Short Question 1 – 3

Posted on April 23, 2020 by user

Question 1:
A group of 4 men and 3 ladies are to be seated in a row for a photographing session. If the men and ladies want to be seated alternately (man-lady-man-lady...), calculate the number of different arrangements.

Solution:
The arrangements of 4 men and 3 ladies to be seated alternately are as follow:

M L M L M L M

The number of ways to arrange the seat for 4 men = 4!

The number of ways to arrange the seat for 3 ladies = 3!

Total number of different arrangements for men and ladies = 4! × 3! = 144

Question 2:
Ahmad has 6 durians, 5 watermelons and 2 papayas. If he wants to arrange these fruits in a row and the fruits of the same kind have to be grouped together, calculate the number of different arrangements. The sizes of the fruits are different.

Solution:
The number of ways to arrange 3 groups of fruits that are same kind = 3!

DDDDDD WWWWW PP

The number of ways to arrange 6 durians = 6!
The number of ways to arrange 5 watermelons = 5!
The number of ways to arrange 2 papayas = 2!

Therefore, the number of ways to arrange the fruits with same kind of fruits was grouped together
= 3! × 6! × 5! × 2!
= 1036800

Question 3:
Calculate the number of arrangements, without repetitions, of the letters from the word `SOMETHING' with the condition that they must begin with a vowel.

Solution:

$\begin{array}{l} Arrangement of letters begin with vowel \\ = O, M and I \\ =^{3} P_{1} \\ Arrangement for the rest of the letters \\ = 7! \\ ∴ Number of arrangements \\ =^{3} P_{1} \times 7! \\ = 15120 \end{array}$

6.3 Combinations

Posted on April 23, 2020 by user

6.3 Combinations

(1) The number of combinations of r objects chosen from n different objects is given by :

$^{n} C_{r} = \frac{n!}{r! (n - r)!}$

(2) A combination of r objects chosen from n different objects is a selection of a set of r objects chosen from n objects. The order of the objects in the chosen set is not taken into consideration.

$\begin{array}{l} N o t e : \\ (i)^{n} C_{0} = 1 \\ (i i)^{n} C_{n} = 1 \\ (i i i)^{n} C_{r} =^{n} C_{n - r} \end{array}$

Example 1:

$\begin{array}{l} {Calculate the value of}^{7} C_{2} \\ ^{7} C_{2} = \frac{7!}{(7 - 2)! \times 2!} \\ = \frac{7!}{5! \times 2!} \\ = \frac{7 \times 6 \times 5!}{5! \times 2!} \\ = \frac{7 \times 6}{2 \times 1} \\ = 21 \end{array}$

Calculator Computation:

Example 2:

There are 6 marbles, each with different colour, which are to be divided equally between 2 children. Find the number of different ways the division of the marbles can be done.

Solution:

Number of ways giving 3 marbles to the first child =

$^{6} C_{3}$

Number of ways giving the remaining 3 marbles =

$^{3} C_{3}$

So, the number of different ways the division of the marbles

$\begin{array}{l} =^{6} C_{3} \times^{3} C_{3} \\ = 20 \times 1 \\ = 20 \end{array}$

6.1 Permutations Part 2

Posted on April 23, 2020 by user

(C) Permutation of n Different Objects, Taken r at a Time

1. The number of permutations of n different objects, taken r at a time is given by:

2. A permutation of n different objects, taken r at a time, is an arrangement of a set of r objects chosen from n objects. The order of the objects in the chosen set is taken into consideration.

3. The number of permutations of n different objects, taken all at a time, is:

Example 1:

Evaluate each of the following.

