Trigonometry: Multiple Angles and Maximum-Minimum Values

Competitive Mathematics · SSC CGL

5 Part12 worked examples25 practice questions

In the earlier chapters you learnt formulas like $sin (A + B)$ . This chapter uses them to do four jobs: (1) double, triple or halve an angle; (2) collapse a product of sines or cosines into one small number, such as $sin 6^{\circ} sin 5 4^{\circ} sin 6 6^{\circ}$ ; (3) get values from a condition like $A + B = 4 5^{\circ}$ or $A + B + C = 18 0^{\circ}$ ; (4) find the largest and smallest value of an expression, such as that of $5 sin t + 12 cos t$ . No formula here is meant to be memorised blindly — each one is built in front of you, and a formula you have built is a formula you do not forget.

∑ Formula

sin (A + B) cos (A + B) tan (A + B) = sin A cos B + cos A sin B = cos A cos B - sin A sin B = \frac{tan A + tan B}{1 - tan A tan B}

Everything below is built from these three (proved in chapter 2)

$A, B$: any two angles (in degrees)
$sin A, cos A$: sine and cosine of angle A

The question looks like	Which method	Where
$sin t$ , $cos t$ or $tan t$ is given, $2 t$ or $3 t$ is asked	Double / triple angle	Methods 1, 2
A half angle such as $sin 22. 5^{\circ}$ , $tan 1 5^{\circ}$	Half angle	Method 3
Values at $1 8^{\circ}$ , $3 6^{\circ}$ , $5 4^{\circ}$ , $7 2^{\circ}$	2θ = 90° − 3θ	Method 4
Product of the angles $t$ , $60 - t$ , $60 + t$	The triple product	Method 5
$cos x cos 2 x cos 4 x \dots$ or $\frac{s i n 2 x}{s i n ( x /4 )}$	Repeated $2 sin cos$	Method 6
$A + B = 4 5^{\circ}$ or $A + B + C = 18 0^{\circ}$ is given	Angle-sum condition	Method 7
$1 + sin x + sin^{2} x + \dots =$ a number	Infinite GP	Method 8
The “largest / smallest value” or the “range” is asked	The seven max-min methods	Methods 9 to 15

Read the chapter in two halves: methods 1 to 8 are about changing angles and the product/series tricks; methods 9 to 15 are about maximum and minimum values and use the $sin 2 t$ , $cos 2 t$ forms of this chapter.

Get readyAll angles are in degrees. Keep the values at 0°, 30°, 45°, 60°, 90° ready (chapter 1).

Double, triple and half angles

2t · 3t · t/2 · 18°

Finding the value at twice, three times and half an angle from the value at the angle itself.

If $sin t$ is known, what is $sin 2 t$ ? It is not $2 \times sin t$ — that is the most common mistake. The true relation is in the four methods below, and we build every one from a single idea: put $B = A$ in $A + B$ .

Method 1: the double angle $2 t$

Put

B = A

The formulas for $sin 2 t$ , $cos 2 t$ , $tan 2 t$ , where they come from and which form to choose.

The idea

Put $B = A$ in the formula for $sin (A + B)$ : $sin (A + A) = sin A cos A + cos A sin A = 2 sin A cos A$ . That is all — the double-angle formula is born. The same trick works on $cos (A + A)$ and $tan (A + A)$ . Mind this: $sin 2 t$ means “the sine of $2 t$ ”, not $2 \times sin t$ ; at $t = 3 0^{\circ}$ , $sin 6 0^{\circ} = 0.866$ but $2 sin 3 0^{\circ} = 1$ .

◭ Diagram

An isosceles triangle (

A B = A C = 1

, apex angle

2 t

): the height

A M

splits the apex into two angles

t

. Green = the area from the angle

2 t

, saffron = the area from base and height; the two are equal.

∑ Formula

sin 2 t = 2 sin t cos t

Sine of the double angle

$t$: any angle
$2 t$: twice that angle

∑ Formula

cos 2 t = cos^{2} t - sin^{2} t = 2 cos^{2} t - 1 = 1 - 2 sin^{2} t

Three forms of the cosine

$cos^{2} t$: $(cos t)^{2}$
$sin^{2} t$: $(sin t)^{2}$

∑ Formula

tan 2 t = \frac{2 tan t}{1 - tan ^{2} t}

Tangent of the double angle

$tan t$: tangent of $t$ ( $t \neq = 4 5^{\circ}$ )

Why three forms of $cos 2 t$ ? The first comes straight from $cos (t + t)$ . Replace $sin^{2} t = 1 - cos^{2} t$ and you get $2 cos^{2} t - 1$ ; replace $cos^{2} t = 1 - sin^{2} t$ and you get $1 - 2 sin^{2} t$ . All three are equal; choose the one that contains only what you already know.

