Graphing (Method for sin\cos,
cos example given)
Basic Sum\Difference formula usages.
(Using cos)
Graphing (Method for cot)
Basic Half Angle formula usages.
Graphing (Method for tan)
Equation (Basic)
Graphing (sec\csc, use previous cos
graph from example above)
Equation (Not so basic)
Write sin\cos equation given graph
Identities (1)
Tangent Graphing, blast from the past
Identities (2)
Composite functions f(f -1(x))
14
First timers, the buttons above take you to a topic. The home button brings you back.
If you find a mistake, IM me at kimtroymath or e-mail me at sakim@fjuhsd.net
There are MANY things on this test, it’s a big one. This powerpoint does NOT cover
everything. Wait for the review sheet before the test for info, and use this powerpoint
to help you cover some of the materials. MATERIAL FOR CH 6 TEST IN RED
A) Factor out the coefficient of x, and use
even-odd properties to simplify
1) Find Amplitude and period
2) Find Phase Shift, and vertical shift
y  3 cos(x  3 )  2
y  3 cos( ( x  3))  2
y  3 cos( ( x  3))  2
3) Find starting and ending x-coordinates
35
4
2
4) Divide into 4 equal parts
5) Label key points
6) Connect
Amplitude =
T=
2
2
 +
P.S. =
V.S. =
1
3
Remember, cos(x) = cos(-x)
3
3 4 7

2
2
