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# TRIGONOMETRY

UNAWEZA JIPATIA NOTES ZETU KWA KUCHANGIA KIASI KIDOGO KABISA:PIGA SIMU:0787237719

Trigonometry deals with the measurements of triangles and problems involving triangles. The trigonometric functions are usually expressed using the various ratios of the side of right angle. Consider figure 8.1

Sine: sin θ = opposite side =   a
Hypotenuse side   c

Cosine: cos θ = Adjacent side =   b
Hypotenuse side    c

Sin θ =  ÷  =  x  =

Sin θ = tan θ. When cos θ ≠ 0
So, tan θ =   a
b

Other trigonometric ratios

The reciprocals of sine, cosine and tangent are also trigonometrically ratios. The reciprocal of sine is called the cosecant, the reciprocal of cosine is secant while the reciprocal of tangent is called cotangent.

The short form of cosecant, secant and cotangent of an angle θ are cosec θ, sec θ and cot θ respectively. Thus,

Angles of any magnitude

The trigonometric ratios can be studied better if we look at where these angles appear in the quadrant. Angles in the range 0o to 90o are in the first quadrant, 90o – 180o, 1800 – 270and 270o – 360o are in the second, third and fourth quadrants respectively.

Angles 0o – 90o

The trigonometric ratios of angles θ, in the first quadrant are all positive information under figure 8.2

Sin θ1 =  = +ve

Cos θ1 =  = +ve

Tan θ=  = + ve

Note

–          All angles between 0o and 90o are called acute angles

–          The size of the angles used to read through mathematical tables are all the time acute angles relative to the horizontal x – axis, ie the angles ≤ 90o

Angles 90o – 180o

The angles θ in the second quadrant have the same trigonometric ratios as the angle θ2 i.e θ2 = 180o – θ. In this quadrant, only the trigonometric ratios of sine are positive. See figure 8.3 and the information under it.

θ= 180o – θ

Sin θ2 = sin (180 – θ) =  = +ve

Cos θ2 = cos (180 – θ) = – = -ve

Tan θ2 = tan (180 – θ) =  = -ve

Sin θ2 is positive but cos θ2 and tan θ2 are negative

Note: All angles between 90o and 180are called obtuse angles.

Angles 180o – 270o

The angles θ in the third quadrant have the same trigonometric ratios as the angles θ3. In this quadrant, only the tangents are positive.

See figure 8.4 and the information under it.

θ3 = θ – 180o

Sin θ= sin (θ – 180o) = – = -ve

Cos θ3 = cos (θ – 180o) = – = -ve

Tan θ3 = tan (θ – 180o) =  = +ve

Then, Tan θ3 is positive but sin θ3 and cos θ3 are negative

Note

All angles between 1800 – 2700 are called angles of reflection

Angles 270– 360o

The angles θ in the fourth quadrant have the same trigonometric ratios as the angles θ4. i.e θ4 = 360o – θ. In this quadrant only the cosine are positive. See figure 8.5 and the information under it.

θ4 = 360 – θ

Sin θ4 = sin (360 – θ) =  = -ve

Cos θ4 = cos (360 – θ) =  = +ve

Tan θ4 = tan (360 – θ) =  = -ve

Sin θ4 and tan θ4 are all negative but cos θ4 is positive

Note: All angles between 270o – 360o are called angles of reflection.

Generally, the values of trigonometric ratios and their sign are shown in figure 8.6 A stand for all sine, cosine and tangent are positive, S stands for only sine is positive stands for only tangent is positive and C stands for only cosine is positive

Examples  1.1

Evaluate each of the following

i) sin 135o

ii) Cos 230o

iii) Tan 315o

iv) Tan 540o

v) Cos (920O)

Solution

i) Sin 135o = +sin (180o – 135o) = +sin 45o =                                                                                                                                                                                                                        2

ii) Cos 230o = -cos (230o – 180o) = cos 50o = 0. 6428

iii) Tan 315o = -tan (360o – 3150) = tan 45o = -1

iv) Tan 540o = tan (540o – 360o) = tan 180o = 0

v) Cos (920o) = -cos (9200 – 720o) = -cos 200o

= -cos (200o – 180o)

= -cos 20o

= -0.40808206

Special Angles

0o, 30o, 60o, 45o, 90o, 180o, 270o, 360o

Consider an equilateral triangle whose side is 2 units

Then, sin 60o =        and sin 30o =

Cos 600 =        cos 30o =

Tan 60o =      tan 30o =  x  =

Consider an isosceles triangle ABC below

Similarly,
Sin 45o =  x  =

Cos 45o =  x  =

Tan 45o =  = 1

Consider unit circle

From the diagram you will find that;

i) Cos θ =

Cos θ = x

ii) Sin θ =

Sin θ = y

If p (x, y) = p (cos θ, sin θ), then

Sin 0o = 0                               sin 90o = 1                                         sin 180o = 0

