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AQA A-level Mathematics (7357) · Trigonometry

Mini-Lesson

Trigonometry

5 missions. Guess, learn, then check. Your place is saved.

Mission 1Two sides and an angle can make two different triangles. How?
Mission 2How many solutions does $sin x = 0.5$ sin x = 0.5 have from $0^{\circ}$ 0° to $72 0^{\circ}$ 720°?
Mission 3Why do mathematicians measure angles in radians, not degrees?
Mission 4Why isn't $sin^{- 1} (sin 15 0^{\circ})$ sin⁻¹(sin 150°) equal to $15 0^{\circ}$ 150°?
Mission 5Add the waves $3 sin x$ 3sin x and $4 cos x$ 4cos x. How tall is the new wave?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

Two sides and an angle can make two different triangles. How?

Make a guess: Two sides and an angle can make two different triangles. How?

Trigonometry · sine and cosine rules

The sine rule

Label the triangle so that side $a$ a is opposite angle $A$ A, side $b$ b opposite $B$ B and side $c$ c opposite $C$ C.
The sine rule:

\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}

Use it when you know a side and its opposite angle, plus one more side or angle.
To find an angle, use it upside down: $\frac{sin A}{a} = \frac{sin B}{b}$ (sin A)/a = (sin B)/b.

Worked example Trigonometry · sine and cosine rules

Finding a side with the sine rule

Question

In triangle ABC, angle

A = 4 0^{\circ}

, angle

B = 7 5^{\circ}

and

a = 8

cm. Find

b

to 3 significant figures.

1
Pair each side with its opposite angle: $\frac{b}{sin 7 5 ^{\circ}} = \frac{8}{sin 4 0 ^{\circ}}$ b/(sin 75°) = 8/(sin 40°)
2
Multiply by $sin 7 5^{\circ}$ sin 75°: $b = \frac{8 sin 7 5 ^{\circ}}{sin 4 0 ^{\circ}} = 12.02 \dots$ b = (8sin 75°)/(sin 40°) = 12.02…
3
Sense check: $b$ b is opposite the larger angle, so it should be longer than $a$ a.

Answer

b = 12.0

⚠️ Watch out

\frac{8 s i n 4 0 ^{\circ}}{s i n 7 5 ^{\circ}} = 5.32

has the ratio upside down; the check in step 3 catches it.

Trigonometry · sine and cosine rules

The ambiguous case

On the unit circle, $sin θ = sin (18 0^{\circ} - θ)$ sin θ = sin (180° − θ). So $sin B = 0.8$ sin B = 0.8 gives $B = 53. 1^{\circ}$ B = 53.1° or $B = 126. 9^{\circ}$ B = 126.9°.
Your calculator only gives the acute angle. Always ask whether $18 0^{\circ}$ 180° minus it also works.
It can when you know two sides and an angle that is not between them. In the diagram, $A = 3 0^{\circ}$ A = 30°, $b = 10$ b = 10, $a = 6$ a = 6: $sin B = \frac{10 s i n 3 0 ^{\circ}}{6}$ sin B = (10sin 30°)/6, so $B = 56. 4^{\circ}$ B = 56.4° or $123. 6^{\circ}$ 123.6°.
Both are valid if the angles still total less than $18 0^{\circ}$ 180°: $3 0^{\circ} + 123. 6^{\circ} < 18 0^{\circ}$ 30° + 123.6° < 180°.

Quick check

?In triangle ABC,

A B = 9

cm,

BC = 7

cm and angle

A = 4 0^{\circ}

. Find the possible values of angle

C

Trigonometry · sine and cosine rules

The cosine rule

The cosine rule:

a^{2} = b^{2} + c^{2} - 2 b c cos A

Use it for two sides and the angle between them (to find the third side), or for three sides (to find an angle):

cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}

An obtuse angle has a negative cosine, so the cosine rule finds angles with no ambiguity.

Calculate

?A boat sails 12 km from harbour A on a bearing of 050° to a buoy B. It then sails 9 km on a bearing of 160° to a point C. Find the distance AC to 3 significant figures.

Trigonometry · sine and cosine rules

Area of a triangle

For two sides and the angle between them:

Area = \frac{1}{2} ab sin C

The angle must be the included angle, formed by the two sides you use.
Sides 5 cm and 6 cm with $5 0^{\circ}$ 50° between them: area $= \frac{1}{2} (5) (6) sin 5 0^{\circ} = 11.5$ = 1/2(5)(6)sin 50° = 11.5 cm².
With three sides, find an angle with the cosine rule first.

Calculate

?A triangle has sides 7 cm, 8 cm and 13 cm. Find its area to 3 significant figures.

cm²

Mission check

?Two sides and an angle can make two different triangles. How can that be?