What you know	Form of $cos 2 t$
only $sin t$	$1 - 2 sin^{2} t$
only $cos t$	$2 cos^{2} t - 1$
both $sin t$ and $cos t$	$cos^{2} t - sin^{2} t$
only $tan t$	$\frac{1 - tan ^{2} t}{1 + tan ^{2} t}$

∑ Formula

sin 2 t cos 2 t = \frac{2 tan t}{1 + tan ^{2} t} = \frac{1 - tan ^{2} t}{1 + tan ^{2} t}

when only

tan t

is known; made by dividing

\frac{2 s i n t c o s t}{s i n ^{2} t + c o s ^{2} t}

cos^{2} t

$tan t$: tangent of the angle $t$

◭ Diagram

The blue

sin t

makes one hump and one dip per turn; the green

sin 2 t

keeps the same height but makes two of each. The dots move together: by the time

t

has gone half way round,

2 t

has gone all the way.

∑ Formula

1 + cos 2 t 1 - cos 2 t \frac{1 - cos 2 t}{sin 2 t} = 2 cos^{2} t = 2 sin^{2} t = tan t

Read the other way: they remove squares

$2 cos^{2} t$: $cos^{2} t$ can be replaced by $\frac{1 + c o s 2 t}{2}$

Spot itWhen you see

1 + cos (something)

1 - cos (something)

, think at once:

2 cos^{2}

2 sin^{2}

of the half angle.

Steps

See what is given ( $sin t$ , $cos t$ or $tan t$ ) and what is asked (something at $2 t$ ).
Get the missing ratio from a right triangle (chapter 1) and fix its sign from the quadrant.
Pick the formula that contains only known things.
Calculate; at the end check the sign against the quadrant of $2 t$ .

Worked examples

◭ Diagram

sin t = \frac{5}{13}

: perpendicular 5, hypotenuse 13, base 12 (Pythagoras). Green = base, rose = perpendicular, violet = hypotenuse.

Example 1 · Basic

sin t = \frac{5}{13}

and

t

is acute. Find

sin 2 t

cos 2 t

and

tan 2 t

Solution, step by step

1 $sin t$ is given, but $sin 2 t = 2 sin t cos t$ also needs $cos t$ . So draw the right triangle of the figure: perpendicular 5, hypotenuse 13.
2Base $= 1 3^{2} - 5^{2} = 144 = 12$ , and $t$ is acute, so $cos t = \frac{12}{13}$ (positive).
3 $sin 2 t = 2 \cdot \frac{5}{13} \cdot \frac{12}{13} = \frac{120}{169}$ .
4From $sin t$ alone, $cos 2 t = 1 - 2 sin^{2} t = 1 - \frac{50}{169} = \frac{119}{169}$ . Check: $cos^{2} t - sin^{2} t = \frac{144 - 25}{169} = \frac{119}{169}$ ✓.
5 $tan 2 t = \frac{s i n 2 t}{c o s 2 t} = \frac{120}{119}$ .

Answer $sin 2 t = \frac{120}{169}, cos 2 t = \frac{119}{169}, tan 2 t = \frac{120}{119}$

The new triangle

120, 119, 169

is also a Pythagorean triple (

12 0^{2} + 11 9^{2} = 16 9^{2}

) — a ready-made check on your answer.

◭ Diagram

tan t = 2

: base 1, perpendicular 2, hypotenuse

5

Example 2 · Standard

tan t = 2

. Find

sin 2 t

and

cos 2 t

, and say in which quadrant

2 t

lies.

Solution, step by step

1Only $tan t$ is known, so use the tangent forms — they need no square roots at all.
2 $sin 2 t = \frac{2 t a n t}{1 + t a n ^{2} t} = \frac{2 \cdot 2}{1 + 4} = \frac{4}{5}$ .
3 $cos 2 t = \frac{1 - t a n ^{2} t}{1 + t a n ^{2} t} = \frac{1 - 4}{1 + 4} = - \frac{3}{5}$ .
4 $sin 2 t$ is positive and $cos 2 t$ negative, so $2 t$ lies in the second quadrant. It agrees: $tan t = 2$ means $t \approx 63. 4^{\circ}$ , so $2 t \approx 126. 9^{\circ}$ .

