5.5.1a Proving Trigonometric Identities Using Addition Formula And Double Angle Formulae (Part 2)

Posted on May 22, 2020 by Myhometuition

Example 3:

(a) Given that

\sin P = \frac{3}{5} and \sin Q = \frac{5}{13},

such that P is an acute angle and Q is an obtuse angle, without using tables or a calculator, find the value of cos (P + Q).

(b) Given that

\sin A = - \frac{3}{5} and \sin B = \frac{12}{13},

such that A and B are angles in the third and fourth quadrants respectively, without using tables or a calculator, find the value of sin (A – B).

Solution:

(a)

\begin{array}{l} \sin P = \frac{3}{5}, \cos P = \frac{4}{5} \\ \sin Q = \frac{5}{13}, \cos Q = - \frac{12}{13} \\ \cos (P + Q) \\ = \cos A \cos B - \sin A \sin B \\ = (\frac{4}{5}) (- \frac{12}{13}) - (\frac{3}{5}) (\frac{5}{13}) \\ = - \frac{48}{65} - \frac{15}{65} \\ = - \frac{63}{65} \end{array}

(b)

\begin{array}{l} \sin A = - \frac{3}{5}, \cos A = - \frac{4}{5} \\ \sin B = - \frac{5}{13}, \cos B = \frac{12}{13} \\ s i n (A - B) \\ = s i n A \cos B - c o s A \sin B \\ = (- \frac{3}{5}) (\frac{12}{13}) - (- \frac{4}{5}) (- \frac{5}{13}) \\ = - \frac{36}{65} - \frac{20}{65} \\ = - \frac{56}{65} \end{array}

5.3.3 Sketching Graphs of Trigonometric Functions (Part 2)

Posted on May 22, 2020 by Myhometuition

5.3.3 Sketching Graphs of Trigonometric Functions

Example 1:

Sketch the graph of each of the following trigonometric functions for π ≤ x ≤ 2π.

(g) y = –sin x
(h) y = –cos x

(i) y = | sin x |
(j) y = | cos x |

(k) y = tan 2x

Solution:

(g)

(h)

(i)

(j)

(k)

Short Question 19

Posted on May 18, 2020 by Myhometuition

Question 19 (4 marks):
It is given that cos α = t where t is a constant and 0^o≤ α ≤ 90^o.
Express in terms of t
(a) sin (180^o + α),
(b) sec 2α.

Solution:
(a)

\begin{array}{l} sin (180^{o} + α) \\ = \sin 180 \cos α + \cos 180 \sin α \\ = 0 - \sin α \\ = - \sin α \\ = - \sqrt{1 - t^{2}} \end{array}

(b)

\begin{array}{l} \sec 2 α = \frac{1}{\cos 2 α} \\ = \frac{1}{2 \cos^{2} α - 1} \\ = \frac{1}{2 t^{2} - 1} \end{array}

Long Question 6

Posted on May 18, 2020 by Myhometuition

Question 6 (10 marks):
(a) Prove that 2 tan x cos² x = sin 2x.

(b) Hence, solve the equation 4 tan x cos² x = 1 for 0 ≤ x ≤ 2π.

(c)(i) Sketch the graph of y = sin 2x for 0 ≤ x ≤ 2π.

(c)(ii) Hence, using the same axes, sketch a suitable straight line to find the number of solutions for the equation 4π tan x cos² x = x – 2π for 0 ≤ x ≤ 2π.
State the number of solutions.

Solution:
(a)

\begin{array}{l} 2 \tan x \cos^{2} x = \sin 2 x \\ Left hand side \\ = 2 \tan x \cos^{2} x \\ = 2 \times \frac{\sin x}{\cos x} \times \cos^{2} x \\ = 2 \sin x \cos x \\ = \sin 2 x \\ = Right hand side (Proven) \end{array}

(b)

\begin{array}{l} 4 \tan x \cos^{2} x = 1, 0 \leq x \leq 2 π \\ 2 (2 \tan x \cos^{2} x) = 1 \\ 2 \sin 2 x = 1 \\ \sin 2 x = \frac{1}{2} \\ Basic angle = \frac{π}{6} \\ 2 x = \frac{π}{6}, (π - \frac{π}{6}), (2 π + \frac{π}{6}), (3 π - \frac{π}{6}) \\ 2 x = \frac{π}{6}, \frac{5 π}{6}, \frac{13 π}{6}, \frac{17 π}{6} \\ x = \frac{π}{12}, \frac{5 π}{12}, \frac{13 π}{12}, \frac{17 π}{12} \end{array}

(c)(i)
y = sin 2x, 0 ≤ x ≤ 2π.

