Trigo functions

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TRIGONOMETRIC
FUNCTIONS
Name
........................................................................................

zefry@sas.edu.my 2
TRIGONOMETRIC FUNCTIONS
5.1 Positive Angle and Negative Angle
Positive Angle Negative Angle
Represent each of the following angles in a Cartesian plane and state the quadrant of the angle.
Example
60
Quadrant 1
1(a) 70 (b) 150
Example
215
Quadrant III
2(a) 195 (b) 345
Example
395
Quadrant I
3(a) 415 (b) 480
Example
5
4

Quadrant III
4(a)
3
4
 (b)
5
3

Example
45
Quadrant IV
5(a) 130
(b)
1
3
 
2 radian = 360
 radian = 180
60
y
xO
y
xO
y
xO
Quadrant
II
Quadrant
I
Quadrant
III
Quadrant
IV
y
x
90
2
 
 
 
0
360 (2)
180 ()
270
3
2
 
 
 
215
y
xO
y
xO
y
xO
A positive angle is measured
in an anticlockwise direction
from the positive x-axis.
45
y
xO
y
xO
y
xO
y
xO
y
xO
5
4

y
xO
y
xO
y
xO
60
y
xO
Anticlockwise
direction
45
y
xO
Clockwise
direction
395
y
xO
35
360
A negative angle is measured
in a clockwise direction
from the positive x-axis.

zefry@sas.edu.my 3
5.2 Six Trigonometric Functions of any Angle (1)
5.2.1 Define sine, cosine and tangent of any angle in a Cartesian plane
1  
 
 
 
 
 
sin
cos
tan

 

Conclusion :
2
r2
= 32
+ 42
r = 2 2
3 4
r = 5
Conclusion :
Pythagoras’ Theorem :
2 2 2 2 2
2 2 2 2 2
2 2 2 2 2
c a b c a b
a c b a c b
b c a b c a
    
    
    
3. Find the length of OA and the values of sine, cosine and tangent of .
(a)  in quadrant I
OP =
=
sin  =
 
5
cos  =
 
12
tan  =
 
5
(b)  in quadrant II
OP =
=
sin  =
 
6
=
cos  =
 
8
=
tan  =
 
6
=

x
r
y
Opposite
sin
Hypotenuse
Adjacent
cos
Hypotenuse
Opposite
tan
Adjacent
 
 
 

3
r
4
b
ca
P (12, 5)
5
12O x
y

P (8, 6)
x
y
O

8
6
Opposite to 
Hypotenuse
Adjacent to 


zefry@sas.edu.my 4
(c)  in quadrant III
OP =
=
sin  =
 
3
cos  =
 
4
tan  =
 
3
=
(d)  in quadrant IV
OP =
=
sin  =
 
12
cos  =
 
5
tan  =
 
12
(e) Conclusion:
Sin  is positive for  in quadrant ……. and …….
Cos  is positive for  in quadrant ……. and …….
Tan  is positive for  in quadrant ……. and …….
Sin  is negative for  in quadrant ……. and …….
Cos  is negative for  in quadrant ……. and …….
Tan  is negative for  in quadrant ……. and …….
4. Find the corresponding reference angle of .
(a)
Reference angle = 55
(b)
Reference angle =  110
= 70
(c)
Reference angle = 215 
= 35
(d)
Reference angle =  300
= 60
y
x
90
0
360
180
270
Sin 
Cos 
Tan 
Sin 
Cos 
Tan 
Sin 
Cos 
Tan 
+
Sin 
Cos 
Tan 
P (4, 3)
x
y
4
O

