1. The Values of sin , cos , tan 

2. The Values of sin , cos , tan 
Quadrants and angles in the unit circle y
Cartesian plane can be divided into four parts called quadrants.
Quadrants are named in the anticlockwise direction. 0°
90°
180°
270°
360°
Quadrant I
Quadrant II
Quadrant III
Quadrant IV x

3. The Values of sin , cos , tan 
Quadrants and angles in the unit circle y
Angle  is measured by rotating the line OP in the anticlockwise direction from the positive x-axis at the origin, O. P  x O

4. The Values of sin , cos , tan 
Verify sin  = y-coordinate in quadrant I of the unit circle y
sin  = =
= y P
(x, y) 1 y x  x O Q
 sin  = y-coordinate

5. The Values of sin , cos , tan 
Verify cos  = x-coordinate in quadrant I of the unit circle y
cos  = =
= x P
(x, y) 1  y x O x Q
 cos  = x-coordinate

6. The Values of sin , cos , tan 
Verify tan  = in quadrant I of the unit circle y
tan  = = P
(x, y) 1  y x O x Q
 tan  =

7. The Values of sin , cos , tan 
Determine whether the value is positive or negative y x
Quadrant I = All positive
sin
All
Quadrant II = sin  positive II I
III IV
Quadrant III = tan  positive
tan
cos
Quadrant IV = cos  positive

8. The Values of sin , cos , tan 
Determine whether the value is positive or negative
Example 1: sin 213°
Sin  is positive in quadrant II. y
The angle 213° lies in quadrant III.
213° x O
Therefore, the value of sin 213° is negative.
Not
quadrant II

9. The Values of sin , cos , tan 
Determine whether the value is positive or negative
Example 2: cos 321°
Cos  is positive in quadrant IV. y
The angle 321° lies in quadrant IV.
321° x O
Therefore, the value of cos 321° is positive.
It is
quadrant IV.

10. The Values of sin , cos , tan 
Determine whether the value is positive or negative
Example 3: tan 123°
Tan  is positive in quadrant III. y
The angle 123° lies in quadrant II.
123° x O
Therefore, the value of tan 123° is negative.
Not
quadrant III

11. The Values of sin , cos , tan 
Determine whether the value is positive or negative
Example 4: sin 32°
All positive in quadrant I. y
The angle 32° lies in quadrant I.
32° x O
Therefore, the value of sin 32° is positive.
It is
quadrant I.

12. The Values of sin , cos , tan 
Determine the values of sine, cosine and tangent for special angles
sin 45° =
45° 1 
cos 45° =
tan 45° = 1

13. The Values of sin , cos , tan 
Determine the values of sine, cosine and tangent for special angles
sin 30° =
30°
60° 1 2
cos 30° =
tan 30° =

14. The Values of sin , cos , tan 
Determine the values of sine, cosine and tangent for special angles
sin 60° =
30°
60° 1 2
cos 60° =
tan 60° =

15. The Values of sin , cos , tan 
Determine the values of sine, cosine and tangent for special angles  0°
90°
180°
270°
360°
sin  1 –1
cos 
tan  y x O
(0, 1)
(–1, 0)
(1, 0)
(0, –1)

16. The Values of sin , cos , tan 
Determine the values of sine, cosine and tangent for special angles
Summary:  0°
30°
45°
60°
90°
180°
270°
360°
sin  1 –1
cos 
tan 

17. The Values of sin , cos , tan 
Determine the values of sine, cosine and tangent for special angles
Question 1: Calculate the values of the following:
7 sin 90° + 4 cos 180 °
Solution:
7 sin 90° + 4 cos 180 ° =
7 × (1) + 4 × (–1)
= 7 – 4
= 3

18. The Values of sin , cos , tan 
Values of angles in quadrant II y x O P  
 = corresponding angle in quadrant I
between x-axis and line OP

