Form 4 Mathematics – TRIGONOMETRY

TRIGONOMETRY

Trigonometrical ratios in a unit circle

The three trigonometrical ratios of sine, cosine and tangent have been defined earlier, using the sides of a right-angled triangle as follows

If A is an angle as shown

Consider a circle of units subdivided into four congruent sectors of the coordinate axes whose origin is at the center of the circle as shown below.

Let ø be any acute angle (0 < ø < 90⁰) and let P with coordinates (x, y) be the point where OP intersect with the circle then

SIGNS OF THE TRIGONOMETRICAL RATIOS
The trigonometrical ratios can be positive or negative depending on the size of the angle and the quadrant in which the angle is found.

The results obtained are illustrated below. There results will be a help in determining whether sine, cosine and tangent of an angle is positive or negative.

Obtuse angle 90⁰<θ<180⁰

Sin (180-θ)⁰ = y

Cos (180-θ)⁰ = – x

Tan (180-θ)⁰ = –

Sin θ = =

Cos θ = =

Tan θ = =

Tan θ =

Tan θ =

Tan θ =


EXERCISE

1. Write the signs of each of the following

a. Cos 160⁰ = negative

b. Cos 310⁰ = positive

c. Cos 75⁰ = positive

d. Sin 220⁰ = negative

e. Cos 335⁰ = positive

f. Tan 190⁰ = positive

2. Express the following in terms of sine, cosine or tangent of an acute angle
a). Cos 308⁰
=360⁰– 308⁰
= 52⁰
=Cos 52⁰

b). Sine 217 ⁰
= ( -217 – 180)⁰
= -( 37⁰)
= Sin -37⁰

c). Tan 175⁰
= -(180⁰ – 175⁰)
= -5⁰
Tan -5⁰

d). Tan 333⁰
= -(360 -333)⁰
= -27⁰
= sin-27⁰
e). Cos 103⁰
= – (180 -103)
= -77⁰
= cos -77⁰
3. Express the following in terms if sine
a). Sin 130 ⁰
= (180- 130)⁰
= sin 50⁰
b). Sin 230⁰
= -(230 -180) ⁰
= -sin 50⁰
c). Sin 310 ⁰
= – ( 360 – 310)⁰
= -sin 50⁰

Examples
1. Let θ be any angle and P with coordinates (-4, 3) be a point in the terminal point of op, see the figure below. Find

a. sinθ

b. cosθ

c. tanθ

(OP)²= (-4)²+ (3)²
OP= 5
a. Sin θ = sin ( 180- θ) = 3/5
b. Cos θ = cos ( 180 – θ) = -4/5
c. Tan θ = tan (180-θ) = – ¾

EXERCISE
1. Find the cos θ and tan if θ is the angle made by the positive x- axis, from the line from the origin to each of the following points.

a. ( 2,6)

Solution

Cos θ =

Sin θ =

b. (-12 , 5)

Cos θ = –

Tan θ = –
Sin θ =

c. (-4,-3)

Cos θ = –

Tan θ = –
Sin θ = –

POSITIVE AND NEGATIVE ANGLES

NOTE;

1. If θ is positive, the negative angle corresponding to θ is (-360⁰ +θ)

2. If θ is negative, the positive angle corresponding to θ is (360^{0 ­}+θ)

Example
1. Find the positive or negative angles corresponding to each of the following angles.
a. 304⁰= ( -360⁰ + θ) = (- 360⁰ + 304 ) = -56 ⁰
b. -115 ⁰ = ( 360 ⁰+ θ) = 360⁰ + -115⁰ = 245⁰

2. Find the sine, cosine and tangent of each of the following angles.
a. 144⁰

Solution
144⁰ = 180⁰ – 144⁰ = 36 ⁰
= sin 36⁰ = 0.5878
= cosine 36⁰ = -0.8090
= tan 36⁰ = -0. 7265

b). -231⁰

= 360⁰ + θ = 360⁰ + -231
= 129⁰
=180⁰– 129⁰
= 51⁰
Sin 51⁰= 0.7771
Cosine 51⁰= 0.6293
Tan 51⁰= -1.2349

c). 310 ⁰

= 360 ⁰ –310⁰ = 50⁰
Sin310⁰= sin50⁰ = 0.7660
Cosine 310 ⁰ = cosine 50⁰= 0.6428
Tan 310 ⁰ = tan50⁰ = 1.1918

RELATIONSHIP BETWEEN TRIGONOMETRIC RATIOS

Consider a triangle A, B, C in which angles A and C are complimentary angles. ie A+C = 90⁰

Sin A = a/b
Cos C = a/b
Sin A = Cos C = a/b
C = 90-A
Sin A = Cos (90-A)⁰ = Cos C = a/b
Cos A = c/b
(Sin A)² = sin² A
Sin ² A + Cos ² A = = 1
 

Exercise

1. Given that Sin θ = 4/9 . find Cos θ

Solution
Sin²θ + cos² θ = 1
(4/9 )² + cos²θ = 1
Cos ² θ = 1- 16/81
Cos θ =
2. If Sin θ = 0.9397 and Cos θ = 0.3420. Find without using tables tan θ.

