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

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

Short Question 6 & 7

Posted on April 22, 2020 by user

Question 6:
The points P, Q and R are collinear. It is given that $\vec{P Q} = 4 \underset{˜}{a} - 2 \underset{˜}{b}$ and

\vec{Q R} = 3 \underset{˜}{a} + (1 + k) \underset{˜}{b}

, where k is a constant. Find
(a) the value of k,
(b) the ratio of PQ : QR.

Solution:
(a)

\begin{array}{l} Note: If P, Q and R are collinear, \\ \vec{P Q} = m \vec{Q R} \\ 4 \underset{˜}{a} - 2 \underset{˜}{b} = m [3 \underset{˜}{a} + (1 + k) \underset{˜}{b}] \\ 4 \underset{˜}{a} - 2 \underset{˜}{b} = 3 m \underset{˜}{a} + m (1 + k) \underset{˜}{b} \\ Comparing vector: \\ \underset{˜}{a} : 4 = 3 m \\ m = \frac{4}{3} \\ \underset{˜}{b} : - 2 = m (1 + k) \\ - 2 = \frac{4}{3} (1 + k) \\ 1 + k = - \frac{6}{4} \\ k = - \frac{3}{2} - 1 \\ k = - \frac{5}{2} \end{array}

(b)

\begin{array}{l} \vec{P Q} = m \vec{Q R} \\ \vec{P Q} = \frac{4}{3} \vec{Q R} \\ \frac{\vec{P Q}}{\vec{Q R}} = \frac{4}{3} \\ ∴ P Q : Q R = 4 : 3 \end{array}

Question 7:
Given that

\underset{˜}{x} = 3 \underset{˜}{i} + m \underset{˜}{j}

and

\underset{˜}{y} = 4 \underset{˜}{i} - 3 \underset{˜}{j}

, find the values of m if the vector

\underset{˜}{x}

is parallel to the vector

\underset{˜}{y}

.

Solution:

\begin{array}{l} If vector \underset{˜}{x} is parallel to vector \underset{˜}{y} \\ \underset{˜}{x} = h \underset{˜}{y} \\ (3 \underset{˜}{i} + m \underset{˜}{j}) = h (4 \underset{˜}{i} - 3 \underset{˜}{j}) \\ 3 \underset{˜}{i} + m \underset{˜}{j} = 4 h \underset{˜}{i} - 3 h \underset{˜}{j} \\ Comparing vector: \\ \underset{˜}{i} : 3 = 4 h \\ h = \frac{3}{4} \\ \underset{˜}{j} : m = - 3 h \\ m = - 3 (\frac{3}{4}) = - \frac{9}{4} \end{array}

4.2 Multiplication of Vector by a Scalar and the Parallel Condition of Two Vectors

Posted on April 22, 2020 by user

4.2 Multiplication of Vector by a Scalar and the Parallel Condition of Two Vectors

1. When a vector

\underset{˜}{a}

is multiplied by a scalar k, the product is

k \underset{˜}{a}

. Its magnitude is k times the magnitude of the vector

\underset{˜}{a}

2. The vector

\underset{˜}{a}

is parallel to the vector

\underset{˜}{b}

if and only if

\underset{˜}{b} = k \underset{˜}{a}

, where k is a constant.

3. If the vectors

\underset{˜}{a}

and

\underset{˜}{b}

are not parallel and

h \underset{˜}{a} = k \underset{˜}{b}

, then h = 0 and k = 0.

Example 1:

If vectors

\underset{˜}{a} and \underset{˜}{b}

are not parallel and

(k - 7) \underset{˜}{a} = (5 + h) \underset{˜}{b}

, find the value of k and of h.

Solution:

The vectors

\underset{˜}{a} and \underset{˜}{b}

are not parallel, so

k – 7 = 0 → k = 7

5 + h = 0 → h = –5

4.3 Addition and Subtraction of Vectors

Posted on April 22, 2020 by user

4.3.2 Subtraction of Vectors

The subtraction of the vector

\underset{˜}{b}

from the vector

\underset{˜}{a}

is written as

\underset{˜}{a} - \underset{˜}{b}

. This operation can be considered as the addition of the vector

\underset{˜}{a}

with the negative vector of

\underset{˜}{b}

. Therefore

\underset{˜}{a} - \underset{˜}{b} = \underset{˜}{a} + (- \underset{˜}{b}) .

