5.6d Solving Trigonometric Equations (Involving Addition Formulae And Double Angle Formulae)

Posted on April 22, 2020 by user

(D) Solving Trigonometric Equations (Involving Addition Formulae and Double Angle Formulae)

Example 1 (Addition Formulae):
Solve the following equation for 0^o ≤ x ≤ 360^o:

(a) sin ( x – 25^o) = 3 sin ( x + 25^o)

(b) 3 cos ( 2x + 10^o) = 2

Solution:

(a)
sin ( x – 25^o) = 3 sin ( x + 25^o)

sin x cos 25^o – cos x sin 25^o = 3 (sin x cos 25^o + cos x sin 25^o)

sin x cos 25^o – cos x sin 25^o = 3 sin x cos 25^o + 3 cos x sin 25^o

– sin x cos 25^o = 4 cos x sin 25^o

\frac{\sin x}{\cos x} = - \frac{4 \sin 25^{\circ}}{2 \cos 25^{\circ}}

tan x = – 2 tan 25^o

tan x = – 2 (0.4663)

tan x = – 0.9326

basic angle x = 43^o

The reference angles of x = 43^o are in the second and fourth quadrants.

Hence, x = 180^o– 43^o, 360^o– 43^o

x = 137^o, 317^o

(b)

3 cos ( 2x + 10^o) = 2 ← (Take the angles in the range of 0^o ≤ x ≤ 720^o, which in 2 complete revolutions)

cos ( 2x + 10^o) = ⅔

basic angle ( 2x + 10^o) = 48.19^o

2x + 10^o = 48.19^o, 360^o– 48.19^o, 360^o+ 48.19^o, 720^o– 48.19^o

2x + 10^o = 48.19^o, 311.81^o, 408.19^o, 671.81^o

2x = 38.19^o, 301.81^o, 398.19^o, 661.81^o

x = 19.10^o, 150.91^o, 199.10^o, 330.91^o

Example 2 (Double Angle):

Find all the angles that satisfy the equation 5 cos 2A + 9 sin A = 7,
0° < A < 360°.

Solution:

5 cos 2A + 9 sin A = 7

5 (1 – 2 sin²A) + 9 sin A = 7 ← (substitution of cos 2A = 1 – 2 sin²A is used. Then the whole equation will be in terms of sin A)

5 – 10 sin²A + 9 sin A – 7 = 0

– 10 sin²A + 9 sin A – 2 = 0

10 sin²A – 9 sin A + 2 = 0

(2 sin A – 1)(5 sin A – 2) = 0 ← (Factorise)

sin A = ½ = 0.5 or sin A=

\frac{2}{5}

= 0.4

When sin A = 0.5,

Basic angle = 30º

A = 30º, 180º – 30º

A = 30º, 150º

When sin A = 0.4,

Basic angle = 23.58º

A = 23.58º, 180º – 23.58º

A = 23.58º, 156.42º

Hence A = 23.58º, 30º, 150º, 156.42º.

Example 3 (Double Angle):

Find all the possible values of θ between 0 and 2π rad for the equation sin 2θ = sin θ

Solution:

sin 2θ = sin θ

2 sin θ cos θ = sin θ ← (sin 2θ = 2 sin θ cos θ)

2 sin θ cos θ – sin θ = 0

sin θ (2 cos θ – 1) = 0 ← (Factorise)

sin θ = 0 or 2 cos θ – 1 = 0

When sin θ = 0

θ = 0, π, 2π

When 2 cos θ – 1= 0

cos θ = ½

\begin{array}{l} θ = \frac{1}{3} π, \frac{5}{3} π \\ Hence, θ = 0, \frac{1}{3} π, π, \frac{5}{3} π, 2 π . \end{array}