3
P (5, 12)
x
y
5
O 
12
55
y
x
Fill in with  or + sign.
110
y
x180 360
215
y
x180 360
300
y
x180 360

zefry@sas.edu.my 5
(e) Conclusion:
Reference angle (RA) is the acute angle formed between the rotating ray of the angle and the
______________________________
In Quadrant II: In Quadrant III In Quadrant IV
sin  = sin (180  ) sin  = sin (  180) sin  = sin (360  )
cos  = cos (180  ) cos  = cos (  180) cos  = cos (360  )
tan  = tan (180  ) tan  = tan (  180) tan  = tan (360  )
5. Given that cos 51 = 0.6293, find the trigonometric ratios of cos 231 without using a calculator or
mathematical tables.
Reference angle of 231 = 231 
=
cos 231 =
=
6. Given that sin 70 = 0.9397, find the trigonometric ratios of sin 610 without using a calculator or
Reference angle of 610 = 610  
=
sin 610 =
=
7. Given that tan 25 = 0.4663, find the trigonometric ratios of tan 335 without using a calculator or
Reference angle of 335 =  335
=
tan 335 =
=
R.A = R.A =  
R.A =   R.A =  
y
x
y
xO 180

y
xO
180   
y
x
O
360  

y
x
y
x
y
x

zefry@sas.edu.my 6
5.2.2 Define cotangent, secant and cosecant of any angle in a Cartesian plane.
1
 
 
 
 
 
 
sin
cos
tan
 
 
 
2
 
 
 
 
 
 
 
 
 
 
1 1 r
sin y
1 1
cos
1 1
tan
 

 

 

3. Definition of cotangent , secant  and cosecant .
1
cosec
sin
1
sec
cos
1
cot
tan
 

 

 

4. Since
sin
tan
cos

 

, then
cot  
5.
 
 
 
sin sin 90
cos cos 90
tan tan 90
y x
r r
x y
r r
y x
x y
    
    
    
6.
Complementary angles:
sin  = cos (90  )
cos  = sin (90  )
tan  = cot (90  )
cosec  = sec (90  )
sec  = cosec (90  )
cot  = tan (90  )
7. Given that sin 48 = 0.7431, cos 48 = 0.6991 and tan 48 = 1.1106, evaluate the value of cos 42.
cos 42 =
= =
8. Given that sin 67 = 0.9205, cos 67 = 0.3907 and tan 67 = 2.3559, evaluate the value of cot 23.
cot 23 =
= =
9. Given that sin 37 = 0.6018, cos 37 = 0.7986 and tan 37 = 0.7536, evaluate the value of sec 53.
sec 53 =
= =

x
r
y

x
r
y
90 

x
r
y
48
90  48
67
90  67
37
90  37

zefry@sas.edu.my 7
5.2.3 Find values of six trigonometric functions of any angle
1. Complete the table below.
30 45 60
sin 
1
2
cos 
1
2
tan  1
2. Use the values of trigonometric ratio for the special angles, 30, 45 and 60, to find the value of the
trigonometric functions below
Example: Evaluate sin 210 a. Evaluate tan 300
Draw diagram to determine positive or negative
 sin
Draw diagram to determine positive or negative
Find reference angle
Reference angle of 210 = 210  180
= 30
Find reference angle
Solve
sin 210 =  sin 30
=
1
2

Solve
b. Evaluate cos 150 c. Evaluate sec 135
Draw diagram to determine positive or negative Draw diagram to determine positive or negative
Find reference angle Find reference angle
Solve Solve
60 60
60
2
2
2
60
30 2
1
2 2
2 1
3