Mission 2

How many solutions does $sin x = 0.5$ sin x = 0.5 have from $0^{\circ}$ 0° to $72 0^{\circ}$ 720°?

Make a guess: How many solutions does $sin x = 0.5$ sin x = 0.5 have from $0^{\circ}$ 0° to $72 0^{\circ}$ 720°?

Trigonometry · graphs and equations

Sine, cosine and tangent for any angle

Turn a radius of a unit circle through $θ$ θ anticlockwise from the positive $x$ x-axis. Its end is $P = (cos θ, sin θ)$ P = (cos θ, sin θ).
This defines sine and cosine for any angle, including negative angles and angles over $36 0^{\circ}$ 360°.
$tan θ = \frac{sin θ}{cos θ}$ tan θ = (sin θ)/(cos θ), undefined when $cos θ = 0$ cos θ = 0. Pythagoras in the triangle under $P$ P gives $sin^{2} θ + cos^{2} θ = 1$ sin²θ + cos²θ = 1.
Know these exact values:

	$0^{\circ}$ 0°	$3 0^{\circ}$ 30°	$4 5^{\circ}$ 45°	$6 0^{\circ}$ 60°	$9 0^{\circ}$ 90°	$18 0^{\circ}$ 180°
$sin$ sin	$0$ 0	$\frac{1}{2}$ 1/2	$\frac{2}{2}$ (√2)/2	$\frac{3}{2}$ (√3)/2	$1$ 1	$0$ 0
$cos$ cos	$1$ 1	$\frac{3}{2}$ (√3)/2	$\frac{2}{2}$ (√2)/2	$\frac{1}{2}$ 1/2	$0$ 0	$- 1$ −1
$tan$ tan	$0$ 0	$\frac{3}{3}$ (√3)/3	$1$ 1	$3$ √3	undefined	$0$ 0

Trigonometry · graphs and equations

Graphs, symmetry and period

$y = sin x$ y = sin x and $y = cos x$ y = cos x have period $36 0^{\circ}$ 360° and take values from $- 1$ −1 to $1$ 1. $y = tan x$ y = tan x has period $18 0^{\circ}$ 180° and asymptotes at $x = 9 0^{\circ}, 27 0^{\circ}, \dots$ x = 90°, 270°, …
Symmetry gives a second solution in each cycle: $sin (18 0^{\circ} - x) = sin x$ sin (180° − x) = sin x, $cos (36 0^{\circ} - x) = cos x$ cos (360° − x) = cos x, $tan (18 0^{\circ} + x) = tan x$ tan (180° + x) = tan x.
Then add or subtract $36 0^{\circ}$ 360° ( $18 0^{\circ}$ 180° for tan) to find every solution in the interval.

Worked example Trigonometry · graphs and equations

Solving an equation in an interval

Question

Solve

cos x = - 0.4

for

0^{\circ} \leq x < 36 0^{\circ}

. Give your answers to 1 decimal place.

1
Use the calculator for the principal value: $x = cos^{- 1} (- 0.4) = 113. 6^{\circ}$ x = cos⁻¹(−0.4) = 113.6°
2
Use the cosine symmetry: $x = 36 0^{\circ} - 113. 6^{\circ} = 246. 4^{\circ}$ x = 360° − 113.6° = 246.4°
3
Adding or subtracting $36 0^{\circ}$ 360° leaves the interval, so there are no more.

Answer

x = 113. 6^{\circ}

x = 246. 4^{\circ}

⚠️ Watch out

18 0^{\circ} - 113. 6^{\circ} = 66. 4^{\circ}

is the sine symmetry, and

cos 66. 4^{\circ}

+ 0.4

Quick check

?Solve

sin x = - 0.6

for

0^{\circ} \leq x < 36 0^{\circ}

Trigonometry · graphs and equations

Multiples of the angle

To solve $cos 3 x = k$ cos 3x = k for $0^{\circ} \leq x \leq 18 0^{\circ}$ 0° ≤ x ≤ 180°, let $u = 3 x$ u = 3x. The interval becomes $0^{\circ} \leq u \leq 54 0^{\circ}$ 0° ≤ u ≤ 540°.
Find every $u$ u in the new interval, then divide by 3.
Change the interval first: tripling the angle fits three times as many cycles.
For $sin (x + 3 0^{\circ})$ sin (x + 30°), let $u = x + 3 0^{\circ}$ u = x + 30°: $0^{\circ} \leq x < 36 0^{\circ}$ 0° ≤ x < 360° becomes $3 0^{\circ} \leq u < 39 0^{\circ}$ 30° ≤ u < 390°.
At A-level, equations may be set in radians; the method is the same.

