Trigonometry | The Maths Tailor — UK Methods

Trigonometry

8 methods · 29 questions

When an equation or function mixes even powers of $sin$ and $cos$ , replace $sin^{2}$ by $1 - cos^{2}$ (or $cos^{2}$ by $1 - sin^{2}$ ) so that only ONE trig ratio remains, then treat that ratio as a single algebraic unknown and factorise the resulting polynomial.

Trigger: the expression contains $sin^{2}$ and $cos^{2}$ (or $sin^{4}$ , $cos^{4}$ ), or a single squared ratio plus its linear term.

Instances:
(i) substitute to get a quadratic in $cos$ , then factorise (e.g. $(u - 1) (u - 2) = 0$ );
(ii) substitute to get $sin^{2} =$ constant and read off the base angle;
(iii) treat a degree-4 equation as a quadratic in $cos^{2}$ and factorise;
(iv) clear $sin^{2}$ from a rational function so the numerator and denominator share a factor that cancels.

Linked questions (7)

TMUA 2016 Paper 1 — Q8

Find the maximum angle

x

in the range

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

which satisfies the equation

cos^{2} (2 x) + 3 sin (2 x) - \frac{7}{4} = 0.

(A)

3 0^{\circ}

(B)

6 0^{\circ}

(C)

12 0^{\circ}

(D)

15 0^{\circ}

(E)

21 0^{\circ}

(F)

24 0^{\circ}

(G)

30 0^{\circ}

(H)

33 0^{\circ}

Voir la correction

Substitute

cos^{2} (2 x) = 1 - sin^{2} (2 x)

:

(1 - sin^{2} 2 x) + 3 sin 2 x - \frac{7}{4} = 0

, so

sin^{2} 2 x - 3 sin 2 x + \frac{3}{4} = 0

. By the quadratic formula:

sin 2 x = \frac{3 \pm 3 - 3}{2} = \frac{3}{2}

. So

sin 2 x = \frac{3}{2}

for

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

, i.e.

0^{\circ} \leq 2 x \leq 72 0^{\circ}

. Solutions for

2 x

:

6 0^{\circ}, 12 0^{\circ}, 42 0^{\circ}, 48 0^{\circ}

. Corresponding

x

values:

6 0^{\circ}, 12 0^{\circ}, 42 0^{\circ}, 48 0^{\circ}

. The maximum is

x

. Hence the answer is F.

TMUA 2016 Paper 2 — Q3

What is the value, in radians, of the largest angle

x

in the range

0 \leq x \leq 2 π

that satisfies the equation

8 sin^{2} x + 4 cos^{2} x = 7

?

(A)

\frac{2 π}{3}

(B)

\frac{5 π}{6}

(C)

\frac{4 π}{3}

(D)

\frac{5 π}{3}

(E)

\frac{7 π}{4}

(F)

\frac{11 π}{6}

Voir la correction

Use

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

:

8 sin^{2} x + 4 (1 - sin^{2} x) = 7 \Rightarrow 4 sin^{2} x + 4 = 7 \Rightarrow 4 sin^{2} x = 3 \Rightarrow sin^{2} x = \frac{3}{4}

, so

sin x = \pm \frac{3}{2}

. The solutions to

sin x = \frac{3}{2}

in

[0, 2 π]

are

x = \frac{π}{3}

and

x = \frac{2 π}{3}

. The solutions to

sin x = - \frac{3}{2}

in

[0, 2 π]

are

x = π + \frac{π}{3} = \frac{4 π}{3}

and

x = 2 π - \frac{π}{3} = \frac{5 π}{3}

. The largest of all four is

\frac{5 π}{3}

. Hence the answer is D.

TMUA 2020 Paper 1 — Q18

Find the number of solutions and the sum of the solutions of the equation

1 - 2 cos^{2} x = ∣ cos x ∣

where

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

.

(A) Number of solutions

= 2

, Sum of solutions

= 18 0^{\circ}

(B) Number of solutions

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

, Sum of solutions

= 24 0^{\circ}

(C) Number of solutions

= 3

, Sum of solutions

= 18 0^{\circ}

(D) Number of solutions

= 3

, Sum of solutions

= 36 0^{\circ}

(E) Number of solutions

= 4

, Sum of solutions

= 24 0^{\circ}

(F) Number of solutions

= 4

, Sum of solutions

= 36 0^{\circ}

Voir la correction

Case 1:

= 24 0^{\circ}

(so

∣ cos x ∣ = cos x

). The equation becomes

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

, i.e.

2 cos^{2} x + cos x - 1 = 0

, factorising as

(2 cos x - 1) (cos x + 1) = 0

. So

cos x = \frac{1}{2}

(giving

x = 6 0^{\circ}

, valid since

cos 6 0^{\circ} = \frac{1}{2} > 0

) or

cos x = - 1

(rejected, since we need

cos x \geq 0

). Case 2:

cos x < 0

(so

∣ cos x ∣ = - cos x

). The equation becomes

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

, i.e.

