FORM FOUR MATHEMATICS STUDY NOTES TOPIC FIVE & SIX

TOPIC 5: TRIGONOMETRY

Trigonometric Ratios

The Sine, Cosine and Tangent of an Angle Measured in the Clockwise and Anticlockwise Directions

Determine the sine, cosine and tangent of an angle measured in the clockwise and anticlockwise directions

The basic three trigonometrical ratios are sine, cosine and tangent which are written in short as Sin, Cos, and tan respectively.

Consider the following right angled triangle.

Also we can define the above triangle ratios by using a unit Circle centered at the origin.

If θis an obtuse angle (900<θ<1800) then the trigonometrical ratios are the same as the trigonometrical ratio of 1800-θ

If θis a reflex angle (1800< θ<2700) then the trigonometrical ratios are the same as that of θ- 1800

If θis a reflex angle (2700< θ< 3600), then the trigonometrical ratios are the same as that of 3600 -θ

We have seen that trigonometrical ratios are positive or negative depending on the size of the angle and the quadrant in which it is found.

The result can be summarized by using the following diagram.

Trigonometric Ratios to Solve Problems in Daily Life

Apply trigonometric ratios to solve problems in daily life

Example 1

Write the signs of the following ratios

1. Sin 1700
2. Cos 2400
3. Tan 3100
4. sin 300

Solution

a)Sin 1700

Since 1700 is in the second quadrant, then Sin 1700 = Sin (1800-1700) = Sin 100

∴Sin 1700 = Sin 100

b) Cos 2400 = -Cos (2400-1800)= -Cos 600

Therefore Cos 2400= -Cos 600

c) Tan 3100 = -Tan (3600-3100) = - Tan 500

Therefore Tan 3100= -Tan 500

d) Sin 3000= -sin (3600-3000) = -sin 600

Therefore sin 3000= - Sin 600

Relationship between Trigonometrical ratios

The above relationship shows that the Sine of angle is equal to the cosine of its complement.

Also from the triangle ABC above

Again using the ΔABC

b2 = a2+c2 (Pythagoras theorem)

And

Example 2

Given thatA is an acute angle and Cos A= 0.8, find

1. Sin A
2. tan A.

Example 3

If A and B are complementary angles,

Solution

If A and B are complementary angle

Then Sin A = Cos B and Sin B = Cos A

Example 4

Given that θand βare acute angles such that θ+ β= 900 and Sinθ= 0.6, find tanβ

Solution

Exercise 1

For practice

Sine and Cosine Functions

Sines and Cosines of Angles 0 Such That -720°≤ᶿ≥ 720°

Find sines and cosines of angles 0 such that -720°≤ᶿ≥ 720°

Positive and Negative angles

An angle can be either positive or negative.

Definition:

Positive angle: is an angle measures in anticlockwise direction from the positive X- axis

Negative angle: is an angle measured in clockwise direction from the positive X-axis

Facts:

1. From the above figure if is a positive angle then the corresponding negative angle to is (- 3600) or (+ - 3600)
2. .If is a negative angle, its corresponding positive angle is (360+)

Example 5

Find thecorresponding negative angle to the angle θif ;

1. θ= 580
2. θ= 2450

Example 6

What is the positive angle corresponding to - 46°?

SPECIAL ANGLES

The angles included in this group are 00, 300, 450, 600, 900, 1800, 2700, and 3600

Because the angle 00, 900, 1800, 2700, and 3600, lie on the axes then theirtrigonometrical ratios are summarized in the following table.

The ∆ ABC is an equilateral triangle of side 2 units

For the angle 450 consider the following triangle

The following table summarizes the Cosine, Sine, and tangent of the angle 300 , 450 and 600

NB: The following figure is helpful to remember the trigonometrical ratios of special angles from 0°to 90°

If we need the sines of the above given angles for examples, we only need to take the square root of the number below the given angle and then the result is divided by 2.

Example 7

Find the sine,cosine and tangents of each of the following angles

1. -1350
2. 1200
3. 3300

Example 8

Find the value of θif Cos θ= -½ and θ≤ θ≤ 360°

Solution

Since Cos θis – (ve), then θlies in either the second or third quadrants,

Now - Cos (180 –θ= - Cos (θ+1800) = -½= -Cos600

So θ= 1800-600 = 1200 or θ= 1800 + 600 = 2400

θ= 1200 0r θ=2400

Example 9

Consider below

Exercise 2

Solve the Following.

