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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