Cos 0o = 1                             cos 90= 0                                         cos 180o = -1

Tan 0o = 0                             tan 90o = ∞ undefined                          tan 180o = 0

Sin 270o = 0                           sin 360o = 0

Cos 270o = 0                          cos 360o = 1

Tan 270o = ∞ (undefined)       tan  360= 0

Negative angels

Consider a unit circle in figure 8.10

Let θ be any angle then

Cos θ =  = x

Sin θ =  = y

a) Sin (-θ) =  = -y

But y = sin θ

-y = -sin θ

Sin (-θ) = -sin θ

b) Cos (-θ) =  = x

But x = cos θ

Cos (-θ) = cos θ

c) Tan (-θ) = sin (-θ) = -sin θ = -tan θ
cos (-θ)     cos θ

∴ Tan (-θ) = – tan θ

Example 1.2

Evaluate sin θ and cos θ for the given θ

a) -45o     b) -180o  c) -90o      d) -240o

Solution

a) Sin (-45o) = -sin 45o =

Cos (-45o) = cos 45o =

b) Sin (-180o) = -sin (180o) = 0

Cos (-180o) = -sin 180o = 1

c) Sin (-90o) = -sin 90o = -1

Cos (-90o) = cos 90 = 0

d) Sin (-240o) = -sin (240o)

= -sin (240o – 180o)

= sin 60o

= +

Cos (240o) = cos (240o) = cos (240o)

= -cos (240o – 180o)

=  ½

Definition

1 radian is an angle subtended at the centre of the circle of the circle by an are equal in length to that radius.

i.e

Circumference = 2

=

=  =  =

=

Note: It is customary to omit the unit radian

= 180o

Exercise 1.3

1.   Convert the following into degrees

a)       b)     c)     d)

e) 5     f)    g)    h)

j)

2.   Convert the following into radians

a) 120o           b) 4o                c) -90o            d) 15o

e) 60o              f) 240           g) -315           h) -90o

i) 357o            j) 300o            k) -630o

3. Solve for x such that 0o ≤ x ≥ 360o

a) Cos x = -1

b) Cos x = –

c) Sin x = –

d) Sin x = 0

e) Cos x =

f) Sin x =

g) Tan x = undefined

h) Tan x = 0

i) Tan x = -1

j) Tan x = 1

4. Evaluate sin θ if the terminal side of θ contains the given point

a) P (10, -4)               b) P (3, 4)       c) P (3, -5)     d) P (-1, 3)

e) P (-8, -13)

5. If cos θ = -2/5 and sin θ < 0, find sin θ

6. If cos θ = 5/3 and sin θ < 0, find sin θ.

Sin A = h/c             h=c sin A

Sin C = h/a             h=a sin C

Sin A / a = sin C / c

Similarly,  by drawing the altitude from A to BC.

b sin C = c sin B

Sin C / c= sin B / b

Hence; sin A /a = sin B / b = sin C / c

It is called the sine rule.

Note:

a) When two angles and one side of any triangle are known, the sine rule can be used to solve the triangle

b) When two sides and an angle opposite to any of the two given sides are known, the sine rule can be used to solve the triangle.

Example

7. In triangle ABC, B = 390, C = 820, a = 6.73 cm find C

Solution

From

=  =

But A = 180o – (39o + 82o)

= 59o

=  =

C = 6.73cm x  = 7.78 cm

Example 1.4

Find the remaining angles of the AABC in which a = 12.5 cm , c = 17.7 cm and C = 1160

Solution

=  =  =

= sin A =

A = sin-1(0.6347)

A = 390.24

B = 180o – 155 = B = 24o 36

Exercise 1.5

1.  Solve completely the triangles in which

a) b = 6m, A= 80o, C = 46o

b) A = 50o, B = 60o, a = 4m

c) a = 400m, A = 31o20, B = 70o40

d) A = 115o, a = 65m, b = 32m

Cosine rule

Consider the triangle ABC

By using Pythagoras theorem

C2 = (x – a) 2 + h2

C= x2 – 2ax + a2 + h2

C2 = x2 + h2 + a2

C2 = a2 + b2 – 2ax

But  = cosC

C2 = a2 + b– 2ab Cos C

For angle B

b2 = a2 + c2 – 2ac Cos B

For angle A

a= b2 + c2 – 2bc Cos A

Note

a) When two sides and an angle between the two sides are included then the angle between the two lines of the cosine rule can be used to solve the triangle by using the cosine rule.

b) We can also solve the triangle by cosine rule if all the three sides are given

Example 1.6

Given a triangle with a = 50cm, c =60cm, and B = 100.Find b

Solution

b2 = a2 + c2 – 2ac Cos B

= (50) 2 + (60) – 2 (50) (60) Cos 100o

= 3600 + 2500 – 6000 (-0.1736)

= 7141.6 cm

Example 1.7

Given a = 8cm, b = 5cm, and c = 4cm, find the smallest and largest angles.