The Graphs of Sine and Cosine

Draw the graphs of sine and cosine

Consider the following table of value for y=sinθ where θranges from - 360°to 360°

For cosine consider the following table of values

From the graphs for the two functions a reader can notice that sinθand cosθboth lie in the interval -1 and 1 inclusively, that is -1≤sinθ1 and -1≤cosθ≤1 for all values of θ.

The graph of y= tanθis left for the reader as an exercise

NB: -∞≤ tanθ≤∞the symbol ∞means infinite

Also you can observe that both Sinθnd cosθrepeat themselves at the interval of

360°, which means sinθ= sin(θ+360) = sin(θ+2x3600) etc

and Cosθ=(Cosθ+3600)= Cos(θ+2x3600)

Each of these functions is called a period function with a period 3600

1. Usingtrigonometrical graphs in the interval -3600≤θ≤3600

Find θsuch that

1. Sin= 0.4
2. Cos= 0.9

solution

Example 10

Use the graph of sinθto find the value ofθif

4Sinθ= -1.8 and -3600 ≤θ≤3600

Solution

4Sinθ= -1.8

Sinθ= -1.8÷4 = -0.45

Sinθ= -0.45

So θ= -1530, -270, 2070, 3330

The graphs of sine and cosine functions

Interpret the graphs of sine and cosine functions

Example 11

Use thetrigonometrical function graphs for sine and cosine to find the value of

1. Sin (-400)
2. Cos (-400)

Solution

1. Sin (-400)= - 0.64
2. Cos (-400)= 0.76

Sine and Cosine Rules

The Sine and Cosine Rules

Derive the sine and cosine rules

Consider the triangle ABC drawn on a coordinate plane

From the figure above the coordinates of A, B and C are (0, 0), (c, 0) and(bCosθ, bSinθ) respectively.

Now by using the distance formula

SINE RULE

Consider the triangle ABC below

From the figure above,

Note that this rule can be started as “In any triangle the side are proportional to the Sines of the opposite angles”

The Sine and Cosine Rules in Solving Problems on Triangles

Apply the sine and cosine rules in solving problems on triangles

Example 12

Find the unknown side and angle in a triangle ABC given that

a= 7.5cm

c= 8.6cm and C= 80°

Find the unknown sides and angle in a triangle ABC in which a= 22.2cmB= 86°and A= 26°

Solution

By sine rule

Sin A= sin B= Sin C

Example 13

Find unknown sides and angles in triangle ABC

Where a=3cm, c= 4cm and B= 30°

Solution

By cosine rule,

Example 14

Find the unknown angles in the following triangle

Exercise 3

1. Given thata=11cm, b=14cm and c=21cm, Find the Largest angle of ΔABC

2. If ABCD is a parallelogram whose sides are 12cm and 16cm what is the length of the diagonal AC if angle B=119°?

3. A and B are two ports on a straight Coast line such that B is 53km east of A. A ship starting from A sails 40km to a point C in a direction E65°N. Find:

1. The distance a of the ship from B
2. The distance of the ship from the

Compound Angles

The Compound of Angle Formulae or Sine, Cosine and Tangent in Solving Trigonometric Problems

Apply the compound of angle formulae or sine, cosine and tangent in solving trigonometric problems

The aim is to express Sin (α±β) and Cos (α±β) in terms of Sinα, Sinβ, Cosαand Cosβ

Consider the following diagram:

From the figure above <BAD=αand <ABC=βthus<BCD=α+β

From ΔBCD

For Cos(α±β) Consider the following unit circle with points P and Q on it such that OP,makes angleα with positive x-axis and OQ makes angle βwith positive x-axes.

From the figure above the distance d is given by

In general

Example 15

1. Withoutusing tables find the value of each of the following:

1. Sin 75°
2. Cos105

Solution:

Example 16

Find:

1. Sin150°
2. Cos 15°

Exercise 4

1. Withoutusing tables, find:

1. Sin15°
2. Cos 120°

2. FindSin 225° from (180°+45°)

3. <!--[endif]-->Verify that

1. Sin90° = 1 by using the fact that 90°=45°+45°
2. Cos90°=0 by using the fact that 90°=30°+60°

4. <!--[endif]-->Express each of the following in terms of sine, cosine and tangent of acute angles.

