A-Level Maths (Edexcel 9MA0) · Pure Year 1 · Chapter 10: Trigonometric identities and equations · Lesson 4

Equations and identities

76 marks

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Date

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Quick questions to start.

A1
Needed today
Solve $3 x^{2} + 7 x - 6 = 0$ .
[2 marks]
Answer
A2
Last lesson
Solve $tan x = - 2.5$ for $0^{\circ} \leq x < 36 0^{\circ}$ . Give your answers to 1 decimal place.
[2 marks]
Answer
A3
Needed today
Write $4 cos^{2} x + 3 sin x - 2$ in terms of $sin x$ only.
[2 marks]
Answer
A4
From chapter 2
Use the discriminant to show that $x^{2} - 3 x + 5 = 0$ has no real roots.
[2 marks]
Answer

Example, then your turn

A-Level

Your teacher works through each example with you. Then try the one beside it. Give angles to 1 decimal place where they are not exact.

Example 1

Solve $3 cos^{2} θ + 5 cos θ - 2 = 0$ for $0^{\circ} \leq θ < 36 0^{\circ}$ .

Hint This is

3 c^{2} + 5 c - 2 = 0

with

c = cos θ

. Factorise, then solve each factor.

B1
Solve $4 sin^{2} x + 11 sin x - 3 = 0$ for $- 18 0^{\circ} \leq x \leq 18 0^{\circ}$ .
[3 marks]
Answer $x = \dots\dots$

Example 2

Solve $tan^{2} (θ + 2 0^{\circ}) = 3$ for $0^{\circ} \leq θ < 36 0^{\circ}$ .

Hint Square root both sides: there are two equations,

tan (θ + 2 0^{\circ}) = 3

and

tan (θ + 2 0^{\circ}) = - 3

. The interval for

θ + 2 0^{\circ}

2 0^{\circ}

38 0^{\circ}

B2
Solve $4 cos^{2} (x - 1 5^{\circ}) = 3$ for $0^{\circ} \leq x < 36 0^{\circ}$ . Give your answers exactly.
[4 marks]
Answer $x = \dots\dots$

Example 3

Solve $2 cos^{2} θ + 3 sin θ = 3$ for $0^{\circ} \leq θ < 36 0^{\circ}$ .

Hint Replace

cos^{2} θ

with

1 - sin^{2} θ

to get a quadratic in

sin θ

B3
Solve $6 sin^{2} x + cos x - 4 = 0$ for $0^{\circ} \leq x < 36 0^{\circ}$ .
[5 marks]
Answer $x = \dots\dots$

Example 4

Solve $5 sin θ cos θ = 2 cos θ$ for $0^{\circ} \leq θ < 36 0^{\circ}$ .

Hint Do not divide by

cos θ

: it can be

0

. Move everything to one side and take out the common factor

cos θ

B4
Solve $7 sin x tan x = 3 tan x$ for $0^{\circ} \leq x < 36 0^{\circ}$ .
[4 marks]
Answer $x = \dots\dots$

Practice

A-Level

C1
Solve $sin^{2} θ + 3 sin θ - 1 = 0$ for $0^{\circ} \leq θ < 36 0^{\circ}$ .
[4 marks]
Answer $θ = \dots\dots$

C2
Find the exact solutions of $3 tan θ = 2 cos θ$ for $- 18 0^{\circ} \leq θ \leq 18 0^{\circ}$ .
[4 marks]
Answer $θ = \dots\dots$
C3
Find the exact solutions of $2 cos^{2} 2 x = cos 2 x$ for $0^{\circ} \leq x < 18 0^{\circ}$ .
[4 marks]
Answer $x = \dots\dots$
C4
Solve $\frac{3 sin ^{2} θ}{1 - cos θ} = 2$ for $0^{\circ} < θ < 36 0^{\circ}$ .
[4 marks]
Answer $θ = \dots\dots$

Exam-style questions

Edexcel exam style

D1
Ravi is asked to solve $5 sin^{2} (x + 2 5^{\circ}) = 2$ for $0^{\circ} \leq x < 36 0^{\circ}$ . Ravi writes: " $sin (x + 2 5^{\circ}) = 0.4 = 0.6325$ , so $x + 2 5^{\circ} = 39. 2^{\circ}$ or $140. 8^{\circ}$ . So $x = 14. 2^{\circ}$ or $x = 115. 8^{\circ}$ ."
Ravi has not found every solution.
(a)
Explain the mistake Ravi has made.
[2 marks]
(b)
Hence, or otherwise, find all the solutions, giving your answers to 1 decimal place.
[4 marks]
Answer $x = \dots\dots$
Total for D1: 6 marks

D2
The angle $θ$ is in the interval $0^{\circ} \leq θ < 36 0^{\circ}$ .
(a)
Show that the equation $5 cos^{2} θ + 3 sin θ = 4$ can be written as $5 sin^{2} θ - 3 sin θ - 1 = 0$ .
[2 marks]
(b)
Hence show that $sin θ = \frac{3 \pm 29}{10}$ .
[2 marks]
(c)
Hence solve $5 cos^{2} θ + 3 sin θ = 4$ for $0^{\circ} \leq θ < 36 0^{\circ}$ , giving your answers to 1 decimal place.
[4 marks]
Answer $θ = \dots\dots$
Total for D2: 8 marks
D3
The angle $x$ is in the interval $0^{\circ} \leq x \leq 18 0^{\circ}$ .
(a)
Show that the equation $4 sin 2 x tan 2 x = 15$ can be written as $4 cos^{2} 2 x + 15 cos 2 x - 4 = 0$ .
[3 marks]
(b)
Hence solve $4 sin 2 x tan 2 x = 15$ for $0^{\circ} \leq x \leq 18 0^{\circ}$ , giving your answers to 1 decimal place.
[5 marks]
Answer $x = \dots\dots$
Total for D3: 8 marks
D4
The angle $θ$ is in the interval $0^{\circ} \leq θ < 36 0^{\circ}$ .
(a)
Show that the equation $3 cos^{2} θ - 2 sin θ = 5$ has no solutions.
[3 marks]
(b)
Solve $3 cos^{2} θ - 2 sin θ = 2$ for $0^{\circ} \leq θ < 36 0^{\circ}$ , giving your answers to 1 decimal place where they are not exact.
[5 marks]
Answer $θ = \dots\dots$
Total for D4: 8 marks

Extension

Stretch

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

E1
Find every solution of $sin^{4} θ - cos^{4} θ = \frac{1}{2}$ in the interval $0^{\circ} \leq θ < 36 0^{\circ}$ . Give your answers exactly.
[6 marks]
Hint 1 · What to try
$sin^{4} θ - cos^{4} θ$ is a difference of two squares.
Hint 2 · The first line
It is $(sin^{2} θ + cos^{2} θ) (sin^{2} θ - cos^{2} θ)$ , and the first bracket is $1$ .
Hint 3 · The full method
Replace $cos^{2} θ$ with $1 - sin^{2} θ$ : $2 sin^{2} θ - 1 = \frac{1}{2}$ , so $sin θ = \pm \frac{3}{2}$ .
Answer $θ = \dots\dots$

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can solve a quadratic in sin, cos or tan, and reject a value that is impossible.
I can use sin²θ + cos²θ = 1 or tan θ = sin θ / cos θ to get an equation in one ratio.
I can take out a common factor instead of dividing, so no solutions are lost.
I can show that an equation has no solutions.

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