Edexcel IAL Maths

P2.6 Trigonometric identities and equations

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What this page covers
Overview

The identities tan θ≡sin θ/cos θ and sin²θ+ cos²θ≡1: where they come from and how to rearrange them

Exact values of sin, cos and tan from one given ratio and the quadrant of the angle

Proving simple identities by working from one side to the other

Solving sin θ= k, cos θ= k and tan θ= k in a given interval, in degrees or radians

Equations containing both sin and cos: dividing to get tan, and factorising instead of cancelling

Equations of the form sin(nθ+ α) = k and the cos and tan versions

Quadratic equations in sin θ, cos θ or tan θ, including those that need sin²θ+ cos²θ= 1 first

In one minute
Key points

1. The two identities tan θ≡sin θ/cos θ and sin²θ+ cos²θ≡1 let you write any expression in terms of one function, and the symmetry of the sine, cosine and tangent graphs turns the calculator's single answer into every solution in an interval.

2. To prove an identity, start from one side and transform it, step by step, into the other. Never work on both sides at once.

3. sin θ= k: α and 180° - α. cos θ= k: α and 360° - α (or -α). tan θ= k: α and α+ 180°. Then add or subtract 360° (or 2π) until you leave the interval.

4. For sin(nθ+ α) = k, change the interval first: if 0 ≤θ< 360°, then α≤nθ+ α< 360n° + α. Solve for the whole bracket, then convert back.

5. Never divide both sides by sin θ or cos θ: factorise, or you lose the solutions where that factor is zero.

In one minute (continued)
Key points

6. A quadratic in sin or cos can give impossible roots: sin θ= 2 or cos θ= -3 has no solutions, because -1 ≤sin θ, cos θ≤1.

Where do the two identities come from?
Section 1 of 13
−11−11OθP(cos θ, sin θ)cos θsin θ1
The point P on a unit circle has coordinates (cos θ, sin θ). The shaded right-angled triangle has hypotenuse 1, so sin²θ + cos²θ = 1; the gradient of OP is sin θ ÷ cos θ = tan θ. For θ = 40°: 0.766² + 0.643² = 1 to the accuracy shown.

Draw a circle of radius 1 centred at the origin. A radius turned through an angle θ anticlockwise from the positive x-axis meets the circle at a point P. For any angle, cos θ is defined as the x-coordinate of P and sin θ as its y-coordinate. For acute angles this agrees with SOH CAH TOA, because the hypotenuse is 1; for larger angles it is what extends sin and cos to every angle.

Pythagoras. P is 1 unit from O, so x² + y² = 1, which is sin²θ+ cos²θ≡1. Here sin²θ means (sin θ)².

Gradient. The gradient of OP is y/x, which is the definition of tan θ, so tan θ≡sin θ/cos θ. It is undefined where cos θ= 0 (at 90°, 270°, …).

The sign ≡ means "identically equal": true for every value of θ, not just for particular solutions.

Where do the two identities come from? (continued)
Section 1 of 13
0901802703600.51θ (degrees)sin²θ + cos²θ = 1sin²θcos²θ
sin²θ (blue) and cos²θ (red) rise and fall in opposite ways: wherever one is large the other is small, and at every angle their sum is exactly 1 (the teal line).

The Pythagorean identity rearranges into two forms that you will use constantly: sin²θ≡1 - cos²θ and cos²θ≡1 - sin²θ.

The two identities

(not in the formulae booklet: learn both, and both rearrangements of the second):
tan θ≡sin θ/cos θ   sin²θ+ cos²θ≡1
Valid for every angle in degrees or radians; the first needs cos θ≠0.

How do I find exact values from one known ratio?
Section 2 of 13

If you know one ratio, sin²θ+ cos²θ= 1 gives the size of the other, and the quadrant of θ decides its sign. The signs follow from the unit circle: in the first quadrant (0° to 90°) all three are positive; in the second (90° to 180°) only sin is positive; in the third (180° to 270°) only tan; in the fourth (270° to 360°) only cos. Then tan θ= sin θ÷cos θ. "Exact" means surds and fractions, so rationalise the denominator if asked for simplest form.

Given tan θ= k instead, write sin θ= kcos θ and substitute into the Pythagorean identity. A right-angled triangle with opposite side k and adjacent side 1 is an equally good method, as long as you attach the signs afterwards.

