5.2.2 Six Trigonometric Functions of Any Angle

Posted on April 22, 2020 by user

5.2b Six Trigonometric Functions of Any Angle

(B) Special Angles

(1) Value of Special Angle 30° and 60°

$\begin{array}{l} (a) \sin 30^{o} = \frac{1}{2} (b) c o s 30^{o} = \frac{\sqrt{3}}{2} (c) t a n 30^{o} = \frac{1}{\sqrt{3}} \\ (d) \sin 60^{o} = \frac{\sqrt{3}}{2} (e) c o s 60^{o} = \frac{1}{2} (f) t a n 60^{o} = \sqrt{3} \end{array}$

(2) Value of Special Angle 45°

$(a) \sin 45^{o} = \frac{1}{\sqrt{2}} (b) c o s 45^{o} = \frac{1}{\sqrt{2}} (c) t a n 45^{o} = 1$

(3) Value of Special Angle 0°, 90°, 180°, 270°, 360°

(a) y = sin x

x	0^o	90^o	180^o	270^o	360^o
sin	0	1	0	-1	0

(b) y = cos x

(c) y = tan x

x	0^o	90^o	180^o	270^o	360^o
tan	0	∞	0	∞	0

Short Question 4 – 6

Posted on April 22, 2020 by user

Question 4:
Solve the equation 3 sin A cos A – cos A = 0 for 0° ≤ A ≤ 360°.

Solution:
3 sin A cos A – cos A = 0
cos A (3 sin A – 1) = 0
cos A = 0 or sin A = ⅓

cos A = 0
A = 90°, 270°

sin A = ⅓
Basic angle = 19°28'
A = 19°28', 180° – 19°28'
A = 19°28', 160°32'

Hence A = 19°28', 90°, 160°32', 270°.

Question 5:

Solve the equation 4 sin (x – π) cos (x – π) = 1 for 0^o ≤ x ≤ 360^o.

Solution:

4 sin (x – π) cos (x – π) = 1

2 [2 sin (x – π) cos (x – π)] = 1

2 sin (x – π) cos (x – π) = ½

sin 2(x – π) = ½ ← (sin 2x= 2 sinx cosx)

sin 2(x – 180^o) = ½ ← (π rad = 180^o)

sin (2x – 360^o) = ½

sin 2x cos 360^o – cos 2x sin 360^o = ½

sin 2x (1) – cos 2x (0) = ½ ← (cos 360^o = 1, sin 360^o = 0)

sin 2x = ½

basic angle = 30^o ← (special angle, sin 30^o= ½)

2x = 30^o, 150^o, 390^o, 510^o

x = 15^o, 75^o, 195^o, 255^o

Long Question 1 & 2

Posted on April 22, 2020 by user

Question 1:

(a) Sketch the graph of y = cos 2x for 0° ≤ x ≤ 180°.

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

$2 {sin}^{2} x = 2 - \frac{x}{180}$ for 0° ≤ x ≤ 180°.

Solution:
(a)(b)

$\begin{array}{l} 2 {sin}^{2} x = 2 - \frac{x}{180} \\ 1 - 2 {sin}^{2} x = 1 - (2 - \frac{x}{180}) \\ \cos 2 x = \frac{x}{180} - 1 \\ y = \frac{x}{180} - 1 \\ x = 0, y = - 1 \\ x = 180, y = 0 \\ Number of solutions = 2 \end{array}$

Question 2:
(a) Sketch the graph of

$y = \frac{3}{2} \cos 2 x for 0 \leq x \leq \frac{3}{2} π .$
(b) Hence, using the same axes, sketch a suitable straight line to find the number of solutions to the equation

$\frac{4}{3 π} x - \cos 2 x = \frac{3}{2} for 0 \leq x \leq \frac{3}{2} π$
State the number of solutions.

Solution:
(a)(b)

$\begin{array}{l} \frac{4}{3 π} x - \cos 2 x = \frac{3}{2} \\ \cos 2 x = \frac{4}{3 π} x - \frac{3}{2} \\ \frac{3}{2} \cos 2 x = \frac{3}{2} (\frac{4}{3 π} x - \frac{3}{2}) \\ y = \frac{2}{π} x - \frac{9}{4} \\ To sketch the graph of y = \frac{2}{π} x - \frac{9}{4} \\ x = 0, y = - \frac{9}{4} \\ x = \frac{3 π}{2}, y = \frac{3}{4} \\ Number of solutions \\ = Number of intersection points \\ = 3 \end{array}$

5.3.4 Sketching Graphs of Trigonometric Functions (Part 3)

Posted on April 22, 2020 by user

5.3.4 Sketching Graphs of Trigonometric Functions (Part 3)

Example 2:

(a) Sketch the graph y = –½ cos x for 0 ≤ x ≤ 2π.