Solution
Tan θ = = = 2.748

3. If Sin α = find sin(90-α)

Solution
α + β = 90⁰
β = 90⁰ – α
Sin α = Cos (90⁰– α) = Cos β =

Sin²β + Cos ²β =1
Sin²β + ²

Sin²β=

Sin²β=

Sin β = Sin (90-α) =
4. If Sin A = 0.9744 and Cos A = 0.225
Find without using tables Tan A?

Solution
Tan A =

=

Tan A = 4.3307
5. Find without using tables sin α if cos α = 0.9272 and tan α=0.404

Solution
Tan α =

0.404 =

Sin α = 0.3746

APPLICATIONS OF TRIGONOMETRICAL RATIOS

Angles of depression and elevation

Example and exercise

1. From the top of a tower , the angle of depression of a point on the ground 1M away from the base of the tower is 60⁰. How high is the tower?

Solution

Tan60⁰=

Tan60⁰=

BA = tan 60⁰ x 10 m

BA = 17.321m = height of a tower

2. P and O are two pegs on level ground and both lie due west of a flag staff. The angle of elevation of the top of the flagstaff from P is 45⁰ and from Q is 60⁰. Find distance PQ.

Solution

Tan45⁰= AB

24m

AB = 24m x tan45⁰

AB = 24m

Tan 60⁰ =

QB =

QB = 13.85m

PQ = 24m – 13.85m

PQ = 10.15m

3. At a point 182 m from the foot of a tower on a level road, the angle of elevation of the top of the tower is 36⁰44¹. Find the height of the tower.

Solution

Tan 36⁰44’ =

h = 0.7463 x 182 m

h = 135.8266m

4. x and y are two points on opposite banks of a river( figure below) . If PY measures 90m and XPY = 59⁰. Find the width of the river.

Solution

Tan59⁰ = w

90m

W = tan59⁰ x 90m

= 1.6643 x 90m

= 149.787m

TRIGONOMETRIC SPECIAL ANGLES

Cos θ=

Cos θ = x

Sin θ =

Sin θ = y

OTHER SPECIAL ANGLES:

Consider an equilateral triangle ABC

Let each side to have two (2) units

Using Pythagoras theorem

(AB)² = (BX)² + (AX)²
2² = 1² + (AX)²
AX = √3

Sin 60⁰ =

Cos 60⁰ =

Tan 60⁰ =

Sin 30⁰ =
Cos 30⁰ = √3
2
Tan 30 ⁰=

For 45⁰
Consider an isosceles triangle ABC

EXAMPLES

1. Find the sine, cosine and tangent of each of the following angles.

(a) -135⁰

= 360⁰ + -135⁰
= 360⁰– 135⁰
= 225⁰
=225⁰ – 180⁰
Sin 45⁰ =
Cos 45 ⁰ =

Tan (– 135⁰) = tan 45⁰= 1

b. 330⁰

= 360⁰ – 330⁰
= 30⁰
= sin 330 ⁰ = – sin 30⁰ = –

2. Find the values of the following without using tables.

=

= x
=

3.
=
= x
=

SINE RULE

Let us consider a triangle ABC, with its including angle C

Let us find the area of the triangle using its including angle and two sides.

Area of triangle ABC = x a x b x sin C
Area of triangle ABC = x a x c x sin B
Area of triangle ABC = x b x c x sin A
x a x b x sin c = x a x c x sin B = x b x c x sin A
Dive each by x a x c
= =

Examples

1. Find the unknown sides and angle sin each of the following triangles.

Solution
=

=

but A = 180⁰ ( 61+43)⁰
=

=

b =

2. Juma notices that the angle of elevation of a coconut tree is 32⁰. Walking 11 m in the direction towards the tree he notices the angle of elevation to be 45⁰. Find the height of the tree.