$\begin{array}{l} {(a)}^{5} P_{2} \\ {(b)}^{7} P_{3} \\ {(c)}^{9} P_{4} \end{array}$

Solution:

$\begin{array}{l} {(a)}^{5} P_{2} = \frac{5!}{(5 - 2)!} = \frac{5!}{3!} \\ = \frac{5 \times 4 \times 3!}{3!} = 5 \times 4 = 20 \\ {(b)}^{7} P_{3} = \frac{7!}{(7 - 3)!} = \frac{7!}{4!} \\ = \frac{7 \times 6 \times 5 \times 4!}{4!} = 7 \times 6 \times 5 = 210 \\ {(c)}^{9} P_{4} = \frac{9!}{(9 - 4)!} = \frac{9!}{5!} \\ = \frac{9 \times 8 \times 7 \times 6 \times 5!}{5!} = 9 \times 8 \times 7 \times 6 = 3024 \end{array}$

Calculator Computation:

Short Question 5 – 8

Posted on April 23, 2020 by user

Question 4:
A committee that consists of 6 members is to be selected from 5 teachers and 4 students. Find the number of different committees that can be formed if
(a) there is no restriction,
(b) the number of teachers must exceed the number of students.

Solution:
(a)
Total number of committees = 5 + 4 = 9
6 members to be selected from 9 committees with no restriction

$^{= 9} C_{6} = 84$

(b)

$\begin{array}{l} If the number of teachers must exceed the \\ number of students, the combination \\ = 4 teachers 2 students + 5 teachers 1 student \\ =^{5} C_{4} \times^{4} C_{2} +^{5} C_{5} \times^{4} C_{1} \\ = 30 + 4 \\ = 34 \end{array}$

Question 5:
A school prefect committee that consists of 6 persons is to be chosen from 6 Malays, 5 Chinese and 4 Indians. Calculate the number of different committees that can be formed if the number of Malays, Chinese and Indians must be equal.

Solution:
Number of different committees that can be formed for 2 Malays, 2 Chinese and 2 Indians

$\begin{array}{l} =^{6} C_{2} \times^{5} C_{2} \times^{4} C_{2} \\ = 900 \end{array}$

Question 6:
There are 10 different flavour candies in a plastic bag.
Find
(a) the number of ways 3 candies can be chosen from the plastic bag.
(b) the number of ways at least 8 candies can be chosen from the plastic bag.

Solution:
(a)
Number of ways choosing 3 candies out of 10 candies

$\begin{array}{l} =^{10} C_{3} \\ = 120 \end{array}$

(b)
Number of ways choosing 8 candies =

$=^{10} C_{8}$
Number of ways choosing 9 candies =

$^{10} C_{9}$
Number of ways choosing 10 candies =

$^{10} C_{10}$

Hence, number of ways of choosing at least 8 candies

$\begin{array}{l} =^{10} C_{8} +^{10} C_{9} +^{10} C_{10} \\ = 56 \end{array}$

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

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

Short Question 1 – 3

Posted on April 22, 2020 by user

Question 1:

Solve the equation 3 cos 2A = 8 sin A – 5 for 0° ≤ A ≤ 360°.

Solution:

3 cos 2A = 8 sin A – 5

3(1–2 sin²A) = 8 sin A – 5

3 – 6 sin²A = 8 sin A – 5

6 sin²A + 8 sin A – 8 = 0

3 sin²A + 4 sin A – 4 = 0

(3 sin A – 2)(sin A + 2) = 0 ←( Factorise the equation )

3 sin A – 2 = 0

$\sin A = \frac{2}{3} \leftarrow (sin A is positive in the 1st and 2nd quadrants .)$

A = 41°49’, 138°11’

sin A + 2 = 0

sin A = –2 (no solution)

Hence A = 41°49’, 138°11’.

Question 2:

Solve the equation 2 cos 2x – cos x – 1 = 0 for 0° ≤ x ≤ 360°.

Solution:

2 cos 2x – cos x – 1 = 0

2 (2 cos² x – 1) – cos x – 1 = 0

4 cos² x– 2 – cos x – 1 = 0

4 cos² x– cos x – 3 = 0

(4 cos x + 3)(cos x – 1) = 0

cos x =

$- \frac{3}{4}$

basic angle = 41°24’

x = 138°36’, 221°24’

cos x = 1,

x = 0°, 360°

Hence x = 0°, 138°36’, 221°24’, 360°

Question 3:

Solve the equation 6 sec² x – 13 tan x = 0, 0° ≤ x ≤ 360°.

