P1 5 Trigonometry | A Level Maths 9709

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🎯What you need to be able to do

Quote the exact values of sine, cosine and tangent for \( 30^{\circ}, 45^{\circ}, 60^{\circ} \) and their radian equivalents.
Sketch the graphs of \( \sin x \), \( \cos x \) and \( \tan x \), and apply transformations to them.
Use the symmetry of the graphs, or the CAST diagram, to find every solution in an interval.
Use the identities \( \tan\theta = \dfrac{\sin\theta}{\cos\theta} \) and \( \sin^{2}\theta + \cos^{2}\theta = 1 \).
Solve trigonometric equations, including quadratics in a trigonometric function.
Handle equations in \( \sin(ax + b) \) by transforming the interval before solving.
Prove simple trigonometric identities.

📚The mathematics

Exact values

These are not in the list of formulae and you are expected to know them. Both triangles below reconstruct the whole table if you forget:

\( \sin 30^{\circ} = \tfrac{1}{2} \)

\( \cos 30^{\circ} = \tfrac{\sqrt3}{2} \)

\( \tan 30^{\circ} = \tfrac{1}{\sqrt3} \)

\( \sin 45^{\circ} = \tfrac{1}{\sqrt2} \)

\( \cos 45^{\circ} = \tfrac{1}{\sqrt2} \)

\( \tan 45^{\circ} = 1 \)

\( \sin 60^{\circ} = \tfrac{\sqrt3}{2} \)

\( \cos 60^{\circ} = \tfrac{1}{2} \)

\( \tan 60^{\circ} = \sqrt3 \)

Draw an equilateral triangle of side 2 and cut it in half: that gives the 30–60 values with sides 1, \( \sqrt3 \), 2. Draw a right-angled isosceles triangle with legs 1: that gives 45 with hypotenuse \( \sqrt2 \). Two sketches, nine values, no memorisation.

The graphs

Know these shapes well enough to sketch them without thinking:

\( y = \sin x \) — starts at 0, period \( 360^{\circ} \) (or \( 2\pi \)), range \( -1 \le y \le 1 \).
\( y = \cos x \) — starts at 1, same period and range; it is the sine graph shifted \( 90^{\circ} \) left.
\( y = \tan x \) — period \( 180^{\circ} \) (or \( \pi \)), unbounded, with vertical asymptotes where \( \cos x = 0 \), that is at \( 90^{\circ}, 270^{\circ}, \dots \)

The transformations from P1 2 apply unchanged: \( a\sin(bx) \) has amplitude \( |a| \) and period \( \dfrac{360^{\circ}}{b} \), and \( \sin x + c \) shifts the whole wave up by \(c\).

Finding every solution

Your calculator returns one value — the principal value. The marks are for the others. Two equivalent tools:

The graph method. Sketch the curve across the required interval, draw the horizontal line \( y = k \), and count the intersections. Then use the symmetry of the curve to get each \(x\) from the first one.

The CAST diagram. In the four quadrants, All are positive, then Sine, Tangent, Cosine, going anticlockwise from the top right. The useful relationships that follow:

\( \sin\theta = \sin(180^{\circ} - \theta) \)

\( \cos\theta = \cos(360^{\circ} - \theta) \)

\( \tan\theta = \tan(180^{\circ} + \theta) \)

So for sine, the second solution is \( 180^{\circ} \) minus the first; for cosine it is \( 360^{\circ} \) minus the first; for tangent, add \( 180^{\circ} \). Then keep adding or subtracting the period until you leave the interval.

The two identities

\( \tan\theta \equiv \dfrac{\sin\theta}{\cos\theta} \)

\( \sin^{2}\theta + \cos^{2}\theta \equiv 1 \)

These are the whole of P1 identity work, and between them they solve almost every equation that is not already in a single function. The strategy is to get everything into one trigonometric function:

An equation mixing \( \sin \) and \( \cos \) linearly, such as \( 3\sin\theta = 2\cos\theta \) — divide by \( \cos\theta \) to get \( \tan\theta = \tfrac{2}{3} \).
An equation mixing \( \sin^{2} \) and \( \cos \) — replace \( \sin^{2}\theta \) with \( 1 - \cos^{2}\theta \), and you have a quadratic in \( \cos\theta \).

