Trigonometry

9 methods · 14 questions

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

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

Instances:
(i) substitute $tan x = \frac{sin x}{cos x}$ and multiply out to expose the $sin^{2} x$ and $cos^{2} x$ terms;
(ii) substitute to get a quadratic in $cos x$ , then factorise;
(iii) apply it inside a bracket after factoring out a common term.

Linked questions (3)

NSAA 2019 Section 1, Part E (Mathematics questions) · Q77

It is given that

7 cos x + tan x sin x = 5

where

0^{\circ} < x < 9 0^{\circ}

What are the possible values of

tan x

A $\frac{1}{2}$ or $\frac{1}{3}$
B $\frac{1}{3}$ or $\frac{1}{2 2}$
C $\frac{3}{2}$ or $\frac{2 2}{3}$
D $3$ or $22$
E $3$ or $2$

Worked solution

Answer:D

Use

tan x = \frac{sin x}{cos x}

, so that

tan x sin x = \frac{sin ^{2} x}{cos x}

. On

0^{\circ} < x < 9 0^{\circ}

we have

cos x \neq = 0

, so multiplying the equation by

cos x

is legitimate:

7 cos^{2} x + sin^{2} x = 5 cos x .

Replace

sin^{2} x

1 - cos^{2} x

to get a quadratic in

cos x

7 cos^{2} x + 1 - cos^{2} x = 5 cos x ⟹ 6 cos^{2} x - 5 cos x + 1 = 0.

This factorises as

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

, so

cos x = \frac{1}{3}

cos x = \frac{1}{2}

. Both values lie strictly between

0

and

1

, so both correspond to an angle in the given range, where

sin x > 0

.

If

cos x = \frac{1}{2}

then

sin x = \frac{3}{2}

and

tan x = 3

.

If

cos x = \frac{1}{3}

then

sin x = 1 - \frac{1}{9} = \frac{2 2}{3}

and

tan x = \frac{2 2 /3}{1/3} = 22

.

The answer is D.

NSAA 2016 Section 1, Part E (Mathematics questions) · Q81

Given that

7 cos θ - 3 tan θ sin θ = 1

, which one of the following is true?

A $cos θ = - \frac{3}{5}$ or $- \frac{1}{2}$
B $cos θ = - \frac{3}{5}$ or $\frac{1}{2}$
C $cos θ = \frac{3}{5}$ or $\frac{1}{2}$
D $cos θ = \frac{3}{5}$ or $- \frac{1}{2}$

Worked solution

Answer:D

Replace

tan θ

\frac{sin θ}{cos θ}

so that the equation involves only

sin θ

and

cos θ

7 cos θ - \frac{3 sin ^{2} θ}{cos θ} = 1.

Now use

sin^{2} θ = 1 - cos^{2} θ

to reach a single trigonometric ratio:

7 cos θ - \frac{3 ( 1 - cos ^{2} θ )}{cos θ} = 1.

Multiply through by

cos θ

, which is non-zero since

tan θ

appears in the original equation, and write

c = cos θ

7 c^{2} - 3 (1 - c^{2}) = c, 7 c^{2} - 3 + 3 c^{2} = c, 10 c^{2} - c - 3 = 0.

This quadratic factorises as

(5 c - 3) (2 c + 1) = 0

, so

c = \frac{3}{5} or c = - \frac{1}{2} .

Both values lie in

[- 1, 1]

, so both are attainable by a cosine and neither has to be discarded.

The answer is D.

ENGAA 2020 Section 1, Part B (Mathematics questions) · Q35

Find the number of solutions of the equation

14 cos^{3} x + 10 sin^{2} x cos x = 13 cos x

in the range

- 2 π \leq x \leq 2 π

A $4$
B $6$
C $8$
D $10$
E $12$
F $14$

Worked solution

Answer:E

Move everything to one side and take out the common factor

cos x

cos x (14 cos^{2} x + 10 sin^{2} x - 13) = 0.

Replacing

sin^{2} x

1 - cos^{2} x

simplifies the bracket:

14 cos^{2} x + 10 (1 - cos^{2} x) - 13 = 4 cos^{2} x - 3,

so the equation is

cos x (4 cos^{2} x - 3) = 0

.

The first factor gives

cos x = 0

, hence

x = \pm \frac{π}{2}

and

x = \pm \frac{3 π}{2}

: four solutions in the range.

The second factor gives

cos^{2} x = \frac{3}{4}

, hence

cos x = \frac{3}{2}

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

. The range

- 2 π \leq x \leq 2 π

covers two full periods, so each of these two values of the cosine is attained four times:

cos x = \frac{3}{2} : x = - \frac{11 π}{6}, - \frac{π}{6}, \frac{π}{6}, \frac{11 π}{6},

cos x = - \frac{3}{2} : x = - \frac{7 π}{6}, - \frac{5 π}{6}, \frac{5 π}{6}, \frac{7 π}{6} .

The twelve values are distinct and all lie strictly inside the range, so the equation has

4 + 4 + 4 = 12

solutions.

The answer is E.