How do I find exact values from one known ratio? (continued)
Section 2 of 13 · Worked example 1 · Medium · Find
12√5α
Part (b) with a triangle: tan α = 2 ÷ 1, so the hypotenuse is √(1² + 2²) = √5. This gives the sizes cos α = 1/√5 and sin α = 2/√5; the fourth quadrant then makes sin negative.

WORKED EXAMPLE · QUESTION

(a) Given that sin θ= (2/7) and that θ is obtuse, find the exact values of cos θ and tan θ. (b) Given that tan x = -2 and 270° < x < 360°, find the exact values of cos x and sin x.

SOLUTION

(a) cos²θ= 1 - sin²θ= 1 - (4/49) = (45/49), so cos θ= ±√45/7 = ±3√5/7. An obtuse angle is in the second quadrant, where cos is negative: cos θ= -3√5/7. Then tan θ= (2/7) ÷(-3√5/7) = -2/3√5 = -2√5/15. cos θ= -3√5/7, tan θ= -2√5/15

(b) sin x = -2cos x, substituted into the identity: 4cos² x + cos² x = 1, so cos² x = ⅕. In the fourth quadrant cos is positive and sin is negative. cos x = √5/5, sin x = -2√5/5

How do I find exact values from one known ratio? (continued)
Section 2 of 13 · Worked example 1 · Medium · Find (continued)

Check (b): sin x ÷cos x = -2 ✓, and (4/5) + ⅕ = 1 ✓.

How do I prove a trigonometric identity?
Section 3 of 13

An identity is a statement that must be true for every angle, so checking one or two values proves nothing: they could agree by chance. A proof is a chain of equal expressions that starts at one side and ends at the other.

Start with the more complicated side and transform it into the other side. Write "LHS =" at the start and finish with "= RHS".

Useful first moves: replace tan θ with sin θ/cos θ; combine fractions over a common denominator; expand brackets; look for sin²θ+ cos²θ and replace it with 1.

Factorising helps too: a² - b² = (a - b)(a + b) is common with fourth powers.

Do not start from the statement you are trying to prove and manipulate both sides: that assumes the result.

How do I prove a trigonometric identity? (continued)
Section 3 of 13 · Worked example 2 · Medium · Prove

WORKED EXAMPLE · QUESTION

Prove that (a) (sin θ+ cos θ)² + (sin θ- cos θ)² ≡2; (b) tan θ+ 1/tan θ ≡1/sin θcos θ; (c) cos⁴θ- sin⁴θ≡2cos²θ- 1.

SOLUTION

(a) Expand both squares from the LHS: (sin²θ+ 2sin θcos θ+ cos²θ) + (sin²θ- 2sin θcos θ+ cos²θ) = 2(sin²θ+ cos²θ) = 2 ×1 = 2. = RHS

(b) Write tan in terms of sin and cos, then use a common denominator:

LHS = sin θ/cos θ + cos θ/sin θ = (sin²θ+ cos²θ)/sin θcos θ = 1/sin θcos θ

= RHS

(c) Difference of two squares: LHS = (cos²θ- sin²θ)(cos²θ+ sin²θ) = cos²θ- sin²θ.

How do I prove a trigonometric identity? (continued)
Section 3 of 13 · Worked example 2 · Medium · Prove (continued)

Replace sin²θ with 1 - cos²θ: cos²θ- (1 - cos²θ) = 2cos²θ- 1. = RHS

In (c) the bracket around 1 - cos²θ matters: subtracting it changes both signs.

How do I prove a trigonometric identity? (continued)
Section 3 of 13

Common error

Writing sin²θ+ cos²θ= 1 as sin θ+ cos θ= 1. Square roots do not split over addition: at θ= 45° the left side is √2, not 1.

How do I find every solution of sin θ = k, cos θ = k or tan θ = k?
Section 4 of 13

The calculator's inverse function returns one angle, the principal value α: sin⁻¹ and tan⁻¹ give an angle from -90° to 90°, and cos⁻¹ gives one from 0° to 180°. That value may not even be in your interval. The graphs repeat and are symmetric, which gives a second angle in each cycle:

sin θ= k: α and 180° - α (the sine curve is symmetric about 90°).

cos θ= k: α and 360° - α, equivalently -α (the cosine curve is symmetric about 0°).

tan θ= k: α and α+ 180° (the tangent repeats every 180°).

