1. Radian
Measure

2. Radians
Radian measure is an alternative to degrees and is based upon the ratio of
arc length (al) radius al θ r
θ- theta

3. If the arc length = the radius θ (radians) = r/r = 1 radian
al = r θ
θ (radians) = r/r = 1 radian r
1 radian = 57∙3o

4. If we now take a semi-circle Here al = ½ of circumference
= ½ of πd
= ½ of 2πr θ
= πr ● r
So θ (radians) = πr /r = π
π radians = 180o
Learn this!!

5. π (radians) = 180o π /2 = 90o 3π /2 = 270o π /3 = 60o 2π /3 = 120o
The connection between radians and degrees is :
π (radians) = 180o
This now gives us
2π = 360o
π /2 = 90o
3π /2 = 270o
π /3 = 60o
2π /3 = 120o
π /4 = 45o
3π /4 = 135o
π /6 = 30o
5π /6 = 150o
NB: Radians are usually expressed as fractional multiples of π.

6. Converting
÷180
then × π
degrees
radians
÷ π
then × 180

7. Converting
÷ 36
÷ 30
72o =
330o =
÷ 36
÷ 30
÷ 18
÷ 18

8. In the days before CDs the most popular format for music was “vinyls”.
Angular Velocity
In the days before CDs the most popular format for music was “vinyls”.
Singles played at 45rpm while albums played at 331/3 rpm.
rpm = revolutions per minute !
Going back about 70 years an earlier version of vinyls played at 78rpm.
Convert these record speeds into “radians per second

9. So 45rpm = 45 × 2π or 90π radians per min
NB: 1 revolution = 360o
= 2π radians
1 min = 60 secs
So 45rpm = 45 × 2π or 90π radians per min
= π/60 or 3π/2 radians per sec
So 331/3rpm = 331/3 × 2π or 662/3 π radians per min
= /3 × π /60 or π /9 radians per sec
So 78rpm = 78 × 2π or 156π radians per min
= π/60 or π /5 radians per sec

10. This triangle will provide exact values for
Some special values of sin, cos and tan are useful left as fractions, We call these exact values
60º 2
60º
30º 2 3 1
This triangle will provide exact values for
sin, cos and tan 30º and 60º

11. ½ ½ 1 ∞ Exact Values x 0º 30º 45º 60º 90º 3 Sin xº Cos xº Tan xº 30º2
30º 3
Exact Values x 0º
30º
45º
60º
90º
Sin xº
Cos xº
Tan xº 1 ∞ 3

12. Consider the square with sides 1 unit
Exact Values
Consider the square with sides 1 unit
45º 2 1 1
45º 1 1
We are now in a position to calculate
exact values for sin, cos and tan of 45o

13. 1
45º 2
Exact Values
Tan xº
Cos xº
Sin xº
90º
60º
45º
30º 0º x 1 1

14. Ratio signs in 4 Quadrants
The diagram shows the quadrants in which the ratios are positive
90o
180o –
All
Sin
180o 0o
, 360o
Tan
Cos
The arrows indicate how each quadrant is entered
360o –
180o +
270o

15. Exact value table and quadrant rules.
tan150o
= tan( ) o
= – tan30o
= – 1/√3
(Q2 so neg)
cos300o
= cos( ) o
= cos60o
= 1/2
(Q4 so pos)
sin120o
= sin( ) o
= sin60o
= √ 3/2
(Q2 so pos)
tan300o
= tan( )o
= – tan60o
= – √ 3
(Q4 so neg)

16. = 3/4 - 1/4 Exact value table and quadrant rules.
Find the exact value of cos2 (5π/6) – sin2 (π/6)
cos (5π/6) =
cos150o
= cos( )o
= – cos30o
= – √3 /2
(Q2 so neg)
sin (π/6)
= sin 30o
= 1/2
cos2 (5π/6) – sin2 (π/6)
= (- √3 /2)2 – (1/2)2
= 3/4 - 1/4
= 1/2

17. Exact value table and quadrant rules.
sin(2π/3) = sin 120o
= sin(180 – 60)o
= sin 60o
= √3/2
cos(2 π /3) = cos 120o
= cos(180 – 60)o
= – cos 60o
= –1/2
tan(2 π /3) = tan 120o
= tan(180 – 60)o
= –tan 60o
= - √3

18. Trig Identities (1) sin2θ + cos2 θ = 1 (2) sin θ cos θ = tan θ
An identity is a statement which is true for all values.
(1) sin2θ + cos2 θ = 1
(2) sin θ cos θ
= tan θ
θ ≠ an odd multiple of π/2 or 90°
since this would mean dividing by 0

19. Why sin2θo + cos2 θo = 1 (1) sin2θo + cos2 θo =b a
a2 + b2 = c2
sinθo = a/c
cosθo = b/c
(1) sin2θo + cos2 θo =

20. Often you will have to rearrange the identities.
sin2θ + cos2 θ = 1
Is the same as:
sin2 θ = 1 – cos2 θ
Is the same as:
cos2 θ = 1 – sin2 θ

21. Since θ is between 0 < θ < π/2
Using Trig Identities
Example1
Given sin θ = 5/13 where 0 < θ < π/2
Find the exact values of cos θ and tan θ .
cos2 θ = 1 - sin2 θ
sinθ cos θ
tan θ =
= 1 – (5/13)2
= 5/13 ÷ 12/13
= 1 – 25/169
= 144/169
= 5/13 × 13/12
= 5/12
cos θ = √(144/169)
= 12/13 or -12/13
tan θ = 5/12
cos θ = 12/13
Since θ is between 0 < θ < π/2

22. Given that cos θ = -2/ √ 5 where π < θ < 3π /2
Find sin θ and tan θ.
sin2 θ = 1 - cos2 θ
sinθ cos θ
tan θ =
= 1 – (-2/ √ 5 )2
= / √ 5 ÷ -2/ √ 5
= 1 – 4/5
= / √ 5 × - √5 /2
= 1/5
sin θ = √(1/5)
Hence tan θ = 1/2
= 1/ √ 5 or - 1/ √ 5
and sinθ = - 1/√5
Since θ is between π < θ < 3π /2

23.   Solving Trig Equations Solve sin 3x = ½, for 0 < x < π
0 < x < π (180o)
0 < 3x < 3π (540o)
sin-1 ½ = 30o, π/6   S A
180 – T C
3x = 30o, 150o
, 390o, 510o
Original angle NOT x
Angles up to 540o, ie add 360
Now change to radians, multiples of original angle, 30o = π/6
3x = π/6, 5π/6, 13π/6, 17π/6
x = π/18, 5π/18, 13π/18, 17π/18

24. The following questions are on
Graphs & Functons
Non-calculator questions will be indicated

25. Determine the values of a, b and c
The diagram shows a sketch of part of the graph of a trigonometric function
Determine the values of a, b and c 2a 1
1 in 
2 in 2 
a is the amplitude:
a = 4
b is the number of waves in 2
b = 2
c is where the wave is centred vertically
c = 1

26. a) Find expressions for: i) f(g(x)) ii) g(f(x))
Functions f and g are defined on suitable domains by f(x)= sin x and g(x) = 2x.
a) Find expressions for:
i) f(g(x)) ii) g(f(x))
b) Solve 2 f(g(x)) = g(f(x)) for 0 ≤ x ≤ 360 a) b)

27. Functions are defined on a suitable set of real numbers.
a) Find expressions for
b) i) Show that
ii) Find a similar expression for
and hence solve the equation a) b)
Now use exact values
equation reduces to