Example 1:

In the diagram above, vector

\vec{O P} = \underset{˜}{p}, \vec{O R} = \underset{˜}{r}

and Q divides PR in the ratio of 2 : 3. Find the following vectors in terms of

\underset{˜}{p} and \underset{˜}{r}

\begin{array}{l} (a) \vec{P R} \\ (b) \vec{O Q} \\ (c) \vec{Q M} if M is the midpoint of O R . \end{array}

Solution:

(a)

\begin{array}{l} \vec{P R} = \vec{P O} + \vec{O R} \\ = - \vec{O P} + \vec{O R} \\ = - \underset{˜}{p} + \underset{˜}{r} \end{array}

(b)

\begin{array}{l} \vec{O Q} = \vec{O P} + \vec{P Q} \\ = \vec{O P} + \frac{2}{5} \vec{P R} \\ = \underset{˜}{p} + \frac{2}{5} (- \underset{˜}{p} + \underset{˜}{r}) \\ = \underset{˜}{p} - \frac{2}{5} \underset{˜}{p} + \frac{2}{5} \underset{˜}{r} \\ = \frac{3}{5} \underset{˜}{p} + \frac{2}{5} \underset{˜}{r} \end{array}

(c)

\begin{array}{l} \vec{Q M} = \vec{Q O} + \vec{O M} \\ = - \vec{O Q} + \vec{O M} \\ = - \vec{O Q} + \frac{1}{2} \vec{O R} \\ = - (\frac{3}{5} \underset{˜}{p} + \frac{2}{5} \underset{˜}{r}) + \frac{1}{2} \underset{˜}{r} \\ = - \frac{3}{5} \underset{˜}{p} - \frac{2}{5} \underset{˜}{r} + \frac{1}{2} \underset{˜}{r} \\ = - \frac{3}{5} \underset{˜}{p} + \frac{1}{10} \underset{˜}{r} \end{array}

Long Question 5

Posted on April 22, 2020 by user

Question 5:
Diagram below shows a triangle KLM.

\begin{array}{l} It is given that K P : P L = 1 : 2, L R : R M = 2 : 1, \vec{K P} = 2 \underset{˜}{x}, \vec{K M} = 3 \underset{˜}{y} . \\ (a) Express in terms of \underset{˜}{x} and \underset{˜}{y}, \\ (i) \vec{M P} \\ (ii) \vec{M R} \\ (b) Given \underset{˜}{x} = 2 \underset{˜}{i} and \underset{˜}{y} = - \underset{˜}{i} + 4 \underset{˜}{j}, find \vec{| M R |} . \\ (c) Given \vec{M Q} = h \vec{M P} and \vec{Q R} = n \vec{K R}, where h and n are constants, \\ find the value of h and of n . \end{array}

Solution:
(a)(i)

\begin{array}{l} \vec{M P} = \vec{M K} + \vec{K P} \\ = - 3 \underset{˜}{y} + 2 \underset{˜}{x} \\ = 2 \underset{˜}{x} - 3 \underset{˜}{y} \end{array}

(a)(ii)

\begin{array}{l} \vec{M R} = \frac{1}{3} \vec{M L} \\ = \frac{1}{3} (\vec{M K} + \vec{K L}) \\ = \frac{1}{3} (- 3 \underset{˜}{y} + 6 \underset{˜}{x}) \\ = 2 \underset{˜}{x} - \underset{˜}{y} \end{array}

(b)

\begin{array}{l} \vec{M R} = 2 (2 \underset{˜}{i}) - (- \underset{˜}{i} + 4 \underset{˜}{j}) \\ = 4 \underset{˜}{i} + \underset{˜}{i} - 4 \underset{˜}{j} \\ = 5 \underset{˜}{i} - 4 \underset{˜}{j} \\ | \vec{M R} | = \sqrt{5^{2} + {(- 4)}^{2}} \\ = \sqrt{41} units \end{array}

(c)

\begin{array}{l} \vec{M Q} + \vec{Q R} = \vec{M R} \\ h \vec{M P} + n \vec{K R} = \vec{M R} \\ h (2 \underset{˜}{x} - 3 \underset{˜}{y}) + n (\vec{K M} + \vec{M R}) = 2 \underset{˜}{x} - \underset{˜}{y} \\ h (2 \underset{˜}{x} - 3 \underset{˜}{y}) + n (3 \underset{˜}{y} + 2 \underset{˜}{x} - \underset{˜}{y}) = 2 \underset{˜}{x} - \underset{˜}{y} \\ 2 h \underset{˜}{x} - 3 h \underset{˜}{y} + 2 n \underset{˜}{x} + 2 n \underset{˜}{y} = 2 \underset{˜}{x} - \underset{˜}{y} \\ (2 h + 2 n) \underset{˜}{x} + (- 3 h + 2 n) \underset{˜}{y} = 2 \underset{˜}{x} - \underset{˜}{y} \\ 2 h + 2 n = 2.......... (1) \\ - 3 h + 2 n = - 1.......... (2) \\ (1) - (2) : 5 h = 3 \\ h = \frac{3}{5} \\ From (1) : h + n = 1 \\ \frac{3}{5} + n = 1 \\ n = 1 - \frac{3}{5} \\ n = \frac{2}{5} \end{array}

Short Question 4 & 5

Posted on April 22, 2020 by user

Question 4:

Diagram below shows a parallelogram ABCD with BED as a straight line.