Then add or subtract the period (360° for sin and cos, 180° for tan) until every solution in the interval has been listed. If |k| > 1, sin θ= k and cos θ= k have no solutions; tan θ= k always has solutions. Give degrees to 1 decimal place unless told otherwise, but carry the full calculator value through any later working.

How do I find every solution of sin θ = k, cos θ = k or tan θ = k? (continued)
Section 4 of 13
θ0°30°45°60°90°
sin θ0½√2/2√3/21
cos θ1√3/2√2/2½0
tan θ0√3/31√3undefined

These exact values must be known. In radians the angles are 0, π/6, π/4, π/3 and π/2.

How do I find every solution of sin θ = k, cos θ = k or tan θ = k? (continued)
Section 4 of 13
Oα180° − α180° + α360° − αASTC
The four angles α, 180° − α, 180° + α and 360° − α (drawn for α = 35°) all make the same acute angle with the x-axis, so sin, cos and tan have the same size at all four; only the signs change. A: all positive, S: sin, T: tan, C: cos.
How do I find every solution of sin θ = k, cos θ = k or tan θ = k? (continued)
Section 4 of 13 · Worked example 3 · Easy · Solve
−9090180270360−1−0.50.51O−23.6°203.6°336.4°y = sin θy = −0.4
(a) The calculator gives −23.6°, outside the interval (hollow dot). Symmetry gives 180° − (−23.6°) = 203.6°, and adding 360° to the calculator value gives 336.4°.

WORKED EXAMPLE · QUESTION

Solve, giving answers to 1 decimal place: (a) sin θ= -0.4 for 0° ≤θ< 360°; (b) cos θ= -0.65 for -180° ≤θ≤180°; (c) tan θ= 2.5 for -180° ≤θ≤180°.

How do I find every solution of sin θ = k, cos θ = k or tan θ = k? (continued)
Section 4 of 13 · Worked example 3 · Easy · Solve (continued)
−180−9090180−1−0.50.51O−130.5°130.5°y = cos θy = −0.65
(b) The cosine curve is symmetric about the vertical axis, so the two solutions are ±130.5°.
How do I find every solution of sin θ = k, cos θ = k or tan θ = k? (continued)
Section 4 of 13 · Worked example 3 · Easy · Solve (continued)
−180−9090180−5−4−3−2−112345O68.2°−111.8°y = 2.5
(c) The tangent graph has period 180° and asymptotes where cos θ = 0 (dashed), so its two solutions in this interval are exactly 180° apart: 68.2° and −111.8°.

SOLUTION

(a) sin⁻¹(-0.4) = -23.58° (not in the interval). Then 180° - (-23.58°) = 203.58° and -23.58° + 360° = 336.42°. θ= 203.6° or 336.4°

(b) The cosine curve is symmetric about θ= 0°: cos⁻¹(-0.65) = 130.54°. θ= ±130.5°

(c) The tangent repeats every 180°: tan⁻¹(2.5) = 68.20°, and 68.20° - 180° = -111.80°. (68.2° + 180° = 248.2° is outside the interval.) θ= 68.2° or -111.8°

Check every answer by putting it back into the calculator.

What do I do when an equation contains both sin and cos?
Section 5 of 13

There are three common shapes, and each has its own safe move.

Linear in both, such as asin θ= bcos θ: divide both sides by cos θ to get tan θ= b/a. This is safe, because cos θ= 0 would force sin θ= 0 as well, which never happens at the same angle.

A common factor, such as sin θcos θ= ksin θ: do not divide by sin θ. Bring everything to one side and factorise: sin θ(cos θ- k) = 0. Then solve each factor = 0.

Squares of one, first powers of the other: use sin²θ+ cos²θ= 1 to reach a quadratic in one function (see the last teaching section).

What do I do when an equation contains both sin and cos? (continued)
Section 5 of 13 · Worked example 4 · Medium · Solve
090180270360−3−2−11230°66.4°180°293.6°y = 5 sin θ cos θy = 2 sin θ
(b) The two sides of the equation as graphs cross four times in 0° ≤ θ < 360°. The crossings at 0° and 180° are the sin θ = 0 solutions that cancelling would throw away.