Calculate

?Solve

cos 3 x = - 0.5

for

0^{\circ} \leq x \leq 18 0^{\circ}

. Give the largest solution.

degrees

Worked example Trigonometry · graphs and equations

A quadratic in cos x

Question

Solve

2 sin^{2} x + 3 cos x = 3

for

0^{\circ} \leq x < 36 0^{\circ}

1
Use $sin^{2} x = 1 - cos^{2} x$ sin² x = 1 − cos² x to get one function: $2 - 2 cos^{2} x + 3 cos x = 3$ 2 − 2cos² x + 3cos x = 3
2
Rearrange to a quadratic in $cos x$ cos x: $2 cos^{2} x - 3 cos x + 1 = 0$ 2cos² x − 3cos x + 1 = 0
3
Factorise: $(2 cos x - 1) (cos x - 1) = 0$ (2cos x − 1)(cos x − 1) = 0, so $cos x = \frac{1}{2}$ cos x = 1/2 or $cos x = 1$ cos x = 1
4
Solve each: $cos x = \frac{1}{2}$ cos x = 1/2 gives $6 0^{\circ}, 30 0^{\circ}$ 60°, 300°; $cos x = 1$ cos x = 1 gives $0^{\circ}$ 0°

Answer

x = 0^{\circ}, 6 0^{\circ}, 30 0^{\circ}

⚠️ Watch out

cos x = 1

gives only

0^{\circ}

here, because

36 0^{\circ}

is outside the interval. It is the solution most often missed.

Calculate

?Solve

3 cos^{2} x = sin x + 1

for

0^{\circ} \leq x < 36 0^{\circ}

. Give the largest solution.

degrees

Mission check

?How many solutions does

sin x = 0.5

have from

0^{\circ}

72 0^{\circ}

Mission 3

Why do mathematicians measure angles in radians, not degrees?

Make a guess: Why do mathematicians measure angles in radians, not degrees?

Trigonometry · radians

What is a radian? (A-level only)

1 radian is the angle at the centre of a circle when the arc length equals the radius.
The circumference is $2 π r$ 2π r, so a full turn is $2 π$ 2π radians: $2 π rad = 36 0^{\circ}$ 2π rad = 360° and $π rad = 18 0^{\circ}$ π rad = 180°.
Degrees to radians: multiply by $\frac{π}{180}$ π/180. Radians to degrees: multiply by $\frac{180}{π}$ 180/π. So 1 rad $\approx 57. 3^{\circ}$ ≈ 57.3°.
Exact values: $3 0^{\circ} = \frac{π}{6}$ 30° = π/6, $4 5^{\circ} = \frac{π}{4}$ 45° = π/4, $6 0^{\circ} = \frac{π}{3}$ 60° = π/3, $9 0^{\circ} = \frac{π}{2}$ 90° = π/2. So $sin \frac{π}{6} = \frac{1}{2}$ sin π/6 = 1/2.

Quick check

?Write

13 5^{\circ}

in radians, and give the exact value of

sin

of that angle.

Trigonometry · radians

Arc length and sector area (A-level only)

With $θ$ θ in radians:

s = r θ A = \frac{1}{2} r^{2} θ

Why: the sector is the fraction $\frac{θ}{2 π}$ θ/2π of the circle, so $s = \frac{θ}{2 π} \times 2 π r = r θ$ s = θ/2π × 2π r = rθ and $A = \frac{θ}{2 π} \times π r^{2} = \frac{1}{2} r^{2} θ$ A = θ/2π × π r² = 1/2r²θ.
In degrees the same results need $\frac{θ}{360}$ θ/360 and a $π$ π: radians cancel them.
Set your calculator to radian mode for trig of angles in radians.

Worked example Trigonometry · radians

Perimeter and area of a sector

Question

A sector of a circle has radius 8 cm and angle 1.2 radians. Find its perimeter and its area.

1
Arc length: $s = r θ = 8 \times 1.2 = 9.6$ s = rθ = 8 × 1.2 = 9.6 cm
2
Perimeter adds the two radii: $8 + 8 + 9.6 = 25.6$ 8 + 8 + 9.6 = 25.6 cm
3
Area: $A = \frac{1}{2} r^{2} θ = \frac{1}{2} \times 64 \times 1.2 = 38.4$ A = 1/2r²θ = 1/2 × 64 × 1.2 = 38.4 cm²

Answer

perimeter 25.6 cm, area 38.4 cm²

⚠️ Watch out

the perimeter of a sector includes both radii, not just the arc.

Calculate

?Chord AB cuts off a segment from a circle of radius 10 cm. Angle AOB at the centre is

\frac{2 π}{3}

radians. Find the area of the segment between the chord and the minor arc AB, to 3 significant figures.

cm²

Calculate

?A sector has area 48 cm² and arc length 12 cm. Find the radius of the circle.

