WEEK 8
TOPIC: Trigonometric functions
The basic trigonometric ratios can be defined in two ways:
(i)traditional definition;
(ii) modern definition.
Traditional Definition
The basic trigonometric ratios can be defined in terms of the sides of a right – angled triangle.
Q
r p
P ѳ
q R
Fig. 14.3
∆PQR in Fig 14.3 in a right angled triangle with QPR = ѳ and PRQ = 90◦. We define the three basic rations as follows:
Cosine of angle ѳ =
Sine of angle ѳ =
Tangent of ѳ =
The cosine of angle ѳ, sine of angle ѳ and the tangent of angle ѳ will be abbreviated cosѳ, sinѳ and tanѳ respectively.
Thus:
Cosѳ =
Sinѳ =
Tanѳ –
Also:
=
tanѳ =
Ratios of the General Angle
First Quadrant
P1
y
Fig 14.7
In Fig. 14.7 ∆OP1N1 is a right – angled triangle constructed from a unit circle.
OP1 = 1
P1N1 = y
ON1 = x
P1ON1=ѳ1
Sinѳ1 = y
Cos̴ѳ1 = x
Tanѳ1 =
Second Quadrant
P2
1
y
(a)
y
P2 P1
Ѳ1 ѳ2 ѳ1
x
(b)
Fig. 14.8
From Fig. 14.8(a)
Sinѳ2 = y
Cosѳ2 = -x
Tanѳ2 =
=
From Fig. 14.8(b)
Ѳ1 + ѳ2 = 180◦
Ѳ2 = 180◦ – ѳ1
: Sinѳ2 = sin(180◦ – ѳ1) = y = sinѳ1
: Sin(180◦ – ѳ) = sinѳ
: Cosѳ2 = cos(180◦ – ѳ1) = -x = -cosѳ1
:Cos(180◦ – ѳ) = -cosѳ
Similarly,
Tanѳ2 = tan (180◦ – ѳ1) = 1
: tan (180◦ – ѳ) = -tanѳ
Hence in the second quadrant:
Sin(180◦ – ѳ) = sinѳ
Cos(180◦ – ѳ) = -cosѳ
Tan(180◦ – ѳ) = -tanѳ
Third Quadrant
N1– x ѳ3 (a)
-y Q
P3 1
y
P1
(b) ѳ1
Ѳ1
Ѳ1
P3
Fig. 14.9
Sinѳ3 = -y
Cosѳ3 = -x
Tanѳ3 = sinѲ3 =
cosѲ3
From Fig. 14.9(b)
Ѳ1 = 180◦ + Ѳ1
SInѳ3 = sin(180◦ + Ѳ) = -y = -sinѲ1
:sin(180◦ + Ѳ) = -sinѲ
Cosѳ3 = cos(180◦ + Ѳ) = -x = -cosѲ1
: cos(180◦ + Ѳ) = -sinѲ
Similarly,
Tanѳ3 = tan (180◦ + Ѳ1)= = tanѳ1
: tan(180◦ + ѳ) = tanѳ
Hence in the third quadrant:
Sin(180◦ + ѳ) = -sinѳ
Cos(180◦ + ѳ) = -cosѳ
Tan(180◦ + ѳ) = tanѳ
Fourth Quadrant
(a)
x
-y
Ѳ1 1
P4
y
(b) P1
Ѳ1
Ѳ1 x
Ѳ1
P2
Fig. 14.10
Sinѳ4 = y
Cosѳ4 = x
Tanѳ4 =
From Fig, 14.10(b)
Ѳ4 + Ѳ1= 360◦
Ѳ4 = 360◦ – ѳ
SInѳ1 = sin(360◦ + Ѳ) = -y = -sinѲ1
: sin(360◦ + Ѳ) = -sinѲ
Cosѳ1 = cos(360◦ + Ѳ) = -x = -cosѲ1
: cos(360◦ + Ѳ) = -sinѲ
Tanѳ4 = tan (360◦ + Ѳ)= = -tanѳ
: tan(360◦ + ѳ) = -tanѳ
Hence in the third quadrant:
Sin(360◦ + ѳ) = -sinѳ
Cos(360◦ + ѳ) = cosѳ
Tan(360◦ + ѳ) = -tanѳ
(a) In the first quadrant, all the ratios are positive.
(b) In the second quadrant, only sin ratio is positive, while the rest are negative.
(c) In the third quadrant, only tangent ratio is positive, while the rest are negative.
(d) In the fourth quadrant, only cosine ratio is positive, while the rest are negative.