Answer $sin 2 t = \frac{4}{5}, cos 2 t = - \frac{3}{5}$

Twice an acute angle is not always acute; when

t > 4 5^{\circ}

cos 2 t

comes out negative.

Example 3 · Advanced

Simplify

\frac{sin 2 x}{1 + cos 2 x}

and then find its exact value at

x = 1 5^{\circ}

Solution, step by step

1Write the numerator $sin 2 x$ as $2 sin x cos x$ .
2The denominator $1 + cos 2 x$ is the “ $1 + cos$ ” pattern, so $1 + cos 2 x = 2 cos^{2} x$ .
3Divide: $\frac{2 s i n x c o s x}{2 c o s ^{2} x} = \frac{s i n x}{c o s x} = tan x$ . So the whole expression is $tan x$ .
4At $x = 1 5^{\circ}$ the value is $tan 1 5^{\circ} = tan (4 5^{\circ} - 3 0^{\circ}) = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{3 - 1}{3 + 1}$ .
5Rationalise: $\frac{( 3 - 1 ) ^{2}}{3 - 1} = \frac{4 - 2 3}{2} = 2 - 3$ .

Answer $tan x$ ; value $2 - 3$

You could also read it directly:

\frac{s i n 2 x}{1 + c o s 2 x} = tan x

means

\frac{s i n 3 0 ^{\circ}}{1 + c o s 3 0 ^{\circ}}

x = 1 5^{\circ}

Practice (solutions hidden)

BasicTry it yourself first

sin t = \frac{8}{17}

and

t

is acute. Find

sin 2 t

BasicTry it yourself first

cos t = \frac{7}{25}

. Find

cos 2 t

StandardTry it yourself first

tan t = \frac{1}{3}

. Find

cos 2 t

StandardTry it yourself first

Find

2 sin 7 5^{\circ} cos 7 5^{\circ}

StandardTry it yourself first

Find

1 - 2 sin^{2} 1 5^{\circ}

AdvancedTry it yourself first

sin t + cos t = \frac{7}{5}

. Find

sin 2 t

AdvancedTry it yourself first

cos t - sin t = \frac{1}{2}

. Find

sin 2 t

Method 2: the triple angle $3 t$

3 t = 2 t + t

The formulas for $sin 3 t$ , $cos 3 t$ , $tan 3 t$ , where they come from and how to use them backwards.

The idea

Break the triple angle as $2 t + t$ and use the double-angle formulas of the last method:

$sin 3 t = sin (2 t + t) = sin 2 t cos t + cos 2 t sin t$ .
Put $sin 2 t = 2 sin t cos t$ and $cos 2 t = 1 - 2 sin^{2} t$ : $= 2 sin t cos^{2} t + (1 - 2 sin^{2} t) sin t$ .
Put $cos^{2} t = 1 - sin^{2} t$ : $= 2 sin t - 2 sin^{3} t + sin t - 2 sin^{3} t = 3 sin t - 4 sin^{3} t$ .
$cos 3 t$ the same way: $cos 2 t cos t - sin 2 t sin t = (2 cos^{2} t - 1) cos t - 2 sin^{2} t cos t = 4 cos^{3} t - 3 cos t$ .

∑ Formula

sin 3 t cos 3 t tan 3 t = 3 sin t - 4 sin^{3} t = 4 cos^{3} t - 3 cos t = \frac{3 tan t - tan ^{3} t}{1 - 3 tan ^{2} t}

The three triple-angle formulas

$sin^{3} t$: $(sin t)^{3}$
$tan 3 t$: tangent of the triple angle ( $1 - 3 tan^{2} t \neq = 0$ )

◭ Diagram

The blue

sin t

and the green

sin 3 t

. In one turn the green has three humps, yet it never goes beyond

\pm 1

— which is why

3 sin t - 4 sin^{3} t

can never pass

\pm 1

RememberTo remember the signs: in

sin 3 t

the first term is positive (

3 sin t

), in

cos 3 t

it is negative (

- 3 cos t

). Check: at

t = 3 0^{\circ}

3 \cdot \frac{1}{2} - 4 \cdot \frac{1}{8} = 1 = sin 9 0^{\circ}

✓.