Trigonometry · radians

Small-angle approximations (A-level only)

When $θ$ θ is small and in radians:

sin θ \approx θ tan θ \approx θ cos θ \approx 1 - \frac{θ ^{2}}{2}

Test with $θ = 0.1$ θ = 0.1: $sin 0.1 = 0.0998 \dots$ sin 0.1 = 0.0998…, $tan 0.1 = 0.1003 \dots$ tan 0.1 = 0.1003…, $cos 0.1 = 0.99500 \dots$ cos 0.1 = 0.99500… against $1 - \frac{0.01}{2} = 0.995$ 1 − 0.01/2 = 0.995.
They only work in radians: $sin 5^{\circ} = 0.0872$ sin 5° = 0.0872, close to $5^{\circ}$ 5° in radians ( $0.0873$ 0.0873), not to 5.
For $sin k θ$ sin kθ, replace $θ$ θ by $k θ$ kθ: $sin 3 θ \approx 3 θ$ sin 3θ ≈ 3θ, $cos 2 θ \approx 1 - 2 θ^{2}$ cos 2θ ≈ 1 − 2θ².

Worked example Trigonometry · radians

Approximating an expression (A-level only)

Question

Given that

θ

is small, find an approximate value for

\frac{1 - cos 4 θ}{θ sin 2 θ}

1
Use $cos 4 θ \approx 1 - \frac{( 4 θ ) ^{2}}{2} = 1 - 8 θ^{2}$ cos 4θ ≈ 1 − ((4θ)²)/2 = 1 − 8θ², so the numerator is $\approx 8 θ^{2}$ ≈ 8θ²
2
Use $sin 2 θ \approx 2 θ$ sin 2θ ≈ 2θ, so the denominator is $\approx θ \times 2 θ = 2 θ^{2}$ ≈ θ × 2θ = 2θ²
3
Divide: $\frac{8 θ ^{2}}{2 θ ^{2}} = 4$ 8θ²/2θ² = 4
4
Check with $θ = 0.1$ θ = 0.1: the expression is $3.97 \dots$ 3.97…

Answer

approximately 4

⚠️ Watch out

square the whole of

4 θ

\frac{( 4 θ ) ^{2}}{2} = 8 θ^{2}

, not

\frac{4 θ ^{2}}{2}

Quick check

?Given that

θ

is small, which is the best approximation to

\frac{cos θ - 1}{θ tan θ}

Mission check

?Why do mathematicians measure angles in radians instead of degrees?

Mission 4

Why isn't $sin^{- 1} (sin 15 0^{\circ})$ sin⁻¹(sin 150°) equal to $15 0^{\circ}$ 150°?

Make a guess: Why isn't $sin^{- 1} (sin 15 0^{\circ})$ sin⁻¹(sin 150°) equal to $15 0^{\circ}$ 150°?

Trigonometry · reciprocal and inverse functions

Three reciprocal functions (A-level only)

Secant, cosecant and cotangent are reciprocals:

sec θ = \frac{1}{cos θ}, cosec θ = \frac{1}{sin θ}, cot θ = \frac{1}{tan θ} = \frac{cos θ}{sin θ}

Memory aid: the third letter of sec, cosec and cot names its partner (c for cos, s for sin, t for tan).
Each is undefined where its partner is zero, e.g. $sec 9 0^{\circ}$ sec 90° does not exist.
Signs follow the partner: $cos 12 0^{\circ} < 0$ cos 120° < 0, so $sec 12 0^{\circ} < 0$ sec 120° < 0.

Worked example Trigonometry · reciprocal and inverse functions

An exact value

Question

Find the exact value of

cosec 30 0^{\circ}

1
Write it as a reciprocal: $cosec 30 0^{\circ} = \frac{1}{sin 30 0 ^{\circ}}$ cosec 300° = 1/(sin 300°)
2
Use the related acute angle $6 0^{\circ}$ 60°; $30 0^{\circ}$ 300° is in the fourth quadrant, where sine is negative: $sin 30 0^{\circ} = - \frac{3}{2}$ sin 300° = −(√3)/2
3
Take the reciprocal: $\frac{1}{- 3 /2} = - \frac{2}{3}$ 1/(−√3/2) = −2/(√3)
4
Rationalise the denominator: $- \frac{2 3}{3}$ −(2√3)/3

Answer

cosec 30 0^{\circ} = - \frac{2 3}{3}

⚠️ Watch out

cosec is the reciprocal of sin, not of cos.