(c)(ii)

\begin{array}{l} 4 π \tan x \cos^{2} x = x - 2 π \\ 2 π (2 \tan x \cos^{2} x) = x - 2 π \\ 2 π \sin 2 x = x - 2 π \\ \sin 2 x = \frac{x}{2 π} - \frac{2 π}{2 π} \\ \sin 2 x = \frac{x}{2 π} - 1 \\ y = \frac{x}{2 π} - 1 \end{array}

Number of solutions = 4

Long Question 5

Posted on May 18, 2020 by Myhometuition

Question 5 (10 marks):

\begin{array}{l} (a) Prove \sin (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) = \cos 3 x \\ (b) Hence, \\ (i) solve the equation s i n (\frac{3 x}{2} + \frac{π}{6}) - \sin (\frac{3 x}{2} - \frac{π}{6}) = \frac{1}{2} for 0 \leq x \leq 2 π \\ and give your answer in the simplest fraction form in terms of π radian . \\ (i i) sketch the graph of y = s i n (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) - \frac{1}{2} for 0 \leq x \leq π . \end{array}

Solution:

\begin{array}{l} (a) Left hand side, \\ \sin (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) \\ = [\sin 3 x \cos \frac{π}{6} + \cos 3 x \sin \frac{π}{6}] - [\sin 3 x \cos \frac{π}{6} - \cos 3 x \sin \frac{π}{6}] \\ = 2 [\cos 3 x \sin \frac{π}{6}] \\ = 2 [\cos 3 x (\frac{1}{2})] \\ = \cos 3 x (right hand side) \end{array}

\begin{array}{l} (b) (i) \\ \sin (\frac{3 x}{2} + \frac{π}{6}) - \sin (\frac{3 x}{2} - \frac{π}{6}) = \frac{1}{2}, 0 \leq x \leq 2 π \\ \cos \frac{3 x}{2} = \frac{1}{2} \\ \frac{3 x}{2} = \frac{π}{3}, (2 π - \frac{π}{3}), (2 π + \frac{π}{3}) \\ \frac{3 x}{2} = \frac{π}{3}, \frac{5 π}{3}, \frac{7 π}{3} \\ x = \frac{2 π}{9}, \frac{10 π}{9}, \frac{14 π}{9} \end{array}

\begin{array}{l} (b) (i i) \\ y = s i n (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) - \frac{1}{2} for 0 \leq x \leq π . \\ y = \cos 3 x - \frac{1}{2} \end{array}

Long Question 4

Posted on May 17, 2020 by Myhometuition

Question 4:

Find all the angles between 0° and 360° which satisfy the following equations:

(a) 2 sin ( 2x – 50^o) = –1

(b) 15 sin²x = sin x + 4 sin 30^o

(c) 7 sin x cos x = 1

Solution:
(a)

2 sin ( 2x – 50^o) = –1

sin ( 2x – 50^o) = – ½

basic angle ( 2x – 50^o) = –30^o ← (sin is negative at third and fourth quadrants)

2x – 50^o = –30^o, 180^o+ 30^o, 360^o – 30^o, 360^o + 180^o+ 30^o

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

2x – 50^o = –30^o, 210^o, 330^o, 570^o

2x = 20^o, 260^o, 380^o, 620^o

x = 10^o, 130^o, 190^o, 310^o

(b)

15 sin²x = sin x + 4 sin 30^o

15 sin²x = sin x + 4 (½) ← (sin 30^o = ½)

15 sin²x = sin x + 2

15 sin²x – sin x – 2 = 0

(5 sin x – 2)(3 sin x + 1) = 0

sin x =

\frac{2}{5}

or sin x = – ⅓

When sin x =

\frac{2}{5}

Basic angle x = 23º 35’

x = 23º 35’, 180º – 23º 35’

x = 23º 35’, 156º 25’

When sin x = – ⅓ ← (sin is negative at third and fourth quadrants)

Basic angle x = 19º 28’

x = 180º + 19º 28’, 360º – 19º 28’

x = 199º 28’, 340º 32’

Hence x = 23º 35’, 156º 25’, 199º 28’, 340º 32’.

(c)

7 sin x cos x = 1

sin x cos x =

\frac{1}{7}

2 sin x cos x =

\frac{2}{7}

← ( × 2 for both sides)

sin 2x =

\frac{2}{7}

sin 2x = 0.2857

Basic angle x = 16º 36’

2x = 16º 36’, 180º – 16º 36’, 360º + 16º 36’, 360º + 180º – 16º 36’

2x = 16º 36’, 163º 24’, 376º 36’, 523º 24’

x = 8º 18’, 81º 42’, 188º 18’, 261º 42’

Hence x = 8º 18’, 81º 42’, 188º 18’, 261º 42’.