These observations can be summarized in the figure below:
y y
SINL ALL S A
T C
(a) (b)
Negative Angles
y
360◦ – 0
x
P
Ps
(a) (b)
Fig. 14.12
Since negative angles are measured in the clockwise sense, the direction of OP when rotated through –ѳ is the same as where it is rotated through 360◦ – ѳ.
Hence in the forth quadrant:
Sin(-ѳ) sin sin(360◦ – ѳ) = -sinѳ
Cos(-ѳ) cos(360◦ – ѳ) = -cosѳ
Tan(-ѳ) tan(360◦ – ѳ) = -tanѳ
Use tables to evaluate each of the following:
(a) sin 143◦ (b) cos 115◦
(c) tan 125◦
Solution
(a) 143◦ is in the second quadrant, so
Sin143◦ = sin(180◦ – 143◦)
= sin37◦
= 0.6018
(b) 115◦ is in the second quadrant, so
Cos115◦ = -cos(180◦ – 115◦)
= -cos65◦
= -0.4226
(c) 125◦ is in the second quadrant, so
Tan125◦ = -tan(180◦ – 125◦)
= -tan55◦
= -1.428
Use tables to evaluate each of the following
(a) sin230◦ (b) cos236◦
(c)tan 242◦
Solution
220◦, 236◦ and 242◦ are all in the third quadrant, hence;
(a) sin220◦ = sin(180◦ + 40◦)
= -sin40◦
= -0.6428
(b) cos236◦ = cos(180◦ + 56◦)
= -cos56◦
= -0.5992
(c) tan242◦ = tan(180◦ + 62◦)
= tan62◦
= 1.881
Use tables to evaluate each of the following:
(a) sin310◦ (b) cos285◦
(c) 334◦
Solution
310◦, 285◦ and 334◦ are all in the fourth quadrant, hence;
(a) sin310◦ = sin(360◦ – 50◦)
= -sin50◦
= -0.7660
(b) cos285◦ = cos(360◦ – 75◦)
= cos75◦
= 0.2588
(c) tan334◦ = tan(360◦ – 26◦)
= -tan26◦
= -0.4877
Use tables to evaluate each of the following
(a) cos(-30◦) (b) sin(-60◦)
(c) tan(-120◦)
Solution
(a) cos(-30◦) = cos330◦
= cos30◦
= 0.8660
(b) sin(-60◦) = sin300◦
= -sin60◦
= -8660
(c) tan(-120◦) = tan240◦
= tan60◦
= 1.732
Use the table to find the value of ѳ between ѳ◦ and 360◦ which satisfy each of the following:
(a) cosѳ = -0.4540
(b) tanѳ = 1.176
(c) sinѳ = -0.9336
Solution
(a) The cosine ratio is negative in the second and third quadrants. First find the acute angle whose cosine is 0.4540
From the tables cos 63◦ = 0.4540
: In the second quadrant
Ѳ = 180◦ – 63◦
= 117◦
In the third quadrant,
Ѳ = 180◦ + 63◦
= 243◦
(b) The tangent ratio is positive in the first and third quadrants.
First find the acute angle whose tangent is 1.176.
From the tables.
Tan49.62◦= 1.176◦
In the first quadrant.
Ѳ = 49.62◦
In the third quadrant.
Ѳ = 180◦+ 49.62◦
= 229.62◦
(c) The sine ratio is negative in the third and fourth quadrant.
First find the acute angles whose sine ratio is 0.9336.
From tables.
Sin69◦ = 0.9336
In the third quadrant
Ѳ = 180◦ + 69◦
= 249◦
In the fourth quadrant.
Ѳ = 360◦ – 69◦
= 291◦
Evaluation
- In what quadrant are the followings ; tan ( -540) , cos (- 1080)
General Evaluation
(1) Prove that (1 –sinѳ) (1 + sinѳ) = cot2ѳ
sin2ѳ
(2) Show that (secѳ – tanѳ) (secѳ + tanѳ) = 1
(3) Find the values of sin (-210) in surd form
Reading Assignment
F/Maths Project 1 pages 225 – 247 Exercise 14 Q1, 3, 5, 7 and 8
Weekend Assignment
Given that sinѳ =
(1)Find cosѳ (a)
(2) Find tanѳ
(3) Find cosecѳ
(4) Find secѳ (a)
(5) Find cotѳ
THEORY
1) Prove that 1/1+cosx + 1/1-cosx = 2 cosec2 x
2) Given that sin x = 5/13 and x is acute find cosec x
, cot x and sec x