\frac{1}{cos 30 0 ^{\circ}}

gives 2, which is

sec 30 0^{\circ}

Quick check

?What is the exact value of

sec 12 0^{\circ}

Trigonometry · reciprocal and inverse functions

Graphs of sec, cosec and cot

$y = sec x$ y = sec x: period $36 0^{\circ}$ 360°; vertical asymptotes where $cos x = 0$ cos x = 0 ( $x = 9 0^{\circ}, 27 0^{\circ}, \dots$ x = 90°, 270°, …); turning points where $cos x = \pm 1$ cos x = ± 1.
$y = cosec x$ y = cosec x: period $36 0^{\circ}$ 360°; asymptotes where $sin x = 0$ sin x = 0 ( $x = 0^{\circ}, 18 0^{\circ}, \dots$ x = 0°, 180°, …).
Range of sec and cosec: $y \leq - 1$ y ≤ −1 or $y \geq 1$ y ≥ 1. They never take values between $- 1$ −1 and $1$ 1.
$y = cot x$ y = cot x: period $18 0^{\circ}$ 180°; asymptotes at $x = 0^{\circ}, 18 0^{\circ}, \dots$ x = 0°, 180°, …; range all real numbers.

Trigonometry · reciprocal and inverse functions

Two new Pythagorean identities

Start from $sin^{2} θ + cos^{2} θ \equiv 1$ sin²θ + cos²θ ≡ 1.
Divide every term by $cos^{2} θ$ cos²θ:

tan^{2} θ + 1 \equiv sec^{2} θ

Divide every term by $sin^{2} θ$ sin²θ:

1 + cot^{2} θ \equiv cosec^{2} θ

Use them to turn an equation into one function, often a quadratic, or to prove other identities.
Proof example: $\frac{1}{1 - sin θ} + \frac{1}{1 + sin θ} \equiv \frac{2}{1 - sin ^{2} θ} \equiv \frac{2}{cos ^{2} θ} \equiv 2 sec^{2} θ$ 1/(1 − sin θ) + 1/(1 + sin θ) ≡ 2/(1 − sin²θ) ≡ 2/cos²θ ≡ 2sec²θ.

Worked example Trigonometry · reciprocal and inverse functions

An equation with sec and tan

Question

Solve

tan^{2} θ = 2 sec θ - 1

for

0^{\circ} \leq θ < 36 0^{\circ}

1
Replace $tan^{2} θ$ tan²θ with $sec^{2} θ - 1$ sec²θ − 1: $sec^{2} θ - 1 = 2 sec θ - 1$ sec²θ − 1 = 2sec θ − 1
2
Collect terms and factorise: $sec^{2} θ - 2 sec θ = 0$ sec²θ − 2sec θ = 0, so $sec θ (sec θ - 2) = 0$ sec θ(sec θ − 2) = 0
3
Reject $sec θ = 0$ sec θ = 0: sec never lies between $- 1$ −1 and $1$ 1. So $sec θ = 2$ sec θ = 2
4
Convert to cosine: $cos θ = \frac{1}{2}$ cos θ = 1/2, giving $θ = 6 0^{\circ}$ θ = 60° or $θ = 36 0^{\circ} - 6 0^{\circ}$ θ = 360° − 60°

Answer

θ = 6 0^{\circ}

θ = 30 0^{\circ}

⚠️ Watch out

dividing both sides by

sec θ

is safe here only because

sec θ

is never 0. With

cos θ

tan θ

it would lose solutions.

Calculate

?Given that

tan θ = \frac{5}{12}

and

18 0^{\circ} < θ < 27 0^{\circ}

, find the value of

sec θ

. Give your answer to 3 decimal places.

Trigonometry · reciprocal and inverse functions

Inverse functions need a restricted domain

Many angles share one sine, so $sin x$ sin x is many-to-one. To have an inverse, restrict it to a one-to-one piece.
$arcsin x$ arcsin x (also written $sin^{- 1} x$ sin⁻¹ x): domain $- 1 \leq x \leq 1$ −1 ≤ x ≤ 1, range $- 9 0^{\circ} \leq y \leq 9 0^{\circ}$ −90° ≤ y ≤ 90°, i.e. $- \frac{π}{2} \leq y \leq \frac{π}{2}$ −π/2 ≤ y ≤ π/2.
$arccos x$ arccos x: domain $- 1 \leq x \leq 1$ −1 ≤ x ≤ 1, range $0 \leq y \leq π$ 0 ≤ y ≤ π.
$arctan x$ arctan x: domain all real $x$ x, range $- \frac{π}{2} < y < \frac{π}{2}$ −π/2 < y < π/2.
The output is the principal value. Each graph is the reflection of its restricted piece in $y = x$ y = x.

Calculate

?Find

arccos (- 0.5)

in radians. Give your answer to 3 significant figures.

Mission check

?Why isn't

sin^{- 1} (sin 15 0^{\circ})

equal to

15 0^{\circ}

Mission 5

Add the waves $3 sin x$ 3sin x and $4 cos x$ 4cos x. How tall is the new wave?