∑ Formula

sin^{3} t cos^{3} t = \frac{3 sin t - sin 3 t}{4} = \frac{3 cos t + cos 3 t}{4}

Backwards form: removes cubes

$sin 3 t$: sine of the triple angle

When to use it

$sin t$ (or $cos t$ ) is given and $sin 3 t$ is asked — without finding the angle.
You see $3 sin □ - 4 sin^{3} □$ or $4 cos^{3} □ - 3 cos □$ : it is $sin 3 □$ or $cos 3 □$ .
An algebra condition like $x + \frac{1}{x} = 2 cos t$ : then $x^{3} + \frac{1}{x ^{3}} = 2 cos 3 t$ .

Worked examples

Example 4 · Basic

sin t = \frac{3}{5}

. Find

sin 3 t

Solution, step by step

1Only $sin t$ is given, so $sin 3 t = 3 sin t - 4 sin^{3} t$ is the direct route; no $cos t$ needed.
2 $sin^{3} t = \frac{27}{125}$ , so $3 \cdot \frac{3}{5} - 4 \cdot \frac{27}{125} = \frac{9}{5} - \frac{108}{125}$ .
3Common denominator 125: $\frac{225 - 108}{125} = \frac{117}{125}$ .

Answer $\frac{117}{125}$

Check:

t \approx 36. 9^{\circ}

3 t \approx 110. 6^{\circ}

and

sin 110. 6^{\circ} \approx 0.936 = \frac{117}{125}

✓.

Example 5 · Standard

Evaluate: (a)

3 sin 1 0^{\circ} - 4 sin^{3} 1 0^{\circ}

, (b)

4 cos^{3} 4 0^{\circ} - 3 cos 4 0^{\circ}

Solution, step by step

1(a) The shape is $3 sin t - 4 sin^{3} t$ , so it is $sin 3 t$ with $t = 1 0^{\circ}$ : $sin 3 0^{\circ} = \frac{1}{2}$ .
2(b) The shape is $4 cos^{3} t - 3 cos t$ , so it is $cos 3 t$ with $t = 4 0^{\circ}$ : $cos 12 0^{\circ}$ .
3 $cos 12 0^{\circ} = cos (18 0^{\circ} - 6 0^{\circ}) = - cos 6 0^{\circ} = - \frac{1}{2}$ (cosine is negative in the second quadrant).

Answer (a) $\frac{1}{2}$ , (b) $- \frac{1}{2}$

Recognising the form is half the work; you never need the value of

sin 1 0^{\circ}

Example 6 · Advanced

x + \frac{1}{x} = 2 cos 2 0^{\circ}

, find

x^{3} + \frac{1}{x ^{3}}

Solution, step by step

1Algebra for the cube: $(x + \frac{1}{x})^{3} = x^{3} + \frac{1}{x ^{3}} + 3 (x + \frac{1}{x})$ . So $x^{3} + \frac{1}{x ^{3}} = (x + \frac{1}{x})^{3} - 3 (x + \frac{1}{x})$ .
2Write $x + \frac{1}{x} = 2 c$ with $c = cos 2 0^{\circ}$ : $x^{3} + \frac{1}{x ^{3}} = 8 c^{3} - 6 c = 2 (4 c^{3} - 3 c)$ .
3The bracket $4 cos^{3} 2 0^{\circ} - 3 cos 2 0^{\circ}$ equals $cos 6 0^{\circ} = \frac{1}{2}$ , so the value is $2 \cdot \frac{1}{2} = 1$ .

Answer $1$

General rule:

x + \frac{1}{x} = 2 cos t \Rightarrow x^{n} + \frac{1}{x ^{n}} = 2 cos n t

(here

2 cos 6 0^{\circ} = 1

Practice (solutions hidden)

BasicTry it yourself first

sin t = \frac{2}{5}

. Find

sin 3 t

BasicTry it yourself first

cos t = \frac{1}{3}

. Find

cos 3 t

StandardTry it yourself first

Find

3 sin 2 5^{\circ} - 4 sin^{3} 2 5^{\circ}

StandardTry it yourself first

tan t = \frac{1}{2}

. Find

tan 3 t

AdvancedTry it yourself first

x + \frac{1}{x} = 1

, find

x^{3} + \frac{1}{x ^{3}}

AdvancedTry it yourself first

sin t = \frac{1}{2}

, find

\frac{sin 3 t}{sin t}

. (Hint:

sin 3 t = sin t (3 - 4 sin^{2} t)

)

Method 3: the half angle $\frac{t}{2}$

cos t = 1 - 2 sin^{2} \frac{t}{2}

When $cos t$ is known and the half angle is wanted, or when $22. 5^{\circ}$ , $1 5^{\circ}$ , $7 5^{\circ}$ turn up.