(b) Hence, using the same axes, sketch a suitable graph to find the number of solutions to the equation

$\frac{π}{2 x} + \cos x = 0$ for 0 ≤ x ≤ 2π.

State the number of solutions.

Solution:

(a)

(b)

$\begin{array}{l} \frac{π}{2 x} + \cos x = 0 \\ \frac{π}{2 x} = - \cos x \\ \frac{π}{4 x} = - \frac{1}{2} \cos x \leftarrow Multiply both sides by \frac{1}{2} \\ y = \frac{π}{4 x} \leftarrow y = - \frac{1}{2} \cos x \end{array}$

The suitable graph to draw is

$y = \frac{π}{4 x} .$

x	$\frac{π}{2}$	π	2π
$y = \frac{π}{4 x}$	½	¼	⅛

From the graphs, there are two points of intersection for 0 ≤ x ≤ 2π.

Number of solutions = 2.

5.3.2 Sketching Graphs of Trigonometric Functions (Part 1)

Posted on April 22, 2020 by user

5.3.2 Sketching Graphs of Trigonometric Functions

Example 1:

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

(a) y = 3 sin x
(b) y = 2 cos x

Solution:
(a)

(b)

(c)

(d)

(e)

(f)

Basic Trigonometric Identities

Posted on April 22, 2020 by user

5.4 Basic Trigonometric Identities

Three basic trigonometric identities are:

sin² x + cos² x = 1

tan² x + 1 = sec² x

cot² x + 1 = cosec² x

[adinserter block="3"]

Example 1 (To Prove Trigonometric Identities which involve the Three Basic Identities)

Prove each of the following trigonometric identities.

(a) sin²x – cos² x = 1 – 2 cos² x

(b) (1 – cosec²x) (1– sec² x) = 1

Solution:

(a)

sin²x– cos² x = 1 – 2 cos²x

LHS: sin²x – cos² x

= 1 – cos² x – cos² x

= 1 – 2 cos² x (RHS)

(b)

$\begin{array}{l} (1 - {cosec}^{2} x) (1 - \sec^{2} x) = 1 \\ LHS: (1 - {cosec}^{2} x) (1 - \sec^{2} x) \\ = (- \cot^{2} x) (- \tan^{2} x) \\ = (\cot^{2} x) (\tan^{2} x) \\ = (\frac{1}{\tan^{2} x}) \tan^{2} x \\ = 1 (RHS) \end{array}$

Example 2 (To Solve Trigonometric Equations which involve the Three Basic Identities)

Solve the following trigonometric equations for 0^o≤ x ≤ 360^o.

(a) sin²x cos x + 1 = cos x

(b) 2 cosec²x – 5 cot x = 0

Solution:

(a)

sin²x cos x + 1 = cos x

(1 – cos²x) cos x + 1 = cos x

cos x – cos³x + 1 = cos x

cos³x = 1

cos x = 1

x = 0^o, 360^o

(b)

2 cosec²x – 5 cot x = 0

2 (1 + cot²x) – 5 cot x = 0

2 + 2 cot²x – 5 cot x = 0

2 cot²x – 5 cot x + 2 = 0

(2 cotx – 1) (cot x – 2) = 0

cotx= ½ or cotx = 2

tan x = 2 tan x = ½

x =63.43^o, 243.43^o x = 26.57^o, 206.57^o

(Note: tangent is positive in the first and third quadrants)

Thus, x = 26.57^o, 63.43^o, 206.57^o, 243.43^o

5.3.1 General Pattern for y = sinx, y = cosx and y = tanx

Posted on April 22, 2020 by user

5.3.1 General pattern for y = sin x , y = cos x and y = tan x

(1) Graph of y = sin x

x	0^o	90^o	180^o	270^o	360^o
sin x	0	1	0	-1	0

(2) Graph of y = cos x

x	0^o	90^o	180^o	270^o	360^o
cos x	1	0	-1	0	1

(3) Graph of y = tan x

x	0^o	90^o	180^o	270^o	360^o
tan x	0	¥	0	¥	0