WORKED EXAMPLE · QUESTION

Solve, for 0° ≤θ< 360°, giving answers to 1 decimal place where appropriate: (a) 3sin θ+ 7cos θ= 0; (b) 5sin θcos θ= 2sin θ.

SOLUTION

(a) Rearrange and divide by cos θ: 3sin θ= -7cos θ⇒tan θ= -(7/3). Tan is negative in the second and fourth quadrants: tan⁻¹(-(7/3)) = -66.80°; add 180° and 360°. θ= 113.2° or 293.2°

(b) One side zero, then factorise: 5sin θcos θ- 2sin θ= 0 ⇒sin θ(5cos θ- 2) = 0.

What do I do when an equation contains both sin and cos? (continued)
Section 5 of 13 · Worked example 4 · Medium · Solve (continued)

Solve each factor: sin θ= 0 ⇒θ= 0°, 180°; cos θ= 0.4 ⇒θ= 66.42° or 360° - 66.42° = 293.58°. θ= 0°, 66.4°, 180°, 293.6°

Dividing (b) by sin θ would give only the two cos θ= 0.4 answers. 360° is excluded because the interval is 0° ≤θ< 360°.

How do I solve sin(nθ + α) = k in a given interval?
Section 6 of 13

Let u = nθ+ α. The graph of sin(nθ+ α) has period 360°/n, so there are about n times as many solutions in the interval as for sin θ= k. The method has three steps:

1. Transform the interval. Apply the same operations to the end points: 0° ≤θ< 360° becomes α≤u < 360n° + α.

2. Solve for u in the new interval exactly as before, adding 360° until you pass the upper limit (and checking below the lower limit too).

3. Convert back: θ= (u - α)/n for every value of u.

Changing the interval first is what stops you missing solutions; dividing by n at the end, using unrounded values, keeps the answers accurate.

How do I solve sin(nθ + α) = k in a given interval? (continued)
Section 6 of 13 · Worked example 5 · Medium · Solve
04590135180225270315360−1−0.50.5188.4°141.6°268.4°321.6°y = cos(2θ − 50°)
y = cos(2θ − 50°) has period 180°, so it completes two cycles between 0° and 360° and meets y = −0.6 four times.

WORKED EXAMPLE · QUESTION

Solve cos(2θ- 50°) = -0.6 for 0° ≤θ< 360°, giving your answers to 1 decimal place.

SOLUTION

New variable and interval: u = 2θ- 50°, so -50° ≤u < 670°.

Principal value and the symmetry of cos: cos⁻¹(-0.6) = 126.87°, and 360° - 126.87° = 233.13°. Add 360° to each while still below 670°: u = 126.87°, 233.13°, 486.87°, 593.13°. (-126.87° is below -50°, and 846.87° is above 670°, so neither is used.)

How do I solve sin(nθ + α) = k in a given interval? (continued)
Section 6 of 13 · Worked example 5 · Medium · Solve (continued)

Convert back with θ= (u + 50°)/2. θ= 88.4°, 141.6°, 268.4°, 321.6°

Four answers: two cycles of the curve, each crossing the line twice. Solutions for u differ by 360°, so the answers for θ come in pairs 180° apart.

How does the same method work in radians?
Section 7 of 13

If the interval is written with π (such as 0 ≤x < 2π), the answers must be in radians, so set the calculator to radian mode. The rules are the same with 180° replaced by π: sin gives α and π- α; cos gives α and 2π- α; tan gives α and α+ π; the period of sin and cos is 2π.

Exact answers are multiples of π, from the exact-value table; other answers are usually given to 3 significant figures. For sin(nx + α) = k, transform the interval exactly as before. It helps to write the end points with a common denominator, e.g. π/4 ≤3x + π/4 ≤13π/4.

How does the same method work in radians? (continued)
Section 7 of 13 · Worked example 6 · Hard · Solve
π/4π/23π/4π−1−0.50.5113π/3617π/36(π, −0.707)y = −√3/2
(a) y = sin(3x + π/4) for 0 ≤ x ≤ π meets y = −√3/2 only at 13π/36 and 17π/36. At the end of the interval the curve has only fallen to −√2/2 ≈ −0.707, above −√3/2 ≈ −0.866, so there is no third solution.