Long Question 3

Posted on May 17, 2020 by Myhometuition

Question 3:

(a) Prove that

\frac{2 \tan x}{2 - \sec^{2} x} = \tan 2 x .

(b)(i) Sketch the graph of y = – tan 2x for 0 ≤ x ≤ π .

(b)(ii) Hence, by drawing a suitable straight line on the same axes, find the number of solutions satisfying the equation

\frac{3 x}{π} + \frac{2 \tan x}{2 - \sec^{2} x} = 0

for 0 ≤ x ≤ π .

State the number of solutions.

Solution:

(a)

\begin{array}{l} \frac{2 \tan x}{2 - \sec^{2} x} = \tan 2 x \\ L H S : \\ \frac{2 \tan x}{2 - \sec^{2} x} = \frac{2 \tan x}{2 - (1 + \tan^{2} x)} \\ = \frac{2 \tan x}{2 - \tan^{2} x} \\ = \tan 2 x (RHS) \end{array}

(b)(i)

(b)(ii)

\begin{array}{l} \frac{3 x}{π} + \frac{2 \tan x}{2 - \sec^{2} x} = 0 \\ \frac{3 x}{π} + \tan 2 x = 0 \leftarrow from part (a) \\ - \tan 2 x = \frac{3 x}{π} \\ y = \frac{3 x}{π} \\ The suitable straight line to sketch is y = \frac{3 x}{π} . \end{array}

When x = 0, y = 0.

When x = π, y = 3.

Number of solutions = 3

Short Question 7 & 8

Posted on April 22, 2020 by user

Question 7:

It is given that

\sin A = \frac{5}{13} and \cos B = \frac{4}{5}

, where A is an obtuse angle and B is an acute angle.

Find

(a) tan A

(b) sin (A + B)

(c) cos (A – B)

Solution:

(a)

\tan A = - \frac{5}{12}

(b)

\begin{array}{l} \sin (A + B) = \sin A \cos B + \cos A \sin B \\ \sin (A + B) = (\frac{5}{13}) (\frac{4}{5}) + (- \frac{12}{13}) (\frac{3}{5}) \leftarrow \begin{array}{l} \cos A = - \frac{12}{13} \\ \sin B = \frac{3}{5} \end{array} \\ \sin (A + B) = \frac{4}{13} - \frac{36}{65} \\ \sin (A + B) = - \frac{16}{65} \end{array}

(c)

\begin{array}{l} \cos (A - B) = \cos A \cos B + \sin A \sin B \\ \cos (A - B) = (- \frac{12}{13}) (\frac{4}{5}) + (\frac{5}{13}) (\frac{3}{5}) \\ \cos (A - B) = - \frac{33}{65} \end{array}

Question 8:

If sin A = p, and 90° < A < 180°, express in terms of p

(a) tan A

(b) cos A

(c) sin 2A

Solution:

\begin{array}{l} Using Pythagoras Theorem, \\ Adjacent side \\ = \sqrt{1^{2} - p^{2}} \\ = \sqrt{1 - p^{2}} \end{array}

(a)

\tan A = - \frac{p}{\sqrt{1 - p^{2}}} \leftarrow \begin{array}{l} tan is negative at \\ second quadrant \end{array}

(b)

\cos A = - \sqrt{1 - p^{2}} \leftarrow \begin{array}{l} \cos is negative at \\ second quadrant \end{array}

(c)

\begin{array}{l} \sin A = 2 \sin A \cos A \\ \sin A = 2 (p) (- \sqrt{1 - p^{2}}) \\ \sin A = - 2 p \sqrt{1 - p^{2}} \end{array}

Short Question 11 – 14

Posted on April 22, 2020 by user

Question 11:

Prove the identity \frac{\cos^{2} x}{1 - \sin x} = 1 + \sin x

Solution:

\begin{array}{l} LHS \\ = \frac{\cos^{2} x}{1 - \sin x} \\ = \frac{1 - \sin^{2} x}{1 - \sin x} \leftarrow \sin^{2} x + \cos^{2} x = 1 \\ = \frac{(1 + \sin x) (1 - \sin x)}{1 - \sin x} \\ = 1 + \sin x \\ = RHS \\ ∴ Proven \end{array}

Question 12:

Prove the identity \sin^{2} x - \cos^{2} x = \frac{\tan^{2} x - 1}{\tan^{2} x + 1}