Solution

i) To find A we use

a2 = b2 + c2 – 2bc Cos A

64 = 25 + 16 – 40 Cos A

40cos A = 41 – 64

40cos A = -23

Cos A =  = -0.5750

A = 125o10

ii) To find C we use

c2 = a2 + b2 – 2abCos C

16 = 82 + 25 – 2 (8) (5) Cos C
16 = 64 + 25 – 2(8)(5)Cos C

80cos C = 89 – 16

Cos C =  = 0.9125

C = 24o10

Trigonometric identities

Figure 8.15 show the right angled triangle with sides a, b, c and angles A, B, C

By the use of Pythagoras theorem

a2 + b2 = c2

Divide throughout by c2

But    = cos θ,  = sin θ

Cos2 θ + sin 2 θ = 1

Exercise 1.8

1. Use the cosine rule to solve the following triangles

a) A = 60o, b = 3cm, c = 4cm

b) A = 100o, b = 8cm, c = 10cm

c) A = 50o, b = 5cm, c = 8cm

d) a = 9cm, b = 10, C = 120o

e) a = 5, b = 10, c = 1200

f) a = 10cm, b = 20cm, c = 120cm

2. Find the largest and smallest angles of triangle whose side area a= 5cm b = 10cm and c = 12cm

3. Two people are on opposite sides of a hill and are 600m apart. If the angles of elevation from these people to the top of the hill are 19o and 21o, how high is the hill?

Compound Angles

The formulae that follow are the ones we refer to as the compound angles

Sin (A + B) = sin A Cos B + Cos A sin B

Sin (A – B) = sin A Cos B – Cos A sin B

Cos (A + B) = Cos A Cos B – sin A sin B

Cos (A – B) = Cos A Cos B + sin A sin B

Solution

Sin (A + B) =  Sin A Cos B + Cos A Sin B

Holder the area of the triangle RST

Area (ΔRST) = area (ΔRNS) + area (ΔRNT)

cd sin (A + B) =  hc sin B +  hd sin B

Multiplied throughout

Sin (A + B) =  sin A +  sin B

But  = cos B and  = cos A

Sin (A + B) = sin A Cos B + Cos A sin B

Cos (A + B)

Apply the cosine rule on the same triangle in figure 8.16

(a + b) 2 = c2 + d2 – 2cd Cos (A + B)

=

=

=

=

=

=   –

=  .

=   = Cos A and  = cos B

=  = sin A and   = sin B

Thus,

Cos (A + B) = Cos A Cos B – sin A sin B

Sin (A – B)

Consider the area of triangle RST

Area (ΔRNT) = area (ΔRST) – area (ΔRSN)

cd sin (A –B) =  hc sin A –  hd sin B

Multiplied throughout by

Sin (A – B) =  ,  sin A – sin B

But   = cos B,   = cos A

Sin (A – B) = Cos B sin A – Cos A sin B

Cos (A – B) use the same figure and apply the cosine rule (a – b) 2 = c2 +    d2 – 2cdCos (A – B)

=

=

=

=

=  +

But,   = Cos A,  = Cos B,  = Sin B

Cos (A – B) = Cos A Cos B + Sin A Sin B

Tan (A + B) =

=

Divide numerator and denominator by Cos A Cos B

Tan (A – B) =

=

Divide numerator and denominator by Cos A Cos B

=

=

∴Tan (A – B) =

Example 1.9

1. Without using tables, evaluate the following

a) Tan (195o)

b) Sin 15o

c) Cos 75o

d) Tan 15o

Solution

a) Tan 195o = tan (135o + 60o) =

b) Sin 15= sin (45o 30o) = sin 45Cos 30o – Cos 45o sin 30o

=     x  –  x

=    –

c) Cos 70o = Cos (45o + 30o) = Cos 45o Cos 30o – sin 45o sin 30o

=  x  –  x

=   –

d) Tan 45o = tan (45– 30o)