Once you have a quadratic in, say, \( \cos\theta \), substitute \( u = \cos\theta \), solve, and then reject any root outside \( -1 \le u \le 1 \) before finding the angles. That rejection is a mark, and it is also a useful signal: if both roots are outside the range you have almost certainly made an algebraic slip.

Equations in \( \sin(ax + b) \)

Transform the interval first, then solve, then transform back. For \( \sin 2x = 0.5 \) with \( 0^{\circ} \le x \le 360^{\circ} \), the quantity you are actually solving for is \( 2x \), whose interval is \( 0^{\circ} \le 2x \le 720^{\circ} \) — twice as long, so twice as many solutions. Find all values of \( 2x \) first, and only then divide each by 2. Solving in \(x\) and doubling at the end loses half the answers, and it is the most expensive single error in P1 trigonometry.

The same applies to \( \sin(x - 30^{\circ}) \): shift the interval by \( 30^{\circ} \), solve, then shift back.

Proving identities

Start with one side — usually the more complicated one — and transform it into the other. Do not work on both sides at once, and do not treat the identity as an equation to be rearranged. Useful moves: write everything in terms of \( \sin \) and \( \cos \), combine fractions over a common denominator, factorise, and look for \( \sin^{2} + \cos^{2} \) to replace with 1. Finish with a statement, not a stranded line of algebra.

✏️Worked example

(a) Solve \( 3\sin^{2}x + 5\cos x - 1 = 0 \) for \( 0^{\circ} \le x \le 360^{\circ} \). (b) Solve \( \tan(2x - 30^{\circ}) = 1 \) for \( 0^{\circ} \le x \le 180^{\circ} \).

(a) The equation mixes \( \sin^{2} \) with \( \cos \), so use the Pythagorean identity to remove the sine: \( \sin^{2}x = 1 - \cos^{2}x \).

\[ 3\left(1 - \cos^{2}x\right) + 5\cos x - 1 = 0 \;\Longrightarrow\; -3\cos^{2}x + 5\cos x + 2 = 0 \]

Multiply by \( -1 \) and factorise: \( 3\cos^{2}x - 5\cos x - 2 = 0 \), so \( (3\cos x + 1)(\cos x - 2) = 0 \). Hence \( \cos x = -\tfrac{1}{3} \) or \( \cos x = 2 \).

Reject \( \cos x = 2 \), since cosine never exceeds 1. From \( \cos x = -\tfrac{1}{3} \), the calculator gives \( x = 109.47^{\circ} \). Cosine is negative in the second and third quadrants, and the second solution is \( 360^{\circ} - 109.47^{\circ} \):

\[ x = 109.5^{\circ} \quad\text{or}\quad x = 250.5^{\circ} \]

(b) Let \( u = 2x - 30^{\circ} \). Transform the interval: when \( x = 0 \), \( u = -30^{\circ} \); when \( x = 180^{\circ} \), \( u = 330^{\circ} \). So we need \( \tan u = 1 \) for \( -30^{\circ} \le u \le 330^{\circ} \).

The principal value is \( u = 45^{\circ} \). Tangent has period \( 180^{\circ} \), so add and subtract 180 repeatedly and keep whatever lands inside the interval: \( 45^{\circ} \), \( 225^{\circ} \), and \( -135^{\circ} \) which is outside. So \( u = 45^{\circ} \) or \( 225^{\circ} \). Converting back with \( x = \dfrac{u + 30^{\circ}}{2} \):

\[ x = 37.5^{\circ} \quad\text{or}\quad x = 127.5^{\circ} \]

Check it. Substitute back into the original equation: \( 2(37.5) - 30 = 45 \) and \( \tan 45^{\circ} = 1 \) ✓; \( 2(127.5) - 30 = 225 \) and \( \tan 225^{\circ} = 1 \) ✓. Also count what you should expect: \( \tan \) has period \( 180^{\circ} \) and the interval for \(u\) spans \( 360^{\circ} \), so exactly two solutions is right.

Two rejections, two different reasons. In (a), \( \cos x = 2 \) is rejected because it is outside the range of cosine — state that, it is a mark. In (b), \( u = -135^{\circ} \) is rejected because it falls outside the transformed interval. Candidates who never write the transformed interval down have no way of knowing which values to keep, and typically report only \( x = 37.5^{\circ} \).

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Solve \( 2\sin x = -1 \) for \( 0^{\circ} \le x \le 360^{\circ} \).

