1. Trigonometric Identities

2. Trigonometric Ratios of (180   )y
The circle is centred at the origin O. It cuts the positive x-axis at A. A x O

3. Trigonometric Ratios of (180   )y
P(3, 4) is a point on the circle. 
P(3, 4)
AOP =  A x O
By the definitions of trigonometric ratios,
sin  =
cos  =
tan  =

4. Trigonometric Ratios of (180   )y
P is reflected about the y-axis to Q.
Q(3, 4) Q 
P(3, 4)
Coordinates of Q =
(3, 4)
180º - q  A x O
AOQ =
180  
By the definitions of trigonometric ratios,
sin (180   ) =
cos (180   ) =
tan (180   ) =

5. We have sin (180   ) = sin  cos (180   ) = cos 
tan  =
tan (180   ) =
sin (180   ) = sin 
cos (180   ) = cos 
tan (180   ) = tan 

6. In fact, the relationships between the trigonometric ratios of  and (180   ) are true for any acute angle .

7. By the definitions of trigonometric ratios,
Suppose P(a, b) is a point on a circle of radius r centred at the origin O. y 
P(a, b)
By the definitions of trigonometric ratios, r B A x O
sin  = r b
cos  = r a
tan  = a b

8. Q is obtained by reflecting P about the y-axis.
Q(a, b) Q 
P(a, b) r r
Coordinates of Q =
(a, b)
180º – q B  A x O
OQ = r
AOQ =
180  

9. By the definitions of trigonometric ratios,
Q(a, b)
P(a, b)
sin (180   ) = r b
= sin  r
180º – q B  A x O r a
cos (180   ) = -
= -cos  a b
tan (180   ) = -
= -tan 
cos  = r a
tan  = b
sin  =
Note:

10. Hence, for any acute angle , we have
sin (180   ) = sin 
cos (180   ) = cos 
tan (180   ) = tan 
If  is an acute angle, then (180   ) lies in quadrant II, and only sin (180   ) is positive.

11. Trigonometric Ratios of (180 +  )y
R is obtained by rotating P(a, b) through 180 about O.
P(a, b) r
180º + q
180
Coordinates of R
= (a, b)  A x O
OR = r r
Reflex AOR
= 180 + 
R(a, b) R

12. Trigonometric Ratios of (180 +  )y
By the definitions of trigonometric ratios,
P(a, b)
sin (180 +  ) = r b 
= -sin 
180º + q  A x O r a
cos (180 +  ) = 
= -cos  r
R(a, b) a b
tan (180 +  ) =
= tan 
cos  = r a
tan  = b
sin  =
Note:

13. Hence, for any acute angle , we have
sin (180 +  ) = –sin 
cos (180 +  ) = –cos 
tan (180 +  ) = tan 
If  is an acute angle, then (180 +  ) lies in quadrant III, and only tan (180  +  ) is positive.

14. Trigonometric Ratios of (360   ) & y
S is obtained by reflecting P(a, b) about the x-axis.
P(a, b)
Coordinates of S =
(a, b) r
360º - q  A x
OS = r O q r
Reflex AOS
= 360   S
(a, b)

15. Trigonometric Ratios of (360   ) & y
By the definitions of trigonometric ratios,
P(a, b)
sin (360 -  ) = r b -
= -sin 
360º - q  A x O r a
cos (360 -  ) =
= cos  r S
(a, b) a b
tan (360 -  ) =
= -tan  -
cos  = r a
tan  = b
sin  =
Note:

16. Hence, for any acute angle , we have
sin (360 –  ) = –sin 
cos (360 –  ) = cos 
tan (360 –  ) = –tan 
If  is an acute angle, then (360 –  ) lies in quadrant IV, and only cos (360 –  ) is positive.

17. What about the trigonometric ratios of  ?
What do you observe about the terminal sides of  and (360   ) ?

18. The terminal sides of - and (360 -  ) are coincident.y
360 -  x O -
The terminal sides of - and (360 -  ) are coincident.

19. Since the terminal sides of  and (360   ) are coincident, we have
sin (360 –  ) = –sin 
cos (– ) =
cos (360 –  ) = cos 
tan (– ) =
tan (360 –  ) = –tan 
If  is an acute angle, then – lies in quadrant IV, and only cos (– ) is positive.

20. Trigonometric Ratios of (360 +  )x y 
360 + 
The terminal sides of  and (360 +  ) are coincident.
We have
sin (360 +  ) = sin 
cos (360 +  ) = cos 
tan (360 +  ) = tan 

21. Let’s summarize using the ‘CAST’ diagram.

22. For any acute angle , we have
Sine is positive.
All are positive. O y x S A
sin (180 –  ) = + sin q
sin (360 +  ) = + sin q
cos (180 –  ) = – cos q
180 - 
cos (360 +  ) = + cos q
180 + 
360 - 
tan (180 –  ) = – tan q 
tan (360 +  ) = + tan q T C
sin (180 +  ) = – sin q
sin (360 –  ) = – sin q
cos (180 +  ) = – cos q
cos (360 –  ) = + cos q
tan (180 +  ) = + tan q
tan (360 –  ) = – tan q
Tangent is positive.
Cosine is positive.
These identities are also true even if  is not an acute angle.