=

Example 1.10

1. Prove that Cosec θ = Sin θ + Cos θ cot θ

Solution

Take the RHS

Using cot θ =

Cosec θ = sin θ + Cos θ x

=  +

=  (common denominator)

=  (sin2 θ + Cos 2θ)

= Cosec θ

RHS = LHS, hence proved

Solution

Take the RHS

=

=

RHS = LHS, hence proved

Exercise 1.11

1.   Simplify a) sin 4θ – Cos 4 θ

b)

c) (Sin θ + cos θ) 2 – 2sin θ cos θ

d)

2.  Provide the identities

a) tan θ cot θ sec θ cos θ = 1

b)  =

c) =   = 2 tan θ sec θ

d) tan θ + cot θ = Sec θ Cos θ  = 1

e) (aCos θ  + b sin θ) 2 + (-asin θ + bCos θ) 2 = a+ b2

f) (Cosec θ – Cot θ) 2 =

g)  tan θ + Cot θ = sec θ Cosec θ

h) (Cos θ – sin θ) 2 + (Cosec θ + sin θ) = 2

3.  Compute without tables or calculators

a) sin 15       b) sin 75      c) tan 15o       d) tan 75o       c) sin 195o

4.  Prove the following identities

a) Sin (A +B) + sin (A – B) = 2sin A cos B

b) Sin (A +B) + sin (A – B) = 2sin A sin B

c) Cos (A +B) + cos (A – B) = 2cos A cos B

d) Cos (A +B) + cos (A – B) = 2sin A sin B

5. Simplify the following

a) Sin (A + 2r)

b) Cos (A + )

c) Sin (A + )

d) Cos (A + 2)

6. Express 6 sin (x + 60o) in the form of p sin x + Q sin x

7.   If cos A =, tan B = , A and B being acute. Evaluate the following

a) Cos (A + B)

b) Tan (A + B)

c) Sin (A + B)

Double angle formulae

By applying the knowledge of compound angles, it is clear that

(1). sin 2x = sin (x + x)

= sinx cosx + cosx sinx

Thus sin 2x = 2sin x cos x

(2). Cos 2x = cos x cos x – sin x sin x

= cos2 x – sin2 x

From cos 2 x + sin2 x = 1

Sin2 x = 1 – cos2 x

Cosx = cos2 x – (1 + cos2x)

= cos2 x + cos2 x – 1

Cos2x =   2cos2x – 1

But cos2 x = 1 – sin2 x

Cosx – sin2 x = cos 2x

1 – Sin2 x – sinx = cos 2x

1 – 2 sin2x = cos2x

∴Cos2x = cos2x – sinx = 2cos2 x – 1 = 1 – 2sin2 x

3.  Tan 2x = tan (x + x)

=

Tan 2x =

Exercise 1.12

1.   a) Express sin 3x in terms of cos x

b) Express cos 3x in terms of cos x

c) Express tan 3x in terms of tan x

2.   Use double angle formulae to prove that

a) Sin 4x = 8cos3x sin x – 4cos x sin x

b) Cos 4x = 8cos4 x – 8 cosx + 1

How to write cos 2A and sin 2A in terms of tan A

From double angle formula, it is clear that,

a) Cos 2A =

From cos2 A + sin2 A = 1

Cos 2A =

Divide the numerator and denominator by cos 2 A we have

b) Similarly, from double angle formula of sine, we see that

Sin 2A =

=

Dividing numerator and denominator by cos2 A, we have

Sin 2A =

Example 1.13

Solve sin (2 + θ) + cos (θ – ) = 1, where 0o ≤ θ ≤ 360o

Solution

Cos 2 = 1, sin 2= 0, cos  = 0 and sin  = 1

Expand the left part of the equation i.e sin (2 + θ) + cos (θ – )

Sin 2cos θ + cos 2sin θ + cos θ cos  + sin θ sin  = 1

Sin θ + sin θ = 1

2sin θ = 1

Sin θ =

θ = 30o, 150o

Exercise 1.14

1.   Solve the following trigonometric equations for 0 ≤ θ ≤ 360o

a) Sin θ – 1 = 0

b) 2cos θ =

c) Tan θ =

d) Tan 3θ = 1

e) 2cos θ

f) Cos θ = sin θ

g) Sin θ = 2

2.  Solve for θ, where 0o ≤ θ ≤ 360o

a) cos2 θ – 2cos θ = 0

b) 2sinθ + sin θ – 1 = 0

c) 2cos 2 + 3cos θ + 1 = 0

d) 4sin2 + 4sin θ = 3

3.  Solve for θ, where 0≤ θ ≤ 360o

a) cos (sin θ) = 1

b) cos 2θ + sin θ – 1

c) 2sin 2θ – tan θ = 0

d) Sin 4θ cos 2θ cos 4θ sin 2θ =

CIRCLE

Length of an Arc

Consider the following

2 (length of arc AB) = θ x 2r

Length of arc AB = rθ

Area of a circular sector

Pg. 141 drawing

=

Where: area of circle = r2 and 360o = 2

=

Area of sector AOB x 2 = θ x r2

Area of sector AOB = r2θ

Where θ is in radians and r is the radius of the circle

Example 1.15

Find the arc length of a circle of radius r which is subtended by a central angle θ for each of the following