Make a guess: Add the waves $3 sin x$ 3sin x and $4 cos x$ 4cos x. How tall is the new wave?

Trigonometry · compound angles and

R cos (θ \pm α)

The compound angle formulae (A-level only)

For any angles $A$ A and $B$ B:

$sin (A \pm B) \equiv sin A cos B \pm cos A sin B$ sin (A ± B) ≡ sin Acos B ± cos Asin B $cos (A \pm B) \equiv cos A cos B \mp sin A sin B$ cos (A ± B) ≡ cos Acos B ∓ sin Asin B $tan (A \pm B) \equiv \frac{tan A \pm tan B}{1 \mp tan A tan B}$ tan (A ± B) ≡ (tan A ± tan B)/(1 ∓ tan Atan B)

In the cos formula the sign flips: $cos (A + B)$ cos (A + B) has a minus.
$sin (A + B)$ sin (A + B) is not $sin A + sin B$ sin A + sin B.
You need to understand the geometrical proof: stack two right-angled triangles with angles $A$ A and $B$ B and read off lengths.

Worked example Trigonometry · compound angles and

R cos (θ \pm α)

An exact value from known angles

Question

Find the exact value of

sin 7 5^{\circ}

1
Split into angles with known exact values: $sin 7 5^{\circ} = sin (4 5^{\circ} + 3 0^{\circ})$ sin 75° = sin (45° + 30°)
2
Expand: $sin 4 5^{\circ} cos 3 0^{\circ} + cos 4 5^{\circ} sin 3 0^{\circ}$ sin 45°cos 30° + cos 45°sin 30°
3
Substitute exact values: $\frac{2}{2} \times \frac{3}{2} + \frac{2}{2} \times \frac{1}{2} = \frac{6}{4} + \frac{2}{4}$ (√2)/2 × (√3)/2 + (√2)/2 × 1/2 = (√6)/4 + (√2)/4

Answer

sin 7 5^{\circ} = \frac{6 + 2}{4}

⚠️ Watch out

sin 4 5^{\circ} + sin 3 0^{\circ} \approx 1.21

is bigger than 1, so it cannot be a sine.

Quick check

?What is the exact value of

cos 1 5^{\circ}

? Use

1 5^{\circ} = 4 5^{\circ} - 3 0^{\circ}

Trigonometry · compound angles and

R cos (θ \pm α)

Double angle formulae

Put $B = A$ B = A in the compound angle formulae:

$sin 2 A \equiv 2 sin A cos A tan 2 A \equiv \frac{2 tan A}{1 - tan ^{2} A}$ sin 2A ≡ 2sin Acos A tan 2A ≡ (2tan A)/(1 − tan² A) $cos 2 A \equiv cos^{2} A - sin^{2} A \equiv 2 cos^{2} A - 1 \equiv 1 - 2 sin^{2} A$ cos 2A ≡ cos² A − sin² A ≡ 2cos² A − 1 ≡ 1 − 2sin² A

Choose the cos form that leaves one function, e.g. $1 - 2 sin^{2} A$ 1 − 2sin² A next to a $sin A$ sin A term.
Halve the angles too: $cos θ \equiv 1 - 2 sin^{2} \frac{θ}{2}$ cos θ ≡ 1 − 2sin²θ/2.
Proof example: $\frac{sin 2 A}{1 + cos 2 A} \equiv \frac{2 sin A cos A}{2 cos ^{2} A} \equiv tan A$ (sin 2A)/(1 + cos 2A) ≡ (2sin Acos A)/(2cos² A) ≡ tan A.

Worked example Trigonometry · compound angles and

R cos (θ \pm α)

An equation with a double angle

Question

Solve

cos 2 x = sin x

for

0^{\circ} \leq x < 36 0^{\circ}

1
Use the form of $cos 2 x$ cos 2x that involves only sine: $1 - 2 sin^{2} x = sin x$ 1 − 2sin² x = sin x
2
Rearrange into a quadratic in $sin x$ sin x: $2 sin^{2} x + sin x - 1 = 0$ 2sin² x + sin x − 1 = 0
3
Factorise: $(2 sin x - 1) (sin x + 1) = 0$ (2sin x − 1)(sin x + 1) = 0, so $sin x = \frac{1}{2}$ sin x = 1/2 or $sin x = - 1$ sin x = −1
4
Solve each in the interval: $sin x = \frac{1}{2}$ sin x = 1/2 gives $3 0^{\circ}, 15 0^{\circ}$ 30°, 150°; $sin x = - 1$ sin x = −1 gives $27 0^{\circ}$ 270°

Answer

x = 3 0^{\circ}, 15 0^{\circ}, 27 0^{\circ}

⚠️ Watch out

using

cos 2 x = 2 cos^{2} x - 1

here leaves a mix of sin and cos that will not factorise.