\( \sin x = -0.5 \). The calculator gives \( -30^{\circ} \), which is outside the interval, so work from the acute angle \( 30^{\circ} \). Sine is negative in the third and fourth quadrants: third gives \( 180^{\circ} + 30^{\circ} = 210^{\circ} \); fourth gives \( 360^{\circ} - 30^{\circ} = 330^{\circ} \). So \( x = 210^{\circ} \) or \( x = 330^{\circ} \).

2. Solve \( 4\sin x = 3\cos x \) for \( 0^{\circ} \le x \le 360^{\circ} \).

Divide both sides by \( \cos x \): \( 4\tan x = 3 \), so \( \tan x = 0.75 \). The principal value is \( 36.87^{\circ} \), and tangent has period \( 180^{\circ} \), so the other solution is \( 216.87^{\circ} \). Hence \( x = 36.9^{\circ} \) or \( x = 216.9^{\circ} \). Dividing by \( \cos x \) is safe here because \( \cos x = 0 \) would force \( \sin x = 0 \) too, which is impossible.

3. Solve \( 2\cos^{2}x + 3\sin x = 3 \) for \( 0^{\circ} \le x \le 360^{\circ} \).

Replace \( \cos^{2}x \) with \( 1 - \sin^{2}x \): \( 2 - 2\sin^{2}x + 3\sin x = 3 \), so \( 2\sin^{2}x - 3\sin x + 1 = 0 \), giving \( (2\sin x - 1)(\sin x - 1) = 0 \). So \( \sin x = \tfrac{1}{2} \) or \( \sin x = 1 \). From \( \sin x = \tfrac12 \): \( x = 30^{\circ} \) or \( 150^{\circ} \). From \( \sin x = 1 \): \( x = 90^{\circ} \). Three solutions altogether — the "nice" root \( \sin x = 1 \) contributes only one because it sits exactly at the maximum.

4. Solve \( \cos 3x = \dfrac{\sqrt3}{2} \) for \( 0^{\circ} \le x \le 180^{\circ} \).

Transform the interval: \( 0^{\circ} \le 3x \le 540^{\circ} \). Now \( \cos u = \dfrac{\sqrt3}{2} \) gives principal value \( u = 30^{\circ} \); cosine is positive in quadrants 1 and 4, so also \( u = 330^{\circ} \). The period is \( 360^{\circ} \), so add it: \( 390^{\circ} \) is inside \( 540^{\circ} \), while \( 690^{\circ} \) is not. So \( u = 30^{\circ}, 330^{\circ}, 390^{\circ} \), and dividing by 3, \( x = 10^{\circ}, 110^{\circ}, 130^{\circ} \).

5. Prove that \( \dfrac{1 - \cos^{2}\theta}{\cos\theta\,\tan\theta} \equiv \sin\theta \).

Start with the left-hand side. The numerator is \( 1 - \cos^{2}\theta = \sin^{2}\theta \). The denominator is \( \cos\theta \times \dfrac{\sin\theta}{\cos\theta} = \sin\theta \). So LHS \( = \dfrac{\sin^{2}\theta}{\sin\theta} = \sin\theta \equiv \) RHS, as required. Note the denominator simplifying to \( \sin\theta \) is the key step — writing \( \tan \) in terms of \( \sin \) and \( \cos \) is almost always the first move in these proofs.

6. Solve \( \sin\!\left(x + 40^{\circ}\right) = 0.6 \) for \( 0^{\circ} \le x \le 360^{\circ} \).

Let \( u = x + 40^{\circ} \), so the interval becomes \( 40^{\circ} \le u \le 400^{\circ} \). From \( \sin u = 0.6 \), the principal value is \( u = 36.87^{\circ} \) — outside the interval, so discard it — and the second-quadrant solution is \( 180 - 36.87 = 143.13^{\circ} \), which is inside. Adding the period gives \( 36.87 + 360 = 396.87^{\circ} \), also inside. So \( u = 143.13^{\circ} \) or \( 396.87^{\circ} \), giving \( x = 103.1^{\circ} \) or \( x = 356.9^{\circ} \). This one rewards writing the transformed interval down: the principal value is rejected and a value beyond \( 360^{\circ} \) is kept, which is the opposite of what instinct suggests.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

GeoGebra — the unit circle with the sine and cosine graphs unrolling beside it
Desmos — plot \( y = \sin 2x \) and a horizontal line to see why the interval doubles
Khan Academy — trigonometric equations and identities