The idea

Put $\frac{t}{2}$ in place of $t$ in the double-angle formulas and they become half-angle formulas: $cos t = 1 - 2 sin^{2} \frac{t}{2}$ and $cos t = 2 cos^{2} \frac{t}{2} - 1$ . Solve them for $sin \frac{t}{2}$ and $cos \frac{t}{2}$ — a square root appears, so a $\pm$ comes with it.

∑ Formula

sin \frac{t}{2} cos \frac{t}{2} = \pm \frac{1 - cos t}{2} = \pm \frac{1 + cos t}{2}

The sign comes from the quadrant of the half angle $\frac{t}{2}$ , not of

t

$t$: the given angle
$\frac{t}{2}$: half of it
$\pm$: a sign chosen from the quadrant

∑ Formula

tan \frac{t}{2} = \frac{sin t}{1 + cos t} = \frac{1 - cos t}{sin t}

The tangent needs no

\pm

— this is the most convenient form

$sin t, cos t$: ratios of the given angle $t$

◭ Diagram

Right triangle

A BC

(angle

t

B

, hypotenuse

A B = 1

). Stretch

CB

beyond

B

so that

B D = 1

. Then

A B D

is isosceles and its base angles are

\frac{t}{2}

, so

tan \frac{t}{2} = \frac{A C}{C D}

Angle	$t$ and its half	Value
$22. 5^{\circ}$	$t = 4 5^{\circ}$	$tan 22. 5^{\circ} = \frac{1 - \frac{2}{2}}{\frac{2}{2}} = 2 - 1$
$1 5^{\circ}$	$t = 3 0^{\circ}$	$tan 1 5^{\circ} = \frac{1 - \frac{3}{2}}{\frac{1}{2}} = 2 - 3$
$7 5^{\circ}$	$t = 15 0^{\circ}$	$tan 7 5^{\circ} = \frac{\frac{1}{2}}{1 - \frac{3}{2}} = 2 + 3$

Here: $t$ is the angle whose sine and cosine we know ( $4 5^{\circ}$ , $3 0^{\circ}$ , $15 0^{\circ}$ ); $tan \frac{t}{2}$ is the angle that was asked. (Heights and distances use $tan 1 5^{\circ} = 2 - 3$ and $tan 7 5^{\circ} = 2 + 3$ again and again.)

∑ Formula

sin t tan t = \frac{2 u}{1 + u ^{2}}, cos t = \frac{1 - u ^{2}}{1 + u ^{2}} = \frac{2 u}{1 - u ^{2}}, u = tan \frac{t}{2}

The reverse road: all ratios of the full angle from the tangent of the half angle

$u$: $tan \frac{t}{2}$ , the tangent of the half angle

CarefulThe sign trap:

t = 22 5^{\circ}

is in the third quadrant but

\frac{t}{2} = 112. 5^{\circ}

is in the second, where

sin

is positive and

cos

negative. Always write down the quadrant of

\frac{t}{2}

first, then choose the

\pm

Steps

Decide which angle the asked one is half of (that is $t$ ) and get $cos t$ (or $sin t$ ).
If the tangent is asked use $\frac{s i n t}{1 + c o s t}$ — no sign question at all.
For $sin$ or $cos$ use the square-root form and choose the $\pm$ from the quadrant of $\frac{t}{2}$ .

Worked examples

◭ Diagram

The triangle of the half angle

\frac{t}{2}

in this example: perpendicular 3, base 4, hypotenuse 5. The full angle

t

is its double, and

cos t = \frac{16 - 9}{25} = \frac{7}{25}

Example 7 · Basic

cos t = \frac{7}{25}

and

t

is acute. Find

sin \frac{t}{2}

cos \frac{t}{2}

and

tan \frac{t}{2}

Solution, step by step

1 $t$ is acute, so $\frac{t}{2}$ is acute too; all three values are positive and we take the $+$ in the $\pm$ .
2 $sin \frac{t}{2} = \frac{1 - \frac{7}{25}}{2} = \frac{18/25}{2} = \frac{9}{25} = \frac{3}{5}$ .
3 $cos \frac{t}{2} = \frac{1 + \frac{7}{25}}{2} = \frac{32/25}{2} = \frac{16}{25} = \frac{4}{5}$ .
4 $tan \frac{t}{2} = \frac{3/5}{4/5} = \frac{3}{4}$ . Check: $(\frac{3}{5})^{2} + (\frac{4}{5})^{2} = 1$ ✓.