19. The Values of sin , cos , tan 
Values of angles in quadrant II
   where  = 180° – 
sin  = + sin 
cos  = – cos 
tan  = – tan  y P   x O X
 =  – 90°

20. The Values of sin , cos , tan 
Values of angles in quadrant III
   where  =  – 180°
sin  = – sin 
cos  = – cos 
tan  = + tan  y  x  P O X
 = 270° – 

21. The Values of sin , cos , tan 
Values of angles in quadrant IV
   where  = 360° – 
sin  = – sin 
cos  = + cos 
tan  = – tan  y  x  O X P
 =  – 270°

22. The Values of sin , cos , tan 
Finding the value of an angle
Question 1: Find the value of sin 231°.
Solution: y x O
231°  quadrant III
sin 231°  negative
sin 231° = – sin (231° – 180°)
231°
= – sin 51° 
= – P

23. The Values of sin , cos , tan 
Finding the value of an angle
Question 2: Find the value of cos 303° 17‘.
Solution: x y O
303° 17'  quadrant IV
cos 303° 17'  positive
cos 303° 17' = cos (360° – 303° 17')
303° 17'
= cos 56° 43'  = P

24. The Values of sin , cos , tan 
Finding the value of an angle
Question 3: Find the value of tan 117° 13'.
Solution: x y O
117° 13'  quadrant II P
tan 117° 13'  negative
tan 117° 13' = – tan (180° – 117° 13') 
117° 13'
= – tan 62° 47'
= – 1.945

25. The Values of sin , cos , tan 
Finding angles between 0° and 360°
Question 1: For sin x = where 0° ≤ x ≤ 360°, find the value of x.
Solution:
72° x P y x y
Corresponding acute angle, x = 72° x O
72° P
 positive
Therefore, the acute angle is in quadrant I or II.
and
Quadrant I: x =
72°
Quadrant II: x =
180° – 72° = 108°

26. The Values of sin , cos , tan 
Finding angles between 0° and 360°
Question 2: For tan x = – where 0° ≤ x ≤ 360°, find the value of x.
Solution:
Corresponding acute angle, x = 60° 12' x y x y
–  negative
60° 12' O P x
60° 12' P x
Therefore, the acute angle is in quadrant II or IV.
and
Quadrant II: x =
180° – 60° 12' = 119° 48'
Quadrant IV: x =
360° – 60° 12' = 299° 48'

27. The Values of sin , cos , tan 
Finding angles between 0° and 360°
Question 3: For cos x = 0.5 where 0° ≤ x ≤ 360°, find the value of x.
Solution: x y x y
Corresponding acute angle, x = 60°
60° P x
60° O P x
0.5  positive
Therefore, the acute angle is in quadrant I or IV.
and
Quadrant I: x =
60°
Quadrant IV: x =
360° – 60° = 300°

28. The Values of sin , cos , tan 
Solve problems involving sine, cosine and tangent
Question: In the diagram below, HMS and JHN are straight lines. H is the midpoint of JN. Given that HM = 12 cm, MN = 13 cm and FJ = 4 cm, calculate:
the length of HN,
the value of cos x°,
the value of tan y°. N H M S J F y° x°

29. The Values of sin , cos , tan 
Solve problems involving sine, cosine and tangent
Solution:
(a) N
Pythagoras’ theorem
HN2 =
132 – 122
13 cm x°
= 169 – 144
= 25 y° S H
12 cm M
 HN = 5 cm F J
4 cm

30. The Values of sin , cos , tan 
Solve problems involving sine, cosine and tangent
Solution:
HMS is a straight line
(b) N
x° = 180° – HMN x°
cos x° =
– cos HMN y° S H M = F J

31. The Values of sin , cos , tan 
Solve problems involving sine, cosine and tangent
Solution:
JHN is a straight line
(c) N
y° = 180° – FHJ x°
tan y° =
– tan FHJ y° S H M = F J

32. The End