2 cos^{2} x - cos x - 1 = 0

, factorising as

(2 cos x + 1) (cos x - 1) = 0

. So

cos x = - \frac{1}{2}

(giving

x = 12 0^{\circ}

, valid since

cos 12 0^{\circ} = - \frac{1}{2} < 0

) or

cos x = 1

(rejected). There are 2 solutions:

x = 6 0^{\circ}

and

x = 12 0^{\circ}

, with sum

18 0^{\circ}

. Hence the answer is A.

TMUA 2021 Paper 1 — Q6

The function f is given by

f (x) = \frac{cos x + 3}{7 + 5 cos x - sin ^{2} x}

Find the positive difference between the maximum and the minimum values of

f (x)

.

(A)

0

(B)

\frac{1}{3}

(C)

\frac{1}{2}

(D)

\frac{2}{3}

(E)

1

(F)

2

Voir la correction

Replace

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

in the denominator:

f (x) = \frac{cos x + 3}{7 + 5 cos x - ( 1 - cos ^{2} x )} = \frac{cos x + 3}{cos ^{2} x + 5 cos x + 6} = \frac{cos x + 3}{( cos x + 2 ) ( cos x + 3 )} .

Since

cos x + 3 \geq 2 > 0

for all

x

, we can cancel:

f (x) = \frac{1}{cos x + 2}

. This is maximised when

cos x = - 1

:

f = 1

. It is minimised when

cos x = 1

:

f = \frac{1}{3}

. The positive difference is

1 - \frac{1}{3} = \frac{2}{3}

. Hence the answer is D.

TMUA 2022 Paper 1 — Q1

How many real solutions are there to the equation

2 cos^{4} θ - 5 cos^{2} θ + 3 = 0

in the interval

0 \leq θ \leq 2 π

?

(A)

1

(B)

2

(C)

3

(D)

4

(E)

5

(F)

6

(G)

7

(H)

8

Voir la correction

Treat the equation as a quadratic in

1

. It factorises as

(2 cos^{2} θ - 3) (cos^{2} θ - 1) = 0

, giving

cos^{2} θ = \frac{3}{2}

or

cos^{2} θ = 1

. Since

cos^{2} θ \leq 1

for all real

θ

, the equation

cos^{2} θ = \frac{3}{2}

has no real solutions. So we need

cos^{2} θ = 1

, i.e.

cos θ = \pm 1

. In the interval

0 \leq θ \leq 2 π

:

cos θ = 1

gives

θ = 0

and

θ = 2 π

;

cos θ = - 1

gives

θ = π

. That is three solutions in total. Hence the answer is C.

TMUA 2022 Paper 1 — Q16

The solutions to

7 x^{4} - 6 x^{2} + 1 = 0

are

\pm cos θ

and

\pm cos β

. Which one of the following equations has solutions

\pm sin θ

and

\pm sin β

?

(A)

7 x^{4} - 8 x^{2} - 5 = 0

(B)

7 x^{4} - 8 x^{2} + 2 = 0

(C)

7 x^{4} - 6 x^{2} - 2 = 0

(D)

7 x^{4} - 6 x^{2} + 1 = 0

(E)

7 x^{4} + 6 x^{2} - 1 = 0

(F)

7 x^{4} + 6 x^{2} + 5 = 0

Voir la correction

The quartic

7 x^{4} - 6 x^{2} + 1 = 0

is quadratic in

x^{2}

with roots

x^{2} = \frac{6 \pm 36 - 28}{14} = \frac{6 \pm 8}{14} = \frac{3 \pm 2}{7}

. So

cos^{2} θ

and

cos^{2} β

are

\frac{3 + 2}{7}

and

\frac{3 - 2}{7}

in some order. Then

sin^{2} θ = 1 - cos^{2} θ

and

sin^{2} β = 1 - cos^{2} β

, giving

sin^{2}

values

\frac{4 - 2}{7}

and

\frac{4 + 2}{7}

. The equation with roots

\pm sin θ

,

\pm sin β

satisfies:

(x^{2} - \frac{4 - 2}{7}) (x^{2} - \frac{4 + 2}{7}) = 0

. Expanding:

x^{4} - \frac{8}{7} x^{2} + \frac{16 - 2}{49} = x^{4} - \frac{8}{7} x^{2} + \frac{2}{7} = 0

. Multiplying by 7:

7 x^{4} - 8 x^{2} + 2 = 0

. Hence the answer is B.

TMUA Specimen Paper 1 — Q2

The number of solutions in the interval

0 \leq θ \leq 4 π

of the equation

sin^{2} θ + 3 cos θ = 3

is

(A)

0

(B)

1

(C)

2

(D)

3

(E)

sin^{2} θ + 3 cos θ = 3

(F)

5

(G)

6

Voir la correction

Replace

sin^{2} θ

with

1 - cos^{2} θ

:

5

, so

cos^{2} θ - 3 cos θ + 2 = 0

. Let

u = cos θ

:

(u - 1) (u - 2) = 0

, giving

u = 1

or

u = 2

. Since

∣ cos θ ∣ \leq 1

, the equation

cos θ = 2

has no solutions. For

cos θ = 1

on

[0, 4 π]

:

∣ cos θ ∣ \leq 1

, giving exactly 3 solutions. Hence the answer is D.