Answer $\frac{3}{5}, \frac{4}{5}, \frac{3}{4}$

Here

\frac{t}{2}

is the 3-4-5 triangle angle and

t

its double (

cos t = \frac{16 - 9}{25}

) — the double-angle method run backwards.

Example 8 · Standard

Find the exact value of

tan 112. 5^{\circ}

Solution, step by step

1What is $112. 5^{\circ}$ half of? $22 5^{\circ}$ . So $t = 22 5^{\circ}$ , with $sin 22 5^{\circ} = - \frac{2}{2}$ and $cos 22 5^{\circ} = - \frac{2}{2}$ (third quadrant: both negative).
2It is a tangent, so $tan \frac{t}{2} = \frac{s i n t}{1 + c o s t} = \frac{- \frac{2}{2}}{1 - \frac{2}{2}} = \frac{- 2}{2 - 2}$ (multiplying top and bottom by 2).
3Rationalise the denominator: $\frac{- 2 ( 2 + 2 )}{( 2 - 2 ) ( 2 + 2 )} = \frac{- 2 2 - 2}{4 - 2} = - (2 + 1)$ .
4Sign check: $112. 5^{\circ}$ is in the second quadrant, where the tangent is negative ✓.

Answer $- (1 + 2)$

You also get it as

tan 112. 5^{\circ} = - cot 22. 5^{\circ} = - \frac{1}{2 - 1} = - (2 + 1)

Example 9 · Advanced

tan \frac{t}{2} = \frac{2}{3}

. Find

sin t

and

cos t

Solution, step by step

1The tangent of the half angle is given; put $u = \frac{2}{3}$ and use the reverse formulas. $u^{2} = \frac{4}{9}$ .
2 $sin t = \frac{2 u}{1 + u ^{2}} = \frac{4/3}{13/9} = \frac{4}{3} \cdot \frac{9}{13} = \frac{12}{13}$ .
3 $cos t = \frac{1 - u ^{2}}{1 + u ^{2}} = \frac{5/9}{13/9} = \frac{5}{13}$ .
4Check: $(\frac{12}{13})^{2} + (\frac{5}{13})^{2} = \frac{144 + 25}{169} = 1$ ✓.

Answer $sin t = \frac{12}{13}, cos t = \frac{5}{13}$

This trick (later called the “

tan \frac{t}{2}

substitution”) turns

sin

and

cos

into a single variable.

Practice (solutions hidden)

BasicTry it yourself first

cos t = \frac{161}{289}

and

t

is acute. Find

cos \frac{t}{2}

BasicTry it yourself first

The exact value of

tan 22. 5^{\circ}

StandardTry it yourself first

sin t = \frac{3}{5}

t

acute. Find

tan \frac{t}{2}

StandardTry it yourself first

The exact value of

tan 7 5^{\circ}

AdvancedTry it yourself first

tan \frac{t}{2} = \frac{1}{2}

. Find

tan t

AdvancedTry it yourself first

cos t = - \frac{7}{25}

and

9 0^{\circ} < t < 18 0^{\circ}

. Find

tan \frac{t}{2}

Method 4: the values at $1 8^{\circ}$ , $3 6^{\circ}$ , $5 4^{\circ}$ , $7 2^{\circ}$

2 θ = 9 0^{\circ} - 3 θ

Building the exact $sin 1 8^{\circ}$ and $cos 3 6^{\circ}$ from the double and triple angle together (no memorising), and their products.

The idea

Take $θ = 1 8^{\circ}$ . Then $5 θ = 9 0^{\circ}$ , i.e. $2 θ + 3 θ = 9 0^{\circ}$ and $2 θ = 9 0^{\circ} - 3 θ$ . Complementary angles have equal sine and cosine, so $sin 2 θ = cos 3 θ$ . The left side is a double angle, the right a triple angle — we have formulas for both, and everything becomes an equation in $sin θ$ .

◭ Diagram

The right angle is split into

2 θ

(green) and

3 θ

(saffron). The height of the point seen from the

x

-axis is

sin 2 θ

; seen from the

y

-axis it is

cos 3 θ

— the same distance.