Calculate

?Angles

A

and

B

are acute, with

sin A = \frac{3}{5}

and

cos B = \frac{5}{13}

. Find the value of

sin (A + B)

. Give your answer to 3 decimal places.

Trigonometry · compound angles and

R cos (θ \pm α)

Harmonic form $R cos (θ - α)$ Rcos (θ − α)

Any $a cos θ + b sin θ$ acos θ + bsin θ is a single wave. Write it as $R cos (θ - α)$ Rcos (θ − α) with $R > 0$ R > 0.
Expand: $R cos (θ - α) = R cos α cos θ + R sin α sin θ$ Rcos (θ − α) = Rcos αcos θ + Rsin αsin θ.
Compare coefficients: $R cos α = a$ Rcos α = a and $R sin α = b$ Rsin α = b. Then

R = a^{2} + b^{2}, tan α = \frac{b}{a}

The wave has maximum $R$ R where $θ = α$ θ = α and minimum $- R$ −R where $θ = α + 18 0^{\circ}$ θ = α + 180°. The forms $R cos (θ + α)$ Rcos (θ + α) and $R sin (θ \pm α)$ Rsin (θ ± α) work the same way.

Worked example Trigonometry · compound angles and

R cos (θ \pm α)

Harmonic form, then solve

Question

Write

cos θ + 3 sin θ

in the form

R cos (θ - α)

, where

R > 0

and

0^{\circ} < α < 9 0^{\circ}

. Hence solve

cos θ + 3 sin θ = 1

for

0^{\circ} \leq θ < 36 0^{\circ}

1
Compare coefficients: $R cos α = 1$ Rcos α = 1, $R sin α = 3$ Rsin α = √3
2
Find $R$ R and $α$ α: $R = 1 + 3 = 2$ R = √(1 + 3) = 2, $tan α = 3$ tan α = √3, so $α = 6 0^{\circ}$ α = 60°
3
Solve $2 cos (θ - 6 0^{\circ}) = 1$ 2cos (θ − 60°) = 1: $cos (θ - 6 0^{\circ}) = \frac{1}{2}$ cos (θ − 60°) = 1/2, with $- 6 0^{\circ} \leq θ - 6 0^{\circ} < 30 0^{\circ}$ −60° ≤ θ − 60° < 300°
4
In that interval $θ - 6 0^{\circ} = - 6 0^{\circ}$ θ − 60° = −60° or $6 0^{\circ}$ 60°

Answer

cos θ + 3 sin θ = 2 cos (θ - 6 0^{\circ})

;

θ = 0^{\circ}

12 0^{\circ}

⚠️ Watch out

shift the interval as well as the angle. Starting from

θ - 6 0^{\circ} = 6 0^{\circ}

30 0^{\circ}

gives

θ = 36 0^{\circ}

, which is outside the interval, and misses

θ = 0^{\circ}

Calculate

?The depth of water in a harbour,

D

metres,

t

hours after midnight is modelled by

D = 6 + 1.5 sin (30 t)^{\circ} + 2 cos (30 t)^{\circ}

. Find the greatest depth predicted by the model.

Mission check

?Add the waves

3 sin x

and

4 cos x

. How tall is the new wave?