WORKED EXAMPLE · QUESTION

(a) Solve sin(3x + π/4) = -√3/2 for 0 ≤x ≤π, giving exact answers. (b) Solve tan 2x = 3 for 0 ≤x < π, giving answers to 3 significant figures.

SOLUTION

(a) New interval: u = 3x + π/4, so π/4 ≤u ≤13π/4. For sin u = -√3/2 the reference angle is π/3, and sin is negative in the third and fourth quadrants: u = π+ π/3 = 4π/3 and u = 2π- π/3 = 5π/3. The next one is 4π/3 + 2π= 10π/3 > 13π/4 (over 12 they are 40π/12 and 39π/12), so stop.

How does the same method work in radians? (continued)
Section 7 of 13 · Worked example 6 · Hard · Solve (continued)

Convert back with x = (u - π/4) ÷3: 13π/12 ÷3 and 17π/12 ÷3. x = 13π/36 or 17π/36

(b) 0 ≤2x < 2π, and tan repeats every π: 2x = tan⁻¹3 = 1.2490 or 1.2490 + π= 4.3906. x = 0.625 or 2.20

How do I solve a quadratic equation in sin, cos or tan?
Section 8 of 13

An equation such as 2sin²θ- sin θ- 1 = 0 is a quadratic in sin θ. Factorise (or use the formula) as if sin θ were a single letter s. If the equation mixes sin²θ with cos θ, or cos²θ with sin θ, first replace the square using sin²θ= 1 - cos²θ or cos²θ= 1 - sin²θ, so that only one function remains.

Each root gives a basic equation of the kind solved above. Reject any value of sin or cos outside -1 to 1, and say why. Expect up to four solutions in a 360° interval: two from each root.

How do I solve a quadratic equation in sin, cos or tan? (continued)
Section 8 of 13 · Worked example 7 · Medium · Solve
090180270360−1−0.50.5130°150°199.5°340.5°y = ½y = −⅓y = sin x
Worked example 7(a): the factorised equation becomes two horizontal lines. y = ½ cuts y = sin x at 30° and 150°; y = −⅓ cuts it at 199.5° and 340.5°.

WORKED EXAMPLE · QUESTION

(a) Solve 6cos² x + sin x - 5 = 0 for 0° ≤x < 360°. (b) Solve tan²θ- 2tan θ- 3 = 0 for 0° ≤θ< 180°. Give answers to 1 decimal place where appropriate.

SOLUTION

(a) Replace cos² x to leave only sin x: 6(1 - sin² x) + sin x - 5 = 0 ⇒6sin² x - sin x - 1 = 0. Factorise: (2sin x - 1)(3sin x + 1) = 0, so sin x = ½ or sin x = -⅓.

sin x = ½: x = 30°, 150°. sin x = -⅓: sin⁻¹(-⅓) = -19.47°, so x = 180° + 19.47° or 360° - 19.47°. x = 30°, 150°, 199.5°, 340.5°

How do I solve a quadratic equation in sin, cos or tan? (continued)
Section 8 of 13 · Worked example 7 · Medium · Solve (continued)

(b) Factorise as a quadratic in tan θ: (tan θ- 3)(tan θ+ 1) = 0, so tan θ= 3 ⇒θ= 71.6°, or tan θ= -1 ⇒θ= -45° + 180° = 135°. θ= 71.6° or 135°

In (b) the interval is only 180° long, one period of tan, so each root gives exactly one solution.

How do I solve a quadratic equation in sin, cos or tan? (continued)
Section 8 of 13 · Worked example 8 · Hard · Show that / Hence
090180270360−1−0.50.511.5275.5°284.5°y = 2: no intersection, cos x ≤ 1y = ¼y = cos x
Worked example 8(b): cos x = ¼ gives two solutions; the line y = 2 lies entirely above the cosine curve, which is why the root cos x = 2 is rejected.

WORKED EXAMPLE · QUESTION

(a) Show that the equation 4sin² x + 9cos x = 6 can be written as 4cos² x - 9cos x + 2 = 0. (b) Hence solve 4sin² x + 9cos x = 6 for 0° ≤x < 360°, giving your answers to 1 decimal place. (c) Hence find the solutions of 4sin² 2θ+ 9cos 2θ= 6 for 0° ≤θ< 180°, to 1 decimal place.

