Trigonometry

9 methods · 35 questions

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For an equation in $sin (k x)$ or $cos (k x)$ asked over an interval (how many solutions, the largest, or the sum), first transform the range of the inner argument (if $0 \leq x \leq L$ then $0 \leq k x \leq k L$ ), solve the base equation across that widened interval, then map each value back by dividing by $k$ . The widened interval is where you actually count.

Trigger: a multiple angle $k x$ inside the trig function and a 'how many / largest / sum of solutions' question.

Instances:
(i) stretch $0 \leq x \leq 2 π$ to $0 \leq 2 x \leq 4 π$ (or $72 0^{\circ}$ ) and list the $2 x$ values before dividing;
(ii) stretch to $0 \leq 3 θ \leq 54 0^{\circ}$ for a triple angle and count crossings of $cos 3 θ = k$ ;
(iii) intersect the $k x$ -solutions of two conditions, keep the common ones, then halve;
(iv) for a wide outer interval, count how many full periods fit and add the partial-period solutions.

Linked questions (3)

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 $33 0^{\circ}$

Worked solution

Answer:F

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:

3 0^{\circ}, 6 0^{\circ}, 21 0^{\circ}, 24 0^{\circ}

. The maximum is

24 0^{\circ}

. Hence the answer is F.

TMUA 2018 Paper 1 · Q18

What is the smallest positive value of

a

for which the line

x = a

is a line of symmetry of the graph of

y = sin (2 x - \frac{4 π}{3})

A $\frac{π}{12}$
B $\frac{5 π}{12}$
C $\frac{7 π}{12}$
D $\frac{11 π}{12}$
E $\frac{19 π}{12}$

Worked solution

Answer:B

Lines of symmetry of

y = sin (2 x - \frac{4 π}{3})

occur where the argument equals

\frac{π}{2} + nπ

for integer

n

(i.e., where

sin = \pm 1

). Solving

2 x - \frac{4 π}{3} = \frac{π}{2} + 2 nπ

2 x = \frac{4 π}{3} + \frac{π}{2} + 2 nπ = \frac{11 π}{6} + 2 nπ ⟹ x = \frac{11 π}{12} + nπ .

The smallest positive value is

\frac{11 π}{12}

(at

n = 0

). Solving

2 x - \frac{4 π}{3} = - \frac{π}{2} + 2 nπ

2 x = \frac{4 π}{3} - \frac{π}{2} + 2 nπ = \frac{5 π}{6} + 2 nπ ⟹ x = \frac{5 π}{12} + nπ .

The smallest positive value is

\frac{5 π}{12}

(at

n = 0

). Comparing:

\frac{5 π}{12} < \frac{11 π}{12}

, so the smallest positive value is

\frac{5 π}{12}

. Hence the answer is B.

TMUA 2018 Paper 2 · Q4

The non-zero real number

c

is such that the equation

cos x = c

has two solutions for

0 < x < \frac{3}{2} π

.

How many solutions of the equation

cos^{2} 2 x = c^{2}

are there in the range

0 < x < \frac{3}{2} π

A $2$
B $3$
C $4$
D $6$
E $7$
F $8$

Worked solution

Answer:D

Since

cos x = c

has two solutions in

(0, \frac{3}{2} π)

and

c \neq = 0

, we need

- 1 < c < 0

(from a sketch of

y = cos x

on this interval).

The equation

cos^{2} 2 x = c^{2}

is equivalent to

cos 2 x = \pm c

. Since

- 1 < c < 0

, we have

0 < - c < 1

and

- 1 < c < 0

, so both

c

and

- c

are in

(- 1, 1)

.

Considering

y = cos 2 x

0 < x < \frac{3}{2} π

(equivalently

0 < 2 x < 3 π

), the graph of

cos

completes

1.5

full periods. In this range:
-

cos 2 x = - c

(a value in

(0, 1)

): the horizontal line

y = - c

intersects

y = cos 2 x

at 3 points.
-

cos 2 x = c

(a value in

(- 1, 0)

): the horizontal line

y = c

intersects

y = cos 2 x

at 3 points.

Total solutions:

3 + 3 = 6

. Hence the answer is D.