23. (a) cos 125 = cos (180  55) = cos 55
Express each of the following trigonometric ratios in terms of the same trigonometric ratio of an acute angle.
(a) cos 125
= cos (180  55)
= cos 55
 cos (180   ) = cos 

24. (b) sin (250) = sin [360 + (250)] = sin 110 = sin (180  70)
Express each of the following trigonometric ratios in terms of the same trigonometric ratio of an acute angle.
(b) sin (250)
= sin [360 + (250)]
 sin (360 +  ) = sin 
= sin 110
= sin (180  70)
= sin 70
 sin (180   ) = sin 

25. (a) sin 210 = sin (180 + 30) = –sin 30 =
Find the values of the following trigonometric ratios.
(a) sin 210
= sin (180 + 30)
= –sin 30
 sin (180 +  ) = sin  =

26. (b) tan 315 = tan (360 – 45) = –tan 45 = –1
Find the values of the following trigonometric ratios.
(b) tan 315
= tan (360 – 45)
= –tan 45
 tan (360   ) = tan 
= –1

27. Follow-up question
Simplify

28. Trigonometric Ratios of (90 +  )
By using the trigonometric ratios of (180 - q) and (90 - q), we have )
90
sin( q + )
90
180
sin( q + - = )]
90 (
180
sin[ q - = )
90
sin( q - =
sin (180 - f) = sin f q
cos = )
90
cos( q + )
90
180
cos( q + - = )]
90 (
180
cos[ q - = )
90
cos( q - =
cos (180 - f) = -cos f q
sin - =

29. Trigonometric Ratios of (90 +  )
By using the trigonometric ratios of (180 - q) and (90 - q), we have )
90
tan( q + )
90
180
tan( q + - = )]
90 (
180
tan[ q - = )
90
tan( q - =
tan (180 - f) = -tan f q
tan 1 - =

30. Hence, for any acute angle , we have
sin (90 +  ) = cos 
cos (90 +  ) = -sin 
tan (90 +  ) =
These identities are also true even if  is not an acute angle.

31. Trigonometric Ratios of (270 –  ) & (270 +  )
By using the trigonometric ratios of (180 + ) and (90 - q), we have
sin (270 – q)
= sin [180 + (90 – )]
= –sin(90 – )
= –cos 
sin (180 + f) = –sin f
cos (270 – q)
= cos [180 + (90 – )]
= –cos (90 – )
= –sin 
cos (180 + f) = –cos f

32. Trigonometric Ratios of (270 –  ) & (270 +  )
By using the trigonometric ratios of (180 + ) and (90 - q), we have
tan (270 – q)
= tan [180 + (90 – )]
= tan (90 – ) q
tan 1
tan (180 + f) = tanf =

33. Trigonometric Ratios of (270 –  ) & (270 +  )
By using the trigonometric ratios of (360 – ) and (90 – q), we have
sin (270 + q)
= sin [360 – (90 – )]
= –sin (90 – )
= –cos 
sin (360 – f) = –sin f
cos (270 + q)
= cos [360 – (90 – )]
= cos (90 – )
= sin 
cos (360 – f) = cos f

34. Trigonometric Ratios of (270 –  ) & (270 +  )
By using the trigonometric ratios of (360 – ) and (90 – q), we have
tan (270 + q)
= tan [360 – (90 – )]
= –tan (90 – )
tan (360 – f) = –tanf q
tan 1 – =

35. Hence, for any acute angle , we have
sin (270 -  ) = -cos 
sin (270 +  ) = -cos 
cos (270 -  ) = -sin 
cos (270 +  ) = sin 
tan (270 -  ) =
tan (270 +  ) =

36. Let’s summarize using the ‘CAST’ diagram.

37. For any acute angle , we have
Sine is positive.
All are positive. O y x S A
sin (90 +  ) = + cos q
sin (90 –  ) = + cos q
cos (90 +  ) = – sin q
cos (90 –  ) = + sin q
90 + 
270 – 
270 + 
tan (90 +  ) = –
90 – 
tan (90 –  ) = + T C
sin (270 –  ) = – cos q
sin (270 +  ) = – cos q
cos (270 –  ) = – sin q
cos (270 +  ) = + sin q
tan (270 –  ) = +
tan (270 +  ) = –
Tangent is positive.
Cosine is positive.
These identities are also true even if  is not an acute angle.

38. Follow-up question
Simplify
cos q
sin q
 tan  =