SOLUTION

(a) Replace sin² x: 4(1 - cos² x) + 9cos x = 6 ⇒4 - 4cos² x + 9cos x - 6 = 0. Multiplying by -1: 4cos² x - 9cos x + 2 = 0 ✓

(b) Factorise:

How do I solve a quadratic equation in sin, cos or tan? (continued)
Section 8 of 13 · Worked example 8 · Hard · Show that / Hence (continued)

(4cos x - 1)(cos x - 2) = 0, so cos x = ¼ or cos x = 2. cos x = 2 has no solutions, since -1 ≤cos x ≤1. cos⁻¹(¼) = 75.52°, and 360° - 75.52° = 284.48°. x = 75.5° or 284.5°

(c) The same equation with x = 2θ, and 0° ≤2θ< 360°: 2θ= 75.52° or 284.48° (the next value, 435.52°, is outside the interval). θ= 37.8° or 142.2°

"Hence" in (c) means reuse the values from (b). Halving the rounded answer 284.5 gives 142.25, which rounds to 142.3 instead of the correct 142.2; halve the unrounded values.

Exam tips for P2.6
Section 9 of 13

Exam tip 1 — sketch before you list

A quick sketch of the graph over the stated interval, with the horizontal line y = k, tells you how many solutions to expect before you find any of them. If your list has a different number, look again.

Exam tip 2 — change the interval first

For sin(nθ+ α) = k, write the new interval for nθ+ α as your first line of working. It shows the method clearly and it is the step that finds all the solutions.

Exam tip 3 — degrees or radians, and how accurate

An interval written with π means radians: switch the calculator mode and give answers in radians, exact (in terms of π) or to 3 s.f. Degrees are normally given to 1 d.p. Never mix the two in one answer.

Exam tips for P2.6 (continued)
Section 9 of 13

Exam tip 4 — factorise, never cancel

When a trigonometric factor appears on both sides, collect everything on one side and factorise. Each factor equal to zero gives its own set of solutions.

Exam tip 5 — proofs read from one side to the other

Start at one side, quote each identity as you use it, and end with the other side written out exactly. In a "show that" equation such as Worked example 8(a), the target line is given, so every step to it must be visible.

Common misconceptions
Section 10 of 13
MisconceptionWhy it is wrongCorrect idea
sin²θ means sin(θ²).The index belongs to the value of the sine, not to the angle.sin²θ= (sin θ)²; at θ= 30° it is ¼.
sin 2θ= 2sin θ, so sin 2θ= 0.6 means sin θ= 0.3.sin is not a multiplier: sin 60° ≠2sin 30°.Let u = 2θ, solve sin u = 0.6, then halve.
The second solution of tan θ= k is 180° - α.That is the rule for sine; tan(180° - α) = -tan α.tan repeats every 180°: α, α+ 180°, …
The calculator answer is the answer.The inverse gives only the principal value, which may lie outside the interval.Use the symmetry of the graph, then add or subtract the period.
sin θ= 1.5 means there was an error earlier.A quadratic can produce a root that is out of range even when all the working is right.State that sin θ= 1.5 has no solutions because -1 ≤sin θ≤1, and continue with the other root.
Checking an identity at θ= 30° proves it.One value can agree by chance; an identity must hold for every angle.Transform one side into the other using known identities.
How this topic connects
Section 11 of 13

Trigonometry in P1

Radians, the exact values and the graphs of sin, cos and tan with their transformations are the starting point here. The ambiguous case of the sine rule is the equation sin θ= k with its two solutions α and 180° - α.

Trigonometry in P3

sec, cosec and cot, the identities 1 + tan²θ≡sec²θ and 1 + cot²θ≡csc²θ, compound and double angle formulae and Rcos(θ- α) all reduce equations to the forms solved on this page. Differentiating and integrating sin and cos (P3, P4) uses radians, and sin² t + cos² t = 1 eliminates the parameter from curves such as x = 3cos t, y = 2sin t.

How this topic connects (continued)
Section 11 of 13

Mechanics and IAL Physics

Resolving forces leads to equations such as tan θ= 0.75 or sin θ= k; oscillations and waves use sin(ωt + φ) = k, solved for t by the interval method on this page.