1. Sin107°
2. Cos300°

5. <!--[endif]-->By using the formula for Sin (A-B), show that Sin (90°-C)=Cos C

TOPIC 6: VECTORS

Displacement and Positions of Vectors

The Concept of a Vector Quantity

Explain the concept of a vector quantity

A vector - is a physical quantity which has both magnitude and direction.

The Difference Between Displacement and Position Vectors

Distinguish between displacement and position vectors

If an object moves from point A to another point say B, there is a displacement

There are many Vector quantities, some of which are: displacement, velocity, acceleration, force, momentum, electric field and magnetic field.

Other physical quantities have only magnitude, these quantities are called Scalars.

For example distance, speed, pressure, time and temperature

Naming of Vectors:

Normally vectors are named by either two capital letters with an arrow above e.g.

Equivalent Vectors:

Therefore two or more vectors are said to be equivalent if and only if they have same magnitude and direction.

Position Vectors;

In the x —plane all vectors with initial points at the origin and their end points elsewhere are called position vectors. Position vectors are named by the coordinates of their end points.

Consider the following diagram.

Components of position vectors:

Example 1

Write the position vectors of the following points: (a) A (1,-1), (b) B (-4,-3)

(c) C= (u, v) where U and V are any real numbers and give their horizontal and vertically components

Example 2

For each of vectors a and b shown in figure below draw a pair of equivalent vectors

Solution:

The following figure shows the vectors a and b and their respective pairs of equivalent vectors

Any Vector into I and J Components

Resolving any vector into I and J components

The unit Vectors i and j.

Definition: A unit vector is a position vector of unit length in the positive direction of x axis or y axis in the xy—plane.

The letters iand jare used to represent unit vectors in the X axis and y – axis respectively.

Consider the following sketch,

Example 3

Write the following vectors in terms of i and j vectors:

Example 4

Write the following vectors as position vectors.

Magnitude and Direction of a Vector

The Magnitude and Direction of a Vector

Calculate the magnitude and direction of a vector

Magnitude (Modules) of a Vector

Definition: The magnitude / modules of a vector is the size of a vector, it is a scalar quantity that expresses the size of a vector regardless of its direction.

Finding the magnitude of given vector.

Normally the magnitude of a given vector is calculated by using the distance formula which is based on Pythagoras theorem.

Using Pythagoras theorem

Example 5

Calculate the magnitude of the position vector v=(- 3 , 4)

What is the magnitude of the vector U if U = 4i – 5j?

Unit Vectors:

Definition: A unit of Vector is any vector whose magnitude or modulus is one Unit.

Example 6

Find a Unit Vector in the direction of Vector U = (12, 5)

Direction of a vector:

The direction of a Vector may be given by using either bearings or direction Cosines.

(a) By Bearings:

Bearings are angles from a fixed direction in order to locate the interested places on the earth's surface.

Reading bearings: There are two method used to read bearings, in the first method all angles are measured with reference from the North direction only where by the North is taken as 0000, the east 0900, the South is 1800 and West 2700

From the figure above, point P is located at a bearing of 0500, while Q is located at a bearing of 1350.

Commonly the bearing of point B from point A is measured from the north direction at point A to the line joining AB and that of A from B is measured from the North direction at point B to the line joining BA.

From the figure above the bearing of B from A, is 0600 while that of A from B is 2400 In the second method two directions are used as reference directions, these are North and south.

In this method the location of places is found by reading an acute angle from the north eastwards or westwards and from the south eastwards or westwards.

From figure above, the direction of point A from O is N 460 E , that of B is N500W while the direction from of C is S200E.

Example 7

Mikumi is 140km at a bearing of 0700f from Iringa. Makambako is 160km at bearing of 2150 from Iringa. Sketch the position of these towns relative to each other, hence calculate the magnitude and direction of the displacement from Makambako to Mikumi.

Sketch

Using the cosine rule

The displacement from Makambakoto Mikumiis 12 286kmBy sine rule

Alternatively by using the scale AB is approximately14.3 cm Therefore AB = 14.3x 20 km = 286km and the bearing is obtained a protractor which is about N510E

(b) Direction cosines

Where Cos A and Cos B are the direction cosines of OP

Example 8

If a = 6i + 8j find the direction cosine of aand hence find the angle made by awith the positive x – axis.