Solution:

6 sec² x – 13 tan x = 0

6 (1 + tan²x) – 13 tan x = 0
6 tan²x – 13 tan x + 6 = 0
(3 tan x – 2)(2 tan x – 3) = 0
tan x = 2/3 or tan x = 3/2

tan x = 2/3
Basic angle = 33.69°
x = 33.69°, 180° + 33.69°
x = 33.69°, 213.69°
Or
tan x = 3/2
Basic angle = 56.31°
x = 56.31°, 180° + 56.31°
x = 56.31°, 236.31°

Hence x = 33.69°, 56.31°, 213.69°, 236.31°.

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)

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

5.6c Solving Trigonometric Equation (Form Quadratic Equation In Sinx/ Cosx/ Tanx/ Cosecx/ Secx/ Cotx)

Posted on April 22, 2020 by user

(C) Solving Trigonometric Equation (Form Quadratic Equation in sinx/ cosx/ tanx/ cosecx/ secx/ cotx)

Example:
Find all the angles between 0° and 360° that satisfy each of the following equations.
(a) 3 sin² x – 2 sin x – 1 = 0
(b) 2 sin x = cosec x + 1
(c) 5 sin² x = 2 (1 + cos x)
(d) 2 sec x = 1 + cos x
(e) 2 cot² x + 8 = 7 cosec x

Solution:
(a)
3 sin² x – 2 sin x – 1 = 0
(3 sin x + 1)(sin x – 1) = 0
sin x = –⅓, sin x = 1
sin x = –⅓
basic angle = 19.47°
x = 180° + 19.47°, 360° – 19.47°
x = 199.47°, 340.53
sin x = 1, x = 90°
Hence x = 90°, 199.47°, 340.53°

(b)
2 sin x = cosec x + 1

$2 \sin x = \frac{1}{\sin x} + 1$

2 sin ² x = 1 + sin x
2 sin ² x – sin x – 1 =0
(2 sin x + 1)(sin x – 1) = 0
sin x = –½, sin x = 1
sin x = –½
basic angle = 30°
x = 180° + 30°, 360° – 30°
x = 210°, 330°
sin x = 1, x = 90°
Hence x = 90°, 210°, 330°

(c)
5 sin² x = 2 (1 + cos x)
5 (1 – cos² x) = 2 + 2 cos x
5 – 5 cos² x – 2 – 2 cos x = 0
– 5 cos² x – 2 cos x + 3 = 0
5 cos² x + 2 cos x – 3 = 0
(5 cos x – 3)(cos x + 1) = 0

$\begin{array}{l} \cos x = \frac{3}{5}, \cos x = - 1 \\ \cos x = \frac{3}{5} \end{array}$

basic angle = 53.13°
x = 53.13°, 360° – 53.13°
x = 53.13°, 306.87°
cos x = – 1
x = 180°
Hence x = 53.13°, 180°, 306.87°

(d)
2 sec x = 1 + cos x

$\frac{2}{\cos x} = 1 + \cos x$

2 = cos x + cos² x

cos² x + cos x – 2 = 0

(cos x – 1)(cos x + 2) = 0

cos x = 1

x = 0°, 360°

cos x = –2 (not accepted)

Hence x = 0°, 360°

(e)
2 cot² x + 8 = 7 cosec x
2 (cosec² x – 1) + 8 = 7 cosec x
2 cosec² x – 2 – 7 cosec x + 8 = 0

2 cosec²x – 7 cosec x + 6 = 0

(2 cosec x – 3)(cosec x – 2) = 0

$\begin{array}{l} \cos e c x = \frac{3}{2}, \cos e c x = 2 \\ \sin x = \frac{2}{3}, \sin x = \frac{1}{2} \\ \sin x = \frac{2}{3} \\ basic angle = {41.81}^{\circ} \\ x = {41.81}^{\circ}, 180 - {41.81}^{\circ} \\ \sin x = \frac{1}{2} \\ {basic angle = 30}^{\circ} \\ x = 30^{\circ}, 180 - 30^{\circ} \\ Hence x = 30^{\circ}, {41.81}^{\circ}, {138.19}^{\circ}, 150^{\circ} \end{array}$