☆

The graph of \(y = \sin x\) (with \(x\) in radians) crosses the line \(y = x\) at how many points?

  1. A\(2\)
  2. B\(3\)
  3. C\(0\)
  4. D\(1\)
  5. Einfinitely many
Worked solution

\(|\sin x| < |x|\) for all \(x \neq 0\), so the only intersection is at the origin.

Full method — how and why
A graphical comparison. The line \(y = x\) grows without bound while \(\sin x\) is trapped between \(-1\) and \(1\).
  1. Note \(x = 0\) is a solution: \(\sin 0 = 0\).
    Both graphs pass through the origin.
  2. For \(x > 0\), \(\sin x < x\) always (the sine curve leaves the origin with gradient \(1\) and then bends away below the line).
    The gradient of \(\sin x\) is at most \(1\), and it is strictly less than \(1\) after the origin, so the line pulls ahead immediately.
  3. For \(x < 0\), by symmetry (both functions are odd), \(\sin x > x\).
    Odd symmetry means the picture is rotated, not reflected - so again no crossing.
  4. Hence exactly one intersection, at the origin.
    The answer is \(1\).
Where it goes wrong
  • Answering 'infinitely many' by confusing \(y = x\) with a horizontal line like \(y = 0.5\).
  • Thinking there must be crossings because the sine oscillates - it oscillates, but only within \([-1,1]\).
Takeaway \(|\sin x| \leq 1\) while \(|x|\) grows: only the origin can satisfy \(\sin x = x\).
Answer D · Trigonometric functions and equations (MM4.4) · question TRG-15
18Trigonometric functions and equations · spec MM4.5 · Paper 2 style · ★★☆

Which of the following is/are true for all real \(x\)?

  1. I\(\sin^2 x + \cos^2 x = 1\)
  2. II\(\sin(2x) = 2\sin x\)
  3. III\(\sin(x + \pi) = -\sin x\)
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I is the Pythagorean identity. II is false (e.g. \(x = \frac\pi2\): \(0 \neq 2\)). III: translating \(\sin\) by half a period negates it, so III is true.

Full method — how and why
Three identity claims. Two are standard, one is a classic false 'linearity' claim.
  1. I: \(\sin^2 x + \cos^2 x = 1\) is the Pythagorean identity, true for every \(x\). True.
    It follows from Pythagoras applied to the unit circle.
  2. II: test \(x = \frac{\pi}{2}\): \(\sin\pi = 0\) but \(2\sin\frac\pi2 = 2\). False.
    Trig functions are not linear: \(\sin(2x) \neq 2\sin x\) (the correct identity, off-specification here, is \(2\sin x\cos x\)).
  3. III: shifting \(\sin\) by half a period reflects it: \(\sin(x + \pi) = -\sin x\). True.
    Check at \(x = 0\): \(\sin\pi = 0 = -\sin 0\) ✓; at \(x = \frac\pi2\): \(\sin\frac{3\pi}{2} = -1 = -\sin\frac\pi2\) ✓.
  4. I and III only.
    Match to the options.
Where it goes wrong
  • Accepting II because it 'looks like' a factor coming out.
  • Doubting III - test it at two values to be sure.
Takeaway Functions are almost never linear: \(f(2x) \neq 2f(x)\) in general. Test suspicious identities at \(x = \frac\pi2\).
Answer F · Trigonometric functions and equations (MM4.5) · question TRG-17
19Trigonometric functions and equations · spec MM4.6 · Paper 1 style · ★★★

How many solutions does the equation \(\sin x = \dfrac{x}{10}\) have?

  1. A\(7\)
  2. B\(6\)
  3. C\(1\)
  4. D\(5\)
  5. E\(9\)
  6. F\(3\)
Worked solution

Solutions need \(|x| \leq 10\). Both sides are odd functions, so count \(x > 0\) and double, then add \(x = 0\). On \((0, \pi)\) the curve starts above the line (since \(\sin x \approx x\)) and ends below it: one crossing. On \((\pi, 2\pi)\), \(\sin x < 0\): none. On \((2\pi, 3\pi)\) the line is between \(0.63\) and \(0.94\) while \(\sin x\) rises to \(1\): two crossings (\(3\pi \approx 9.42 < 10\)). Beyond \(4\pi > 10\) the line exceeds \(1\). So \(3\) positive, \(3\) negative and \(0\): seven solutions.

Full method — how and why
A transcendental equation - it cannot be solved algebraically. Count intersections of \(y = \sin x\) with the straight line \(y = \frac{x}{10}\) by sketching.
  1. Bound the solutions: \(|\sin x| \leq 1\), so \(\left|\frac{x}{10}\right| \leq 1\), i.e. \(-10 \leq x \leq 10\).
    Outside that range the line is beyond the reach of the sine curve, so no intersections are possible.
  2. Note both sides are odd functions, so intersections come in \(\pm\) pairs; count for \(x > 0\) and double, then add \(x = 0\) (which is a solution).
    \(\sin(-x) = -\sin(-x)\)... more simply, if \(x_0\) is a solution so is \(-x_0\), because both sides change sign together.
  3. For \(0 < x \leq 10\): on \((0, \pi)\) the sine hump starts steeper than the line and ends below it, giving one crossing. On \((\pi, 2\pi)\) the sine is negative while the line is positive: none.
    Near \(0\), \(\sin x\) rises much faster than \(\frac{x}{10}\); by \(x = \pi\) it has returned to \(0\), below the line's value \(0.31\).
  4. On \((2\pi, 3\pi)\) (i.e. up to \(9.42\)) the line runs from \(0.63\) to \(0.94\) while the sine hump rises to \(1\) and falls back: two crossings. Beyond \(3\pi\) the next hump starts past \(x = 4\pi > 10\).
    A hump that rises above the line and comes back down crosses it twice.
  5. So \(3\) positive solutions, \(3\) negative and \(x = 0\): 7 in total.
    Doubling the positive count and adding the origin.
Where it goes wrong
  • Forgetting \(x = 0\) and answering \(6\).
  • Missing that the hump near \(x = 8\) crosses the line twice, and answering \(5\).
Takeaway For \(\sin x = kx\), sketch both graphs, bound the region by \(|kx| \leq 1\), and use oddness to halve the counting.
Answer A · Trigonometric functions and equations (MM4.6) · question TRG-10

How this topic has been examined

Real TMUA questions on Trigonometric functions and equations, newest first. Links open the official paper at the question and the online marking page.

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