Key formulae
Section 12 of 13
FormulaUseIn the formulae list?
tan θ≡sin θ/ cos θreplacing tan; turning asin θ= bcos θ into tan θ= b/aNo — learn it
sin²θ+ cos²θ≡1exact values from one ratio; proofs; converting to a quadratic in one functionNo — learn it
sin²θ≡1 - cos²θ, cos²θ≡1 - sin²θreplacing a square to leave one functionNo — derive it from the identity
sin θ= k: θ= α, 180° - α (±360°)all solutions of a sine equationNo — derive it from the graph
cos θ= k: θ= ±α (±360°)all solutions of a cosine equationNo — derive it from the graph
tan θ= k: θ= α (±180°)all solutions of a tangent equationNo — derive it from the graph
Key formulae (continued)
Section 12 of 13
FormulaUseIn the formulae list?
sin 30° = ½, sin 45° = √2/2, sin 60° = √3/2 (and the cos, tan values)exact answers in degrees or multiples of πNo — learn it
Check your understanding
Section 13 of 13

1. Given that cos θ= -0.8 and 180° < θ< 270°, find sin θ and tan θ.

2. Solve sin θcos θ= sin θ for 0° ≤θ≤360°.

3. How many solutions does sin 3θ= 0.2 have for 0° ≤θ< 360°?

4. Solve cos x = ½ for 0 ≤x < 2π, giving exact answers.

5. Why does 2sin²θ- 5sin θ+ 2 = 0 have only two solutions for 0° ≤θ< 360°?

6. Prove that (1 - cos²θ)/sin θcos θ ≡tan θ.

Check your understanding · answers
Section 13 of 13

1. sin²θ= 1 - 0.64 = 0.36, and sin is negative in the third quadrant, so sin θ= -0.6. Then tan θ= -0.6 ÷-0.8 = 0.75 (positive, as expected in the third quadrant).

2. sin θ(cos θ- 1) = 0. sin θ= 0 gives 0°, 180°, 360°; cos θ= 1 gives 0°, 360°. The solutions are 0°, 180° and 360°. Dividing by sin θ would have lost 180°.

3. With u = 3θ, 0° ≤u < 1080°: three full cycles, each meeting y = 0.2 twice. Six solutions.

4. The principal value is π/3; the other solution is 2π- π/3. So x = π/3 or 5π/3.

5. It factorises as (2sin θ- 1)(sin θ- 2) = 0. sin θ= 2 is impossible, so only sin θ= ½ remains, giving θ= 30° and 150°.

6. LHS = sin²θ/sin θcos θ = sin θ/cos θ = tan θ= RHS, using 1 - cos²θ≡sin²θ and then tan θ≡sin θ/cos θ.

Check understanding

Practice questions

The answer and each step appear one click at a time.

Question 1 [5 marks]

Prove each of the following identities, showing every step of your working.

(a) (1 − sin²θ) tan θ ≡ sin θ cos θ [2]

(b) (cos θ + 2 sin θ)² + (2 cos θ − sin θ)² ≡ 5 [3]

Mark scheme click to reveal each step

1 − sin²θ = cos²θ and tan θ = sin θ ⁄ cos θ used

cos²θ × sin θ ⁄ cos θ = sin θ cos θ, conclusion stated

both brackets expanded: cos²θ + 4 sin θ cos θ + 4 sin²θ + 4 cos²θ − 4 sin θ cos θ + sin²θ

5 cos²θ + 5 sin²θ

= 5(sin²θ + cos²θ) = 5, conclusion stated

Question 2 [6 marks]

Figure 1 shows the graph of y = tan θ for −180° ≤ θ ≤ 180°, together with the line y = 1.5. The line meets the curve at P and at Q.

(a) Write down the equations of the two vertical asymptotes shown in Figure 1. [1]

(b) Find the values of θ at P and at Q, giving your answers to 1 decimal place. [3]

(c) Hence solve tan θ = −1.5 for −180° ≤ θ ≤ 180°, giving your answers to 1 decimal place. [2]

−180°−90°90°180°−4−3−2−11234θyOy = 1.5PQy = tan θ

Mark scheme click to reveal each step

θ = −90° and θ = 90°

tan⁻¹ 1.5 = 56.31°

P: θ = 56.3°

Q: θ = 56.31° − 180° = −123.7°

θ = −56.3°

θ = 180° − 56.31° = 123.7° ft

Keep practising

P2.6 Trigonometric identities and equations

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