Recap

Big ideas

Sine rule $\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C}$ a/(sin A) = b/(sin B) = c/(sin C) for a side–opposite-angle pair; check for a second, obtuse angle.
Cosine rule $a^{2} = b^{2} + c^{2} - 2 b c cos A$ a² = b² + c² − 2bccos A for two sides and the included angle, or three sides.
Area $= \frac{1}{2} ab sin C$ = 1/2absin C, with $C$ C the angle between $a$ a and $b$ b.
Sine and cosine come from the unit circle, so they exist for any angle; $tan θ = \frac{s i n θ}{c o s θ}$ tan θ = (sin θ)/(cos θ) and $sin^{2} θ + cos^{2} θ = 1$ sin²θ + cos²θ = 1.
Principal value, then symmetry, then add or subtract the period until you leave the interval.
For $sin (n x)$ sin (nx) or $cos (x + a)$ cos (x + a), change the interval first; for quadratics, use $sin^{2} θ + cos^{2} θ = 1$ sin²θ + cos²θ = 1 to get one function and factorise.
1 radian: arc = radius. $π$ π rad $= 18 0^{\circ}$ = 180°; multiply by $\frac{π}{180}$ π/180 to convert to radians.
With $θ$ θ in radians: arc $s = r θ$ s = rθ, sector $A = \frac{1}{2} r^{2} θ$ A = 1/2r²θ, segment $= \frac{1}{2} r^{2} (θ - sin θ)$ = 1/2r²(θ − sin θ).
For small $θ$ θ in radians: $sin θ \approx θ$ sin θ ≈ θ, $tan θ \approx θ$ tan θ ≈ θ, $cos θ \approx 1 - \frac{θ ^{2}}{2}$ cos θ ≈ 1 − θ²/2.
$sec θ = \frac{1}{c o s θ}$ sec θ = 1/(cos θ), $cosec θ = \frac{1}{s i n θ}$ cosec θ = 1/(sin θ), $cot θ = \frac{1}{t a n θ}$ cot θ = 1/(tan θ); sec and cosec never lie strictly between $- 1$ −1 and $1$ 1.
$1 + tan^{2} θ \equiv sec^{2} θ$ 1 + tan²θ ≡ sec²θ and $1 + cot^{2} θ \equiv cosec^{2} θ$ 1 + cot²θ ≡ cosec²θ turn equations into a single function.
arcsin, arccos and arctan return principal values only: $[- \frac{π}{2}, \frac{π}{2}]$ [−π/2, π/2], $[0, π]$ [0, π] and $(- \frac{π}{2}, \frac{π}{2})$ (−π/2, π/2).
$sin (A \pm B) \equiv sin A cos B \pm cos A sin B$ sin (A ± B) ≡ sin Acos B ± cos Asin B; $cos (A \pm B) \equiv cos A cos B \mp sin A sin B$ cos (A ± B) ≡ cos Acos B ∓ sin Asin B; put $B = A$ B = A for the double angle formulae.
Choose the version of $cos 2 A$ cos 2A that leaves one trig function, then solve as a quadratic.
$a cos θ + b sin θ = R cos (θ - α)$ acos θ + bsin θ = Rcos (θ − α) with $R = a^{2} + b^{2}$ R = √(a² + b²), $tan α = \frac{b}{a}$ tan α = b/a: one wave with maximum $R$ R and minimum $- R$ −R.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Trigonometry for AQA A-level Mathematics (7357). 🎉

Mission 1Two sides and an angle can make two different triangles. How?
Mission 2How many solutions does $sin x = 0.5$ sin x = 0.5 have from $0^{\circ}$ 0° to $72 0^{\circ}$ 720°?
Mission 3Why do mathematicians measure angles in radians, not degrees?
Mission 4Why isn't $sin^{- 1} (sin 15 0^{\circ})$ sin⁻¹(sin 150°) equal to $15 0^{\circ}$ 150°?
Mission 5Add the waves $3 sin x$ 3sin x and $4 cos x$ 4cos x. How tall is the new wave?

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Specification points and sources for this lesson are listed at the bottom of the page.

Sources (5)

The specification points and sources behind this lesson.

1 AQA 7357 A-level Mathematics specification, E1, E9
2 AQA 7357 A-level Mathematics specification, E1, E3, E5, E7
3 AQA 7357 A-level Mathematics specification, E1, E2, E3
4 AQA A-level Mathematics 7357 specification v1.3, E4, E5, E7, E8
5 AQA A-level Mathematics 7357 specification v1.3, E6, E7, E8, E9

Trigonometry

Two sides and an angle can make two different triangles. How?

The sine rule

Finding a side with the sine rule

The ambiguous case

The cosine rule

Area of a triangle

How many solutions does sinx=0.5sin x = 0.5 have from 0∘0° to 720∘720°?

Sine, cosine and tangent for any angle

Graphs, symmetry and period

Solving an equation in an interval

Multiples of the angle

A quadratic in cos x

Why do mathematicians measure angles in radians, not degrees?

What is a radian? (A-level only)

Arc length and sector area (A-level only)

Perimeter and area of a sector

Small-angle approximations (A-level only)

Approximating an expression (A-level only)

Why isn't sin−1(sin150∘)sin⁻¹(sin 150°) equal to 150∘150°?

Three reciprocal functions (A-level only)

An exact value

Graphs of sec, cosec and cot

Two new Pythagorean identities

An equation with sec and tan

Inverse functions need a restricted domain

Add the waves 3sinx3sin x and 4cosx4cos x. How tall is the new wave?

The compound angle formulae (A-level only)

An exact value from known angles

Double angle formulae

An equation with a double angle

Harmonic form Rcos(θ−α)Rcos (θ − α)

Harmonic form, then solve

Big ideas

Mini-lesson complete!

📣 Smashed it? Share your score

How many solutions does $sin x = 0.5$ sin x = 0.5 have from $0^{\circ}$ 0° to $72 0^{\circ}$ 720°?

Why isn't $sin^{- 1} (sin 15 0^{\circ})$ sin⁻¹(sin 150°) equal to $15 0^{\circ}$ 150°?

Add the waves $3 sin x$ 3sin x and $4 cos x$ 4cos x. How tall is the new wave?

Harmonic form $R cos (θ - α)$ Rcos (θ − α)