Put both formulas into $sin 2 θ = cos 3 θ$ : $2 sin θ cos θ = 4 cos^{3} θ - 3 cos θ$ .
$cos θ \neq = 0$ , so divide by it: $2 sin θ = 4 cos^{2} θ - 3 = 4 (1 - sin^{2} θ) - 3 = 1 - 4 sin^{2} θ$ .
Write $s = sin 1 8^{\circ}$ : $4 s^{2} + 2 s - 1 = 0$ . Quadratic formula: $s = \frac{- 2 \pm 4 + 16}{8} = \frac{- 1 \pm 5}{4}$ .
$sin 1 8^{\circ}$ is positive, so $sin 1 8^{\circ} = \frac{5 - 1}{4} \approx 0.309$ (reject the negative root $\approx - 0.809$ ).
Now $cos 3 6^{\circ} = 1 - 2 sin^{2} 1 8^{\circ} = 1 - 2 \cdot \frac{6 - 2 5}{16} = \frac{1 + 5}{4}$ , and $cos 7 2^{\circ} = sin 1 8^{\circ}$ (complementary angles).

Angle	$sin$	$cos$
$1 8^{\circ}$	$\frac{5 - 1}{4}$	$\frac{10 + 2 5}{4}$
$3 6^{\circ}$	$\frac{10 - 2 5}{4}$	$\frac{5 + 1}{4}$
$5 4^{\circ}$	$\frac{5 + 1}{4}$	$\frac{10 - 2 5}{4}$
$7 2^{\circ}$	$\frac{10 + 2 5}{4}$	$\frac{5 - 1}{4}$

Here: $sin 5 4^{\circ} = cos 3 6^{\circ}$ and $cos 5 4^{\circ} = sin 3 6^{\circ}$ (complementary angles), so four values are simple: $sin 1 8^{\circ}$ , $cos 3 6^{\circ}$ , and from them $cos 7 2^{\circ}$ , $sin 5 4^{\circ}$ . The other two ( $cos 1 8^{\circ}$ , $sin 3 6^{\circ}$ ) come from $1 - sin^{2}$ .

CheckIn exams

sin 1 8^{\circ} = \frac{5 - 1}{4}

and

cos 3 6^{\circ} = \frac{5 + 1}{4}

are often taken as known, but if you forget them the five lines above rebuild them in a minute. A check:

cos 3 6^{\circ} - cos 7 2^{\circ} = \frac{1}{2}

and

cos 3 6^{\circ} cos 7 2^{\circ} = \frac{1}{4}

must always come out right.

Worked examples

Example 10 · Basic

Find

cos 3 6^{\circ} - cos 7 2^{\circ}

Solution, step by step

1From the table $cos 3 6^{\circ} = \frac{5 + 1}{4}$ and $cos 7 2^{\circ} = \frac{5 - 1}{4}$ .
2Subtract: $\frac{( 5 + 1 ) - ( 5 - 1 )}{4} = \frac{2}{4} = \frac{1}{2}$ . The $5$ cancels, which is why the result is so neat.

Answer $\frac{1}{2}$

Numerical check:

0.8090 - 0.3090 = 0.5

✓.

Example 11 · Standard

Find

sin^{2} 1 8^{\circ} + cos^{2} 3 6^{\circ}

Solution, step by step

1Write the values before squaring: $sin 1 8^{\circ} = \frac{5 - 1}{4}$ , so $sin^{2} 1 8^{\circ} = \frac{5 - 2 5 + 1}{16} = \frac{6 - 2 5}{16} = \frac{3 - 5}{8}$ .
2 $cos 3 6^{\circ} = \frac{5 + 1}{4}$ , so $cos^{2} 3 6^{\circ} = \frac{5 + 2 5 + 1}{16} = \frac{3 + 5}{8}$ .
3Adding, the $5$ terms cancel: $\frac{3 - 5 + 3 + 5}{8} = \frac{6}{8} = \frac{3}{4}$ .

Answer $\frac{3}{4}$

Check:

0.0955 + 0.6545 = 0.75

✓.

Example 12 · Advanced

Find the exact value of

sin 3 6^{\circ} sin 7 2^{\circ}

Solution, step by step

1From the table $sin 3 6^{\circ} = \frac{10 - 2 5}{4}$ and $sin 7 2^{\circ} = \frac{10 + 2 5}{4}$ .
2Multiply: the denominators give $4 \times 4 = 16$ . Combine the roots, $a b = ab$ : $(10 - 25) (10 + 25)$ .
3 $(10 - 25) (10 + 25) = 100 - 20 = 80$ ( $a^{2} - b^{2}$ ), and $80 = 45$ .
4So the product is $\frac{4 5}{16} = \frac{5}{4} \approx 0.559$ .

Answer $\frac{5}{4}$

Check:

0.5878 \times 0.9511 = 0.559

✓.

