RadiansRadians

Radian measure is an alternative to degrees Radian measure is an alternative to degrees and is based upon the ratio of and is based upon the ratio of

arc length (al) arc length (al) radius radius

rrθθ

alal

θθ- theta- theta

r

aradians

If the arc length = the radius

θr

al = r

θ (radians) = r/r = 1 radian

1 radian = 57∙3o

If we now take a semi-circle

θr

al Here al = ½ of circumference

= ½ of πd

= πr

So θ (radians) = πr /r = π

π radians = 180o

●

= ½ of 2πr

Learn this!!

The connection between radians and degrees is :The connection between radians and degrees is :

ππ (radians) = 180 (radians) = 180oo

This now gives usThis now gives us 22ππ = 360 = 360oo

ππ //22 = 90 = 90oo

π π //33 = 60 = 60oo

33ππ //22 = 270 = 270oo

22ππ //33 = 120 = 120oo

ππ //44 = 45 = 45oo 33ππ //44 = 135 = 135oo

ππ //66 = 30 = 30oo 55ππ //66 = 150 = 150oo

NBNB: Radians are usually expressed as fractional : Radians are usually expressed as fractional multiples of multiples of ππ..

ConvertingConverting

degreesdegrees radiansradians

÷180÷180

then then ×× ππ

÷ ÷ ππ

then then ×× 180 180

72o = 330o =

Converting

18072

÷ 36

52

180330

÷ 36

÷ 30

÷ 30 611

rad9

2

92

1809

360 o40

rad18

23

1823

1801

230 o230

÷ 18

÷ 18

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

NB: 1 revolution = 360o = 2π radians 1 min = 60 secs

So 45rpm = 45 × 2π or 90π radians per min

= 90π/60 or 3π/2 radians per sec

So 331/3rpm = 331/3 × 2π or 662/3 π radians per min

= 662/3 × π /60 or 10 π /9 radians per sec

So 78rpm = 78 × 2π or 156π radians per min

= 156π/60 or 13 π /5 radians per sec

2222

22

60º60º

60º60º60º60º11

60º60º

2230º30º

33

This triangle will provide exact values for This triangle will provide exact values for

sin, cos and tan 30º and 60ºsin, cos and tan 30º and 60º

Exact ValuesExact Values

Some special values of sin, cos and tan are useful left Some special values of sin, cos and tan are useful left as fractions, We call these as fractions, We call these exact valuesexact values

xx 0º0º 30º30º 45º45º 60º60º 90º90º

Sin xºSin xº

Cos xºCos xº

Tan xºTan xº

½½

½½

33

23

23

31

1160º60º

2230º30º33

Exact ValuesExact Values

00

11

00

11

00

∞∞

Exact ValuesExact Values

11 1145º45º

45º45º

22

Consider the square with sides 1 unitConsider the square with sides 1 unit

1111

We are now in a position to calculate We are now in a position to calculate exact values for sin, cos and tan of 45exact values for sin, cos and tan of 45oo

Exact ValuesExact Values11

45º45º

45º45º

22

11

Tan xºTan xº

Cos xºCos xº

Sin xºSin xº

90º90º60º60º45º45º30º30º0º0ºxx

00

00

11

11

00

21

21

23

23

33

1

21

21

11

Ratio signs in 4 Quadrants

AllSin

Tan Cos

0o

90o

180o

270o

, 360o

180o –

180o + 360o –

The diagram shows the quadrants in

which the ratios are positive

The arrows The arrows indicate how each indicate how each

quadrant is quadrant is enteredentered

Exact value table and quadrant rules.

tan150o

(Q2 so neg)

= tan(180 - 30) o = – tan30o = – 1/√3

cos300o

(Q4 so pos)

= cos(360 - 60) o = cos60o = 1/2

sin120o

(Q2 so pos)= sin(180 - 60) o = sin60o = √ 3/2

tan300o

(Q4 so neg)

= tan(360 - 60)o = – tan60o = – √ 3

Find the exact value of cos2 (5π/6) – sin2 (π/6)

cos (5π/6) = cos150o

(Q2 so neg)

= cos(180 - 30)o = – cos30o

= – √3 /2

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

Exact value table and quadrant rules.