Exercise 1

1. Find the magnitude of

Sum and Difference of Vectors

The Sum of Two or More Vectors

Find the sum of two or more vectors

Addition of vectors

The sum of any two or more vectors is called the resultant of the given vectors. ­The sum of vectors is governed by triangle, parallelogram and polygon laws of vector addition.

(1) Triangle law of vector Addition

Adding two vectors involves joining two vectors such that the initial point of the second vector is the end point of first vector and the resultant is obtained by completing the triangle with the vector whose initial point is the initial point of the first vector and whose end points the end point of the second vector.

From the figure above a + b is the resultant of vectors a and b as shown below

(2) The parallelogram law

When two vectors have a common initial point say P, then their resultant is obtained by completing a parallelogram, where the two vectors are the sides of the diagonal through P and with initial point at P

Example 9

Find the resultant of vectorsuandvin the following figure.

Solution

To get the resultant of vectors u and v, you need to complete the parallelogram as shown in the following figure

From the figure above, the result of u and v is PR = U + V = U + V

Note that by parallelogram law of vector addition, commutative property is verified.

Polygon law of vector addition:

If you want to add more than two vectors, you join the end point to the initial point of the vectors one after another and the resultant is the vector joining the initial point of the first vector to the end point of the last vector

Example 10

Find the resultant of vectors a, b, c and d as shown in the figure below.

Solution

In the figure above P is the initial point of a, b has been joined toaat point Q and c is joined to b at R, while d is joined to c at point S and PT = a + b + c + d which is the resultant of the four vectors.

Opposite vectors

Two vectors are said to be opposite to each other if they have the same magnitude but different directions

From the figure above a and b have the same magnitude (3m) but opposite direction.

So a and b are opposite vectors.

Opposite vectors have zero resultant that is if a and b are opposite vectors, then

Example 11

Find the vector p opposite to the vector r = 6i – 2j

The Difference of Vectors

Find the difference of vectors

Normally when subtracting one vector from another the result obtained is the same as that of addition but to the opposite of the other vector.

Therefore the different of two vector is also the resultant vector

Consider the following figure

Multiplication of a Vector by a Scalar

A Vector by a Scalar

Multiply a vector by a scalar

If a vector U has a magnitude m units and makes an angleθwith a positive x axis, then doubling the magnitude of U gives a vector with magnitude 2m.

Generally if U = (u1, u2) and t is any non zero real number while (u1, u2) are also real numbers, then

Example 12

If a = 3i + 3j and b = 5i + 4j

Find – 5a + 3b

Example 13

Given that p = (8, 6) and q = (7, 9). Find 9p – 8q

Application of Vectors

Vectors in Solving Simple Problems on Velocities, Displacements and Forces

Apply vectors in solving simple problems on velocities, displacements and forces

Vector knowledge is applicable in solving many practical problems as in the following examples.

A student walks 40 m in the direction S 450 E from the dormitory to the parade ground and then he walks 100m due east to his classroom. Find his displacement from dormitory to the classroom.

Solution

Consider the following figure describing the displacement which joins the dormitory D. parade ground P and Classroom C.

From the figure above the resultant is DC. By cosine rule

Example 14

Three forces F1 = (3,4), F2 = (5,-2) and F3 = (4,3) measured in Newtons act at point O (0,0)

1. Determine the magnitude and direction of their resultant.
2. Calculate the magnitude and direction of the opposite of the resultant force.

(b) Let the force opposite to F be Fo, then Fo = -F = - (12, 5) = (-12, -5)

Fo= 13N and its bearing is (67.40+1800) = 247.40

So the magnitude and direction of the force opposite to the resultant force is 13N and S67.40W respectively..

Exercise 2

1. Given that U = (3, -4), V= (-4, 3) and W = (1, 1), calculate.

1. The resultant of U + V + W
2. The magnitude and direction of the resultant calculated in part (a) above.

2. A boat moves with a velocity of 10km/h upstream against a downstream current of 10km/h. Calculate the velocity of the boat when moving down steam.

3. Two forces acting at a point O makes angles of 300 and 1350 with their resultant having magnitude 20N as shown in the diagram below.

<!-- [if !supportLists]-->4. <!--[endif]-->Calculate the magnitude and direction of the resultant of the velocities V1=5i + 9j,V2 = 4i + 6j and V3 = 4i – 3j where i and j are unit vectors of magnitude 1m/s in the positive directions of the x and y axis respectively