Solution

Tan45 ⁰ =

h = tan45⁰ ( x – 11)
h = x – 11…..(i)

h = tan 32⁰ X x
h = 0.6249x………….ii)

compare i) and ii) eqns
x-11=0.6249x
x-0.6249x=11
0.3751x= 11
x=0.0341m

COSINE RULE

Consider a triangle ABC whose coordinates are A ( 0,0) , B ( c, 0) and C (b cos A , b sin A)

Cos A =
Sin A =
X = b cos A
Y = b sin A

a²= (b cos A – c)² + ( b sin A – 0)²
a²= b² cos ²A – 2bccos A + c² + b² sin² A
a²= b² sin ²A + b²cos ² A – 2bccosA+ c²
a² =(sin² A+ cos²A)b² + c²-2bccosA
a²= b²+ c² – 2bc cos A

b²= a²+ c²– 2ac cos B cosine rule
c²= b²+ a² – 2 a b cos c

Example

Find the value of angle A.

a²= b²+ c ² – 2bcCos A
(2.8)²= (3.4) ² + (4.5) ² – 2 x 3.4 x 4.5 x cos A
3²= 3² + 5²– 2 x 3 x 5 x cos A
Cos A = 5/6 =0.8333
A = cos^-1 0.8333

A = 33⁰30’

COSINE OF THE SUM AND DIFFERENCE OF TWO ANGLES (A and B)

Cosine ( A + B ) = cos A cos B – sin A sin B
Cosine (A-B) = cos A cos B + sin A sin B

Verify that
Cos ( 90 – 60) = cos 90 cos 60 + sin 90 sin 60
Cos 30= cos 90⁰ cos 60⁰ + sin 90⁰ sin 60⁰
= 0 x +
=

Questions

1. find the cosine of 75⁰ without using mathematical tables.

solution
Cos 75⁰ = cosine (30 + 45)
= cosine 45 cosine 30 – sin45⁰ sin 30⁰
= x – x

=

=

= 0.2588

We use the knowledge of coordinate geometry to find the distance and cosine rule.

Consider a unit circle of radius 1 with points P and Q and angles A and B shown in the figure.

Let the distance from P to Q be d.

By distance formula

d²= ( cos A – cos B )² + ( sin A – Sin B) ²
d² = cos ²A 2 cos A cos B + cos ² B + sin ² A – 2 sin A sin B + sin²B
d²= cos ²A + sin ² A + cos ²B + sin ² B – 2 sin A sin B + sin ² B
d² = 1+1 – 2 (cos A cos B + sin A sin B)
d² = 22(sin A sin B + cos A cos B) ……. (i)

by the cosine rule

d² = 1² + 1² – 1 cos (A-B)
d² = 22 cos ( A-B) ……(ii)

equate equation (i) and (ii)
22 cos (A-B) = 2-2 ( cos A cos B + sin A sin B)
-2 cos (A-B) = -2 ( cos A cos B + sin A sin B)
Cos ( A B) = cos A cos B + sin A sin B …. (iii)

also
Cos (A+B) = cos ( A -B ) = cos A cos–B + sin A sin –B
= cos A cos B – sin A sin B ……( iv)

THE SINE OF THE SUM AND DIFFERENCE OF ANY TWO ANGLES

Consider a triangle ABC and c as a acute angle.

Sin C = Cos (90 – C) = …….. (i)
Now let C = 90 – A

Sin (90 – A) = Cos A = c/b …..(ii)

Also let c be A + B

Sin (A+ B) = cos (90 – (A+B))
Sin ( A+ B) = cos((90 – A) –B)
Sin (A+B) = cos (90- A ) cos B + sin (90- A) Sin B
Sin (A+B) = sin A cos B + sin B ( cos A)
Sine of the difference of any two angles A and B
Refer to the above expression
Sin (A-B) = sin (A + – B) = sin A cos –B + sin –B cos A

Sin A cos B – sin B cos A __ Difference of any two angles.

Exercise

1. (a) Find the truth set of
Sin θ = – in the domain 0≤ θ≤360⁰

(b) verify that for any small angle A⁰
cos (90-A ) = sin A

Solution
a. sin θ = –
sin negative is in the 3^rd and 4^th quadrant
3^rd
Θ-180⁰ = 30⁰
Θ = 180 + 30 ⁰
Θ = 210 ⁰

4^th
360 – θ =30⁰
– θ = 30⁰– 360⁰
– θ =- 330 ⁰
θ = 330 ⁰

The truth set of sin θ = = 0⁰ ≤ θ≤360⁰

b. cos (90-A) = sin A
cos 90 cos A + sin 90 sin A =sin A
0 x cos A + 1 x sin A = sin A
Sin A = sin A

2. Use sin( S + -t ) to help find a formula for sin (s t)

Solution
S t = s + ( -t)
= sin S cos –t + sin (-t) cos S
= sin S cos t – sin t cos S

3 . Verify that sin (+ ) = sin cos + sin sin

θ =

=

=
= 60⁰ x 5
= 300 ⁰
= 360⁰300⁰
Sin ( 120 + 300) = 120 cos 300 + cos 120 sin 300
Sin 420⁰= sin 60 cos 60 + -cos 60 –sin 60
Sin 60 = x + – x
= +
=