Practice (solutions hidden)

BasicTry it yourself first

sin 1 8^{\circ} + sin 5 4^{\circ}

equals?

BasicTry it yourself first

Find

cos 3 6^{\circ} cos 7 2^{\circ}

StandardTry it yourself first

Find

4 sin 1 8^{\circ} cos 3 6^{\circ}

StandardTry it yourself first

Find

cos 7 2^{\circ} + cos 14 4^{\circ}

AdvancedTry it yourself first

Find

tan 3 6^{\circ} tan 7 2^{\circ}

AdvancedTry it yourself first

Find

sin 16 2^{\circ} + cos 7 2^{\circ}

Products that collapse into a number

t, 60 - t, 60 + t

cos x cos 2 x cos 4 x

Turning a product of many sines or cosines into one number with two tricks.

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Method 5: the triple $t, 6 0^{\circ} - t, 6 0^{\circ} + t$

Product

= \frac{1}{4} sin 3 t

The three-angle product, its proof and how to spot it (also for cos and tan).

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Method 6: repeated $2 sin cos$ (the telescoping trick)

cos x cos 2 x cos 4 x \dots

When each angle is double the last: turning the product into $\frac{s i n 2 ^{n} x}{2 ^{n} s i n x}$ , and the ratio forms.

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Conditions and series

A + B = 4 5^{\circ}

A + B + C = 18 0^{\circ}

· infinite GP

When the angles obey a condition, or the sum of an infinite series in sin/cos is given.

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Method 7: an angle-sum condition

Turning

tan (A + B)

around

The ready-made relations from $A + B = 4 5^{\circ}$ , $13 5^{\circ}$ , $9 0^{\circ}$ and $A + B + C = 18 0^{\circ}$ , and where they come from.

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Method 8: an infinite GP in $sin$ , $cos$

S = \frac{a}{1 - r}

Finding the angle when the sum of the series is given: first term, ratio, the condition $∣ r ∣ < 1$ and all the solutions.

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Largest and smallest values

ranges · squares ·

a sin t + b cos t

· AM-GM · powers

Seven methods for the maximum and minimum value of a trigonometric expression.

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Method 9: the range of the six ratios

k + m \cdot f (t)

Which values each ratio can take, why $sec$ and $cosec$ never lie between $- 1$ and $1$ , and the range of $k + m f (t)$ .

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Method 10: squares and quadratics

a sin^{2} t + b cos^{2} t

p s^{2} + q s + r

Writing the expression in one function ( $sin t$ or $cos t$ ) and treating it as a quadratic; checking whether the vertex is inside or outside.

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Method 11: $a sin t + b cos t$

\pm a^{2} + b^{2}

Where $a sin t + b cos t = R sin (t + α)$ comes from, the largest and smallest values from it, forms with a constant or an exponent, and expressions with $sin t cos t$ .

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Method 12: AM-GM for $tan^{2} + cot^{2}$ and $sec^{2} + cosec^{2}$

p + q \geq 2 pq

The minimum of a sum of a positive number and its “reverse”: where it comes from, when equality holds, and why $tan t + cot t$ never lies between $- 2$ and $2$ .

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Method 13: $a sin^{2} t + b cosec^{2} t$ (with a boundary)

a \geq b

a < b

When $sin^{2} t$ and its reciprocal appear together: when AM-GM applies and when the minimum sits on the boundary $u = 1$ .

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Method 14: $sin^{n} t cos^{n} t$ and $sin^{2 p} t + cos^{2 q} t$

When the

4 5^{\circ}

rule is true

Limits of expressions with powers: the $\frac{1}{2 ^{n}}$ rule, $sin^{4} + cos^{4}$ and $sin^{6} + cos^{6}$ , and the myth that the minimum is always at $4 5^{\circ}$ .

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Method 15: the angles of a triangle add to $18 0^{\circ}$

Extreme at equal angles

When $A + B + C = 18 0^{\circ}$ , the largest/smallest values of symmetric expressions: at the equilateral triangle (each angle $6 0^{\circ}$ ).

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Summary, mixed practice, revision

Every method in one place

A table of all fifteen methods, exam-style mixed questions and revision.

Summary of all methods

When · what · how

Every method: how to recognise the question and which formula to use.

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Mixed practice

Exam-style questions · all methods

Mixed questions on all fifteen methods; first decide which method fits.

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Revision

Quick questions and common mistakes

All formulas at a glance and the mistakes that cost marks.

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