\begin{array}{l} Given that \vec{A B} = 7 \underset{˜}{p}, \vec{A D} = 5 \underset{˜}{q} and D E = 3 E B, express, in terms of \underset{˜}{p} and \underset{˜}{q} . \\ (a) \vec{B D} \\ (b) \vec{E C} \end{array}

Solution:
(a)

\begin{array}{l} Note: for parallelogram, \vec{A B} = \vec{D C} = 7 \underset{˜}{p}, \vec{A D} = \vec{B C} = 5 \underset{˜}{q} . \\ \vec{B D} = \vec{B A} + \vec{A D} \\ \vec{B D} = - 7 \underset{˜}{p} + 5 \underset{˜}{q} \end{array}

(b)

\begin{array}{l} \vec{D E} =3 \vec{E B} \\ \frac{\vec{E B}}{\vec{D E}} = \frac{1}{3} \to E B : D E = 1 : 3 \\ ∴ \vec{E B} = \frac{1}{4} \vec{D B} \\ = \frac{1}{4} (- \vec{B D}) \\ = \frac{1}{4} [- (- 7 \underset{˜}{p} + 5 \underset{˜}{q})] \leftarrow From (a) \\ = \frac{7}{4} \underset{˜}{p} + \frac{5}{4} \underset{˜}{q} \\ \vec{E C} = \vec{E B} + \vec{B C} \\ \vec{E C} = \frac{7}{4} \underset{˜}{p} + \frac{5}{4} \underset{˜}{q} + 5 \underset{˜}{q} \\ \vec{E C} = \frac{7}{4} \underset{˜}{p} + \frac{25}{4} \underset{˜}{q} \end{array}

Question 5:

Use the above information to find the values of h and k when r = 2p – 3q.

Solution:

\begin{array}{l} r = 2 p - 3 q \\ (h - 1) \underset{˜}{a} + (h + k) \underset{˜}{b} = 2 (5 \underset{˜}{a} - 7 \underset{˜}{b}) - 3 (- 2 \underset{˜}{a} + 3 \underset{˜}{b}) \\ (h - 1) \underset{˜}{a} + (h + k) \underset{˜}{b} = 10 \underset{˜}{a} - 14 \underset{˜}{b} + 6 \underset{˜}{a} - 9 \underset{˜}{b} \\ (h - 1) \underset{˜}{a} + (h + k) \underset{˜}{b} = 16 \underset{˜}{a} - 23 \underset{˜}{b} \\ Comparing vector: \\ h - 1 = 16 \\ h = 17 \\ h + k = - 23 \\ 17 + k = - 23 \\ k = - 40 \end{array}

Long Question 2

Posted on April 22, 2020 by user

Question 2:
Given that

\vec{A B} = (\begin{matrix} 10 \\ 14 \end{matrix}), \vec{O B} = (\begin{matrix} 4 \\ 6 \end{matrix})

and

\vec{C D} = (\begin{matrix} m \\ 7 \end{matrix})

, find
(a) the coordinates of A,
(b) the unit vector in the direction of

\vec{O A}

.
(c) the value of m if CD is parallel to AB .

Solution:
(a)

\begin{array}{l} \vec{A B} = \vec{A O} + \vec{O B} \\ (\begin{matrix} 10 \\ 14 \end{matrix}) = (\begin{matrix} x \\ y \end{matrix}) + (\begin{matrix} 4 \\ 6 \end{matrix}) \\ (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 10 \\ 14 \end{matrix}) - (\begin{matrix} 4 \\ 6 \end{matrix}) \\ \vec{A O} = (\begin{matrix} 6 \\ 8 \end{matrix}) \\ \vec{O A} = (\begin{matrix} - 6 \\ - 8 \end{matrix}) \\ A = (- 6, - 8) \end{array}

(b)

\begin{array}{l} | \vec{O A} | = \sqrt{{(- 6)}^{2} + {(- 8)}^{2}} \\ | \vec{O A} | = \sqrt{100} = 10 \\ the unit vector in the direction of \vec{O A} \\ = \frac{\vec{O A}}{| \vec{O A} |} \\ = \frac{(\begin{matrix} - 6 \\ - 8 \end{matrix})}{10} = \frac{1}{10} (\begin{matrix} - 6 \\ - 8 \end{matrix}) \\ = (\begin{matrix} - \frac{3}{5} \\ - \frac{4}{5} \end{matrix}) \end{array}

(c)

\begin{array}{l} Given \vec{C D} parallel \vec{A B} \\ ∴ \vec{C D} = k \vec{A B} \\ (\begin{matrix} m \\ 7 \end{matrix}) = k (\begin{matrix} 10 \\ 14 \end{matrix}) \\ (\begin{matrix} m \\ 7 \end{matrix}) = (\begin{matrix} 10 k \\ 14 k \end{matrix}) \\ 7 = 14 k \\ k = \frac{1}{2} \\ m = 10 k = 10 (\frac{1}{2}) = 5 \end{array}