AQA Level 2 Further Maths · Level 2 Certificate · Chapter 6: Geometry I · Lesson 7

Trigonometric identities

About 80 minutesCalculator allowed73 marks

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Date

Do Now

Four questions to start. Two of them use what you need today.

A1
Needed today
Solve $3 y^{2} - 2 y - 1 = 0$ .
[2 marks]
Answer $y = \dots\dots$ or $y = \dots\dots$
A2
Last lesson
Solve $5 sin x = 2$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ . Give your answers to 1 decimal place.
[2 marks]
Answer $x = \dots\dots$ or $x = \dots\dots$
A3
From chapter 6
Do not use a calculator. Work out the exact value of $cos 3 0^{\circ} \times tan 6 0^{\circ}$ .
[2 marks]
Answer
A4
From chapter 2
Simplify fully $\frac{x ^{2} - 9}{x ^{2} - x - 6}$ .
[2 marks]
Answer

Example, then your turn

Level 2

Your teacher works through each example with you. Then try the one beside it. Use the identities $sin^{2} x + cos^{2} x \equiv 1$ and $tan x \equiv sin x \div cos x$ .

Example 1

Prove that $\frac{sin ^{2} x}{1 + cos x} \equiv 1 - cos x$ .

Hint Start with the harder side. Write

sin^{2} x

1 - cos^{2} x

, then factorise it as a difference of two squares.

B1
Prove that $\frac{cos ^{2} x}{1 - sin x} - sin x \equiv 1$ .
[3 marks]

Example 2

Show that $(2 cos x + sin x)^{2} + (cos x - 2 sin x)^{2}$ simplifies to an integer.

Hint Expand both brackets. The

sin x cos x

terms cancel. Collect the

sin^{2} x

and

cos^{2} x

terms.

B2
Show that $(3 sin x - cos x)^{2} + (sin x + 3 cos x)^{2} - 4$ simplifies to an integer.
[3 marks]
Answer $\dots\dots$

Example 3

Write $sin x tan x + cos x$ as a single fraction in terms of $cos x$ only.

Hint Write

tan x

\frac{sin x}{cos x}

, put everything over

cos x

, then use

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

B3
Write $\frac{cos x}{tan x} + sin x$ as a single fraction in terms of $sin x$ only.
[2 marks]
Answer $\dots\dots$

Example 4

Do not use a calculator. Solve $2 sin^{2} x + 3 cos x - 3 = 0$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .

Hint Replace

sin^{2} x

with

1 - cos^{2} x

. This gives a quadratic in

cos x

only. Factorise it.

B4
Solve $6 cos^{2} x + sin x - 5 = 0$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ . Give your answers to 1 decimal place where they are not exact.
[4 marks]
Answer $x = \dots\dots$

Example 5

Solve $3 sin x = 4 cos x$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .

Hint Divide both sides by

3 cos x

. Then

\frac{sin x}{cos x}

tan x

B5
Solve $5 sin x + 2 cos x = 0$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .
[3 marks]
Answer $x = \dots\dots$ or $x = \dots\dots$

Practice

Level 2

C1
Prove that $(sin x + 2 cos x) (2 sin x + cos x)$ is identical to $2 + 5 sin x cos x$ .
[3 marks]
C2
Write $\frac{cos ^{2} x}{1 + sin x}$ in terms of $sin x$ . Give your answer without a fraction.
[2 marks]
Answer $\dots\dots$

C3
Solve $3 sin x = 7 cos x$ for $- 18 0^{\circ} \leq x \leq 18 0^{\circ}$ .
[3 marks]
Answer $x = \dots\dots$ or $x = \dots\dots$
C4
Do not use a calculator. $cos x = - \frac{3}{7}$ and $18 0^{\circ} < x < 27 0^{\circ}$ . Work out the exact value of $sin x$ and the exact value of $tan x$ .
[4 marks]
Answer $sin x = \dots\dots$ $tan x = \dots\dots$
C5
Solve $4 sin x = 3 tan x$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .
[4 marks]
Answer $x = \dots\dots$
C6
Solve $cos^{2} x + 3 sin x = 2$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ . Give your answers to 1 decimal place.
[4 marks]
Answer $x = \dots\dots$ or $x = \dots\dots$

Exam-style questions

AQA exam style

D1
Sam says, " $(sin x + cos x)^{2} \equiv 1$ , because $sin^{2} x + cos^{2} x \equiv 1$ ."
Show that Sam is wrong.
[3 marks]

D2
Show that $\frac{( sin x + cos x ) ^{2} - 1}{2 cos x} - \frac{3 sin ^{2} x - 3}{cos x}$ can be written in the form $a cos x + b sin x$ , where $a$ and $b$ are integers.
[4 marks]
Answer $a = \dots\dots$ $b = \dots\dots$
D3
Do not use a calculator. $x$ is an acute angle and $tan x = 2$ . Work out the exact value of $sin x cos x$ .
[3 marks]
Answer $\dots\dots$
D4
Answer both parts.
(a)
Show that $3 sin^{2} x - cos^{2} x \equiv 4 sin^{2} x - 1$ .
[1 mark]
(b)
Hence solve $3 sin^{2} x - cos^{2} x = 2 sin x$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ . Give your answers to the nearest degree.
[4 marks]
Answer $x = \dots\dots$
Total for D4: 5 marks
D5
Solve $2 sin^{2} x = 3 sin x cos x$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ . You must show your working.
[4 marks]
Answer $x = \dots\dots$
D6
Answer both parts.
(a)
Prove that $tan x + \frac{1}{tan x} \equiv \frac{1}{sin x cos x}$ .
[3 marks]
(b)
Hence, without a calculator, work out the exact value of $tan 3 0^{\circ} + \frac{1}{tan 3 0 ^{\circ}}$ . Give your answer in the form $\frac{a 3}{b}$ , where $a$ and $b$ are integers.
[2 marks]
Answer $\dots\dots$
Total for D6: 5 marks
D7
Which of these is equal to $\frac{1 - cos ^{2} x}{sin x cos x}$ ? Circle your answer.
$sin x cos x tan x 1 \div tan x$
[1 mark]
Answer

Extension

Stretch

No route is given. The hints are at the end of the sheet. Use one at a time.

E1
Do not use a calculator. Solve $sin^{4} x - cos^{4} x = \frac{1}{2}$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .
[5 marks]
Hint 1 · What to try
$sin^{4} x - cos^{4} x$ is a difference of two squares.
Hint 2 · The first line
Factorise it as $(sin^{2} x - cos^{2} x) (sin^{2} x + cos^{2} x)$ . The second bracket is $1$ .
Hint 3 · The full method
$sin^{2} x - (1 - sin^{2} x) = \frac{1}{2}$ , so $sin^{2} x = \frac{3}{4}$ and $sin x = \pm \frac{3}{2}$ : $x = 6 0^{\circ}$ , $12 0^{\circ}$ , $24 0^{\circ}$ , $30 0^{\circ}$ .
Answer $x = \dots\dots$

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can prove an identity by starting with one side and changing it into the other.
I can use sin²x + cos²x = 1 to write an equation in one function, then solve it.
I can turn a sin x = b cos x into an equation in tan x.
I can find sin x and tan x exactly from cos x and the quadrant.

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