a) r = 3cm and θ = 45o

b) r = 6cm and θ = 180o

Use  = 3.14

Solution

a) 1 = rθ

1 = 3 x cm or  = 2.354 cm

b) arc length, 1 = r θ

1 = 6 x

1 = 6 x 3.14

= 18.84cm

Example 1.16

Find the area of the sector of a circle with radius 6cm and subtended by an angle of 60o use  = 3.14

Solution

Area =  r2θ

= ½ x 6 x 6

= 6  cm2

Since = 3.14

Therefore, area = 3.14 x 6

= 18.84cm2

Trigonometric Equation

Example 1.17

Solve sin θ = ½ for 0o ≤ θ ≤ 360o

Solution

Sin θ = ½
θ = Sin-1 ( ½)
θ = 30o, 150o

Example 1.18

Solve 2sin θ –  = 0 for 0 ≤ θ ≤ 360

Solution

2 sin θ =

Sin θ =

θ = 60o, 120o

Example 1.19

2cos2 θ + 3cos θ – 2 = 0

Solution

This equation is quadratic in cos θ

Factorize the equation

2cos2 θ + 3cos θ – 2 = 0

Let cos θ = t

2t2 + 3t – 2 = 0

2t2 + 4t – t – 2 = 0

2t (t + 2) – (t + 2) = 0

(t + 2) (2t – 1) = 0

t = -2 or 2t = 1

t = -2 or t = ½

But, t  = Cos θ

Cos θ = -2

θ = undefined or no solution since the minimum value of cos θ = -1

Cos θ = ½ = θ = 60o, 300o

θ = (60o, 300o)

Example

Solve Cos 2 θ Cos θ + Sin 2θ = 1, where 0≤ θ ≤ 360o

Solution

Use the compound angle formula Cos (A – B) = Cos A Cos B + Sin A Sin B

Thus, Cos 2 θ Cos θ + Sin θ = Cos (2θ – θ) = Cos θ

Cos θ = 1
θ = (0o, 360o)

Trigonometric functions

Let x be angle in degree or radians and f (x) be the image of x. Then the functions defined as f (x) = sin x, f (x) = cos x and f (x) =tan are called trigonometric functions. These functions can be graphed just like any other functions. Values of angles should be on the horizontal axis and images values should be on the vertical axis.

Sine functions f (x) = sin x

To graph sine functions, draw a table of values in which x values will be angles and y = f (x) will be images of these angles. The angles may be in degrees or radians. Table 8.1 has values that suit a function y = sin x, (The y – values are correct to one decimal place)

 x 0 30 45 60 90 120 135 150 180 210 225 240 270 ysinx 0 0.5 0.7 1 1 0.9 0.7 0.5 0 -0.5 -0.7 -0.9 -1

Cosine function y = cos x

To graph cosine function follow the procedure followed in drawing sine functions. Table 8.2 contains values for y = cos x. In table 8.2 values are given to one decimal place.

Note

1.  The domain of y = sin x and y = cos x is set of real numbers and range is     (-1 ≤ y ≤ 1).

2.  These functions are periodic, they repeat after every one complete   revolution (360o)

Graph of y = tan x

To graph y = tan x draw a table values and plot the point required then join the points

with given correct to one decimal places.

 xn 90 60 -45 30 0 30 45 60 90 120 135 150 180 Y tan x 1.7 -30 -1 0 -0.6 1 1.7 -1.7 -1 -0.6 0

Note

1.   The range of y = tan x is a set of real numbers

2.   The domain is (x: x ≠ 90o, x ≠ 270, x ≠ 450…)

3.    The tangent function is periodic too. It repeats itself after every 180o

4.   At x = 90o, x = 450o etc, the functions not defined. These values are its asymptotes. It tends to approach these values but never touches them

Exercise

1.  State the range and period for each of the following

a) y = cos x

b) y = ½ cos 2x

c) y = ½ cos x

d) y = 3 sin x

e) y = 3 – sin x

j) y = 1 + sin

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