Solution:

\begin{array}{l} RHS \\ = \frac{\tan^{2} x - 1}{\tan^{2} x + 1} \\ = \frac{\frac{\sin^{2} x}{\cos^{2} x} - 1}{\frac{\sin^{2} x}{\cos^{2} x} + 1} \leftarrow \tan x = \frac{\sin x}{\cos x} \\ = \frac{\frac{\sin^{2} x - \cos^{2} x}{\cos^{2} x}}{\frac{\sin^{2} x + \cos^{2} x}{\cos^{2} x}} \\ = \frac{\sin^{2} x - \cos^{2} x}{\sin^{2} x + \cos^{2} x} \\ = \sin^{2} x - \cos^{2} x \leftarrow \sin^{2} x + \cos^{2} x = 1 \\ = LHS \\ ∴ Proven \end{array}

Question 13:

Prove the identity \tan^{2} θ - \sin^{2} θ = \tan^{2} θ \sin^{2} θ

Solution:

\begin{array}{l} LHS \\ = \tan^{2} θ - \sin^{2} θ \\ = \frac{\sin^{2} θ}{\cos^{2} θ} - \sin^{2} θ \\ = \frac{\sin^{2} θ - \sin^{2} θ \cos^{2} θ}{\cos^{2} θ} \\ = \frac{\sin^{2} θ (1 - \cos^{2} θ)}{\cos^{2} θ} \\ = \frac{\sin^{2} θ \sin^{2} θ}{\cos^{2} θ} \\ = (\frac{\sin^{2} θ}{\cos^{2} θ}) (\sin^{2} θ) \\ = \tan^{2} θ \sin^{2} θ \\ = RHS \\ ∴ Proven \end{array}

Question 14:

{Prove the identity cosec}^{2} θ (\sec^{2} θ - \tan^{2} θ) - 1 = \cot^{2} θ

Solution:

\begin{array}{l} LHS \\ = {cosec}^{2} θ (\sec^{2} θ - \tan^{2} θ) - 1 \\ = {cosec}^{2} θ (1) - 1 \leftarrow \begin{array}{l} \tan^{2} θ + 1 = \sec^{2} θ \\ \sec^{2} θ - \tan^{2} θ = 1 \end{array} \\ = {cosec}^{2} θ - 1 \\ = \cot^{2} θ \leftarrow \begin{array}{l} 1 + \cot^{2} θ = \cos e c^{2} θ \\ \cos e c^{2} θ - 1 = \cot^{2} θ \end{array} \\ = RHS \\ ∴ Proven \end{array}

5.6b Solving Trigonometric Equation (Factorization)

Posted on April 22, 2020 by user

5.6b Solving Trigonometric Equation (Factorization)

Example:

Find all the angles that satisfy each of the following equations for 0° £ x £ 360°.

(a)

\cot x = - 2 \cos x

(b)

3 \sec x = 4 \cos x

(c)

16 \tan x = \cot x

Solution:
(a)

\begin{array}{l} \cot x = - 2 \cos x \\ \frac{\cos x}{\sin x} = - 2 \cos x \\ \cos x = - 2 \cos x \sin x \\ \cos x + 2 \sin x \cos x = 0 \\ \cos x (1 + 2 \sin x) = 0 \\ \cos x = 0 \\ x = 90^{\circ}, 270^{\circ} \\ 1 + 2 \sin x = 0 \\ \sin x = - \frac{1}{2} \\ Basic ∠ = 3 0^{\circ} \\ x = (180^{\circ} + 30^{\circ}), (360^{\circ} - 30^{\circ}) \\ x = 210^{\circ}, 330^{\circ} \\ ∴ x = 90^{\circ}, 210^{\circ}, 270^{\circ}, 330^{\circ} \end{array}

(b)

\begin{array}{l} 3 \sec x = 4 \cos x \\ \frac{3}{\cos x} = 4 \cos x \\ 3 = 4 \cos^{2} x \\ \cos^{2} x = \frac{3}{4} \\ \cos x = \pm \frac{\sqrt{3}}{2} \\ Basic ∠ = 30^{\circ} \\ x = 30^{\circ}, (180^{\circ} - 30^{\circ}), (180^{\circ} + 30^{\circ}), (360^{\circ} - 30^{\circ}) \\ x = 30^{\circ}, 150^{\circ}, 210^{\circ}, 330^{\circ} \end{array}

(c)

\begin{array}{l} 16 \tan x = \cot x \\ 16 \tan x = \frac{1}{\tan x} \\ \tan^{2} x = \frac{1}{16} \\ \tan x = \pm \frac{1}{4} \\ Basic ∠ = {14.04}^{\circ} \\ x = {14.04}^{\circ}, (180^{\circ} - {14.04}^{\circ}), \\ (180^{\circ} + {14.04}^{\circ}), (360^{\circ} - {14.04}^{\circ}) \\ x = {14.04}^{\circ}, {165.96}^{\circ}, {194.04}^{\circ}, {345.96}^{\circ} \end{array}