1
1
2 2
1 1
2


45
45
1 1
1
1
y
x180 360
cos ( ) = cos 
sin ( ) =  sin 
tan ( ) =  tan 
y
x
O
 

zefry@sas.edu.my 8
5.2.4 Solve trigonometric equations
A. Steps to solve trigonometric equation
1. Determine the range of the angle.
2. Find the reference angle using tables or calculator.
3. Determine the quadrant where the angle of the trigonometric function is placed.
4. Determine the values of angles in the respective quadrants.
1. Solve the following equation for 0    360.
Example: sin  = 0.6428 a. cos  = 0.3420
Range :
0    360 0    360
Reference angle :
 = sin1
0.6428
 = 40
Quadrant :
Quadrant I Quadrant II Quadrant ____ Quadrant ____
Actual angles
 = 40 ,  = 180  40
 = 40 , 140
b. tan  = 1.192 c. cos  =  0.7660
Range :
Reference angle :
Quadrant :
Quadrant ___ Quadrant ___ Quadrant ___ Quadrant ___
Actual angles
y
x
y
x
y
x180 360
S A
T C
y
x180 360
S A
T C
y
x
y
x
y
x180 360
40
S A
T C
y
x180 360
40
S A
T C

zefry@sas.edu.my 9
d. sin  =  0.9397 e. tan  =  0.3640
Range :
Reference angle :
Quadrant :
Actual angles
f. cot  =  1.4826 g. cosec  =  2.2027
Range :
Reference angle :
Quadrant :
Actual angles
2. Solve the following equation for 0    360.
example : sec 2 = 2 a. 2 sin 2 = 1.6248
Range : 0    360
0  2  720
Reference angle :
1
2
2
1
2
2
2 60
cos
cos


 
  
Quadrant :
Actual angles
2 = 60, 360  60, 60 + 360, (36060) + 360
 = 60, 300, 420, 660
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x180 360,720
S A
T C
60
60

zefry@sas.edu.my 10
b. cos 3 =  0.9781
c. tan
2

=  2.05
Range
Reference angle :
Quadrant :
Actual angles
d. sin ( + 10) = 0.7660 e. cos ( + 40) = 0.7071
f. tan ( + 15) = 1 g. cos (  20) = 0.5
h. tan (2  10) =  2.082 i. sin (2  30) = 0.5

zefry@sas.edu.my 11
j. sin  = cos 20 k. cos  =  sin 55
Example : 2 sin x cos x = cos x
2 sin x cos x  cos x = 0
cos x ( 2 sin x  1) = 0
cos x = 0 , 2 sin x  1 = 0
sin x =
1
2
x = 30
x = 90 , 270
x = 30, 150
 x = 30, 90, 150, 270
m. 2sin x cos x = sin x
n. 2 cos 2
 + 3 cos  =  1 o. 2 sin2
 + 5 sin  = 3
p. tan2
 = tan  q. 3 sin  = 2 + cosec 
y
x360
y
x180 360
S A
T C

zefry@sas.edu.my 12
3. Given that px sin and 00
< x < 900
.
Express each of the following trigonometric ratios in terms of p.
(a) sec x = (b) cosec x =
(c) tan x = (d) cot x =
(e) sin ( 900
- x) = (f) cos (900
- x) =
(g) sec (900
- x) = (h) cosec (900
– x) =
(i) tan ( 90o
- x) = (j) cot ( 90o
– x ) =
(k) sin(-x) = (l) cos (-x) =
x

zefry@sas.edu.my 13
4. Given that
17
8
sin x and 2700
< x < 3600
.
Without using tables or calculator, find the values
of.
5. Given that
17
8
-cos x and 1800
< x < 2700
.
Without using tables or calculator, find the values
of
(a) cos x = (a) sin x =
(b) tan x = (b) tan x =
(c) cosec x = (c) cosec x =
(d) sec x = (d) sec x =
(e) cos (900
– x) = (e) sec (900
– x) =
(f) sin ( 900
– x ) = (f) cot ( 900
– x ) =
(g) sin (-x) = (g) sin (-x) =
(h) tan (-x) = (h) cos (-x) =
x x

zefry@sas.edu.my 14
5.4 Basic Identities
5.4.1 Prove Trigonometric Identities using Basic Identities
Three basic trigonometric identities :
sin 2
 + cos 2
 = 1
1 + tan 2
 = sec 2

1 + cot 2
 = cosec 2

Formula of compound angle :
sin (A  B) = sin A cos B  cos A sin B
cos (A  B) = cos A cos B Ŧ sin A sin B
tan (A  B) =
tan tan
1 tan tan
A B
A B

Formula of double angle :
sin 2A = 2 sin A cos A
cos 2A = cos2
A − sin2
A
= 2 cos2
A − 1
= 1 − 2sin2
A
tan 2A =
A
A
2
tan1
tan2

Formula of half angle :
sin A = 2 sin
2
A
cos
2
A
cos A = cos2
2
A
− sin2
2
A
= 2 kos2
2
A
− 1
= 1 − 2sin2
2
A
tan 2A =
2
2
2
1
2
A
tan
A
tan
1. Prove the following identities
Example: cot  + tan  = cosec  sec 
2 2
1
cos sin
cot tan
sin cos
cos sin
sin cos
sin cos
cosec sec
 
   
 
  

 

 
  
a. tan2
 (1  sin2
) = sin2

b.
2
1
1
sin
cos
cos

  
 
c. sin2
 + cot2
 = cosec2
  cos2

cos2
 = 1 – sin2

sin2
 = 1 – cos2


zefry@sas.edu.my 15
d.  2 1
1
sin
sec tan
sin
 
   
 
e.
1
2
1
sin x cos x
sec x
cos x sin x

 

2. Solve the following equations for 0  x  360.
a. 3 sin x + 2 = cosec x
b. 2 cot2
x  5 cot x + 2 = 0 c. cos2
x  3 sin2
x + 3 = 0
d. cot2
x= 1 + cosec x e. 2 tan 2
x = 4 + sec x

zefry@sas.edu.my 16
ANSWERS
5.2.1
5. cos 51= 0.6293
5.2.4
2a. 0  2  720 , 54.33
 = 27.17, 62.83, 207.17, 242.83
3(a)
2
1
1
p
5.(a) 
17
15
6. sin 70= 0.9397 b. 0  3  1080 , 12.01
 = 56, 64, 176, 184, 296, 304
(b)
p
1
(b)
15
8
7. tan 25= 0.4663
c. 0 
2

 180 , 64
 = 232
(c)
2
1 p
p

(c)  17
15
5.2.2
7. sin 48 = 0.7431
d.  = 40, 120
(d)
p
p2
1 (d) 
17
8
8. tan 67 = 2.3559 e.  = 5, 275 (e) 2
1 p (e) 
17
15
9. cosec 37 = 1.6617 f.  = 30, 210 (f) p
(f)
8
15
5.2.3
2a. tan 300 = 3
g.  = 80, 320 (g)
p
1 (g) 15
17
b. cos 150 =
3
2

h.  = 62.83, 152.83, 242.83, 332.83
(h)
2
1
1
p
(h)
17
8

c. sec 135 =
1
2

i.  = 30, 90, 210, 270
(i)
p
p2
1
5.2.4
1a.70 , Quadrant I, IV
 = 70, 290
j.  = 70, 110
(j)
2
1 p
p

b. 0    360 , 50 ,
Quadrant I, IV
 = 30, 330
k.  = 145, 215 (k) -p
c. 0    360 , 40 ,
Quadrant II, III
 = 140, 220
m.  = 60, 180, 300 (l) 2
1 p
d. 0    360 , 70
Quadrant III, IV
 = 250, 290
n.  = 120, 180, 240 4.(a)
17
15
e. 0    360 , 20.01
Quadrant II, IV
 = 159.99, 339.99
o.  = 30, 150
(b) 
15
8
f. 0    360 , 34
Quadrant II, IV
 = 146, 326
p.  = 0, 45, 225 (c) 
17
8
g. 0    360 , 27
Quadrant III, IV
 = 207, 333
q.  = 90, 199.47, 350.53
(d)
15
17
(e)
17
8

(f)
17
15
(g)
17
8
(h)
8
15

Trigo functions

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Trigo functions