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

tan(2 π /3) = tan 120o

= cos(180 – 60)o

= tan(180 – 60)o

= – cos 60o

= –tan 60o

= –1/2

= - √3

3

2tan

32cos

32sin

that Prove

2

12

3

32cos

32sin

LHS

3 RHS

An identity is a statement which is true for all values.

(1) sin2θ + cos2 θ = 1

(2) sin θ cos θ

θ ≠ an odd multiple of π/2 or 90°

since this would mean dividing by 0

Trig Identities

= tan θ

θo

c

b

a

a2 + b2 = c2

sinθo = a/c

cosθo = b/c

(1) sin2θo + cos2 θo =

2 2

2 2

a bc c

2 2

2

a bc

2

2 1cc

Why sin2θo + cos2 θo = 1

sin2θ + cos2 θ = 1

sinsin22 θ θ = 1 – cos = 1 – cos2 2 θθ

coscos22 θ θ = 1 – sin = 1 – sin2 2 θθ

Often you will have to rearrange the identities.

Is the same as:

Is the same as:

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

= 1 – (5/13)2

= 1 – 25/169

= 144/169

cos θ = √(144/169)

= 12/13 or -12/13

Since θ is between 0 < θ < π/2 cos θ = 12/13

tan θ =

= 5/13 ÷ 12/13

= 5/13 × 13/12

tan θ = 5/12

Using Trig Identities

sinθ cos θ

= 5/12

Given that cos θ = -2/ √ 5 where π < θ < 3π /2

Find sin θ and tan θ.

sin2 θ = 1 - cos2 θ

= 1 – (-2/ √ 5 )2

= 1 – 4/5

= 1/5

sin θ = √(1/5)

= 1/ √ 5 or - 1/ √ 5

Since θ is between π < θ < 3π /2

and sinθ = - 1/√5

= - 1/ √ 5 ÷ -2/ √ 5

= - 1/ √ 5 × - √5 /2

Hence tan θ = 1/2

tan θ = sinθ cos θ

Solving Trig EquationsSolving Trig Equations

Solve Solve sinsin 3 3xx = ½ = ½, , for 0 < for 0 < xx < < ππ 0 < 0 < xx < < π π (180(180oo))

0 < 30 < 3xx < 3 < 3ππ (540(540oo))sinsin-1-1 ½½ = 30 = 30oo, , ππ//66 AASS

TT CC

180 180

––

33xx = 30 = 30oo, 150, 150oo Original angle Original angle NOT NOT xx

Angles up to 540Angles up to 540o, o, ie ie add 360add 360

, 390, 390oo, 510, 510oo

Now change to Now change to radians, multiples of radians, multiples of original angle, 30original angle, 30oo = =

ππ//66

33xx = = ππ//66,, 55ππ//66,,

1313ππ//66,, 1717ππ//66

xx = = ππ//1818,, 55ππ//1818,,

1313ππ//1818,, 1717ππ//1818

Graphs & Functons

The following questions are on

Non-calculator questions will be indicated

The diagram shows a sketch of part of the graph of a trigonometric

function

Determine the values of a, b and c

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

2a

1 in

2 in 2

1

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

( ( )) (2 )f g x f xa) sin 2x ( ( )) (sin )g f x g x 2sin x

b) 2sin 2 2sinx x sin 2 sin 0x x

2sin cos sin 0x x x sin (2cos 1) 0x x

1or

2sin 0 cosx x 0 , 180 , 360x

60 , 300x

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

4and( ) sin , ( ) cos ( )f x x g x x h x x

( ( ))f h x ( ( ))g h x1 1

2 2( ( )) sin cos f h x x x

( ( ))g h x

for( ( )) ( ( )) 1 0 2f h x g h x x

4( ( )) ( )f h x f x a) 4

sin( )x 4

( ( )) cos( )g h x x

b) Now use exact values

equation reduces to2sin 12

x 2 1sin

2 2x

3,4 4

x

xxx cos4

sin4

cossin)4

sin(

xx cos2

1sin

2

1

4sinsin

4coscos)

4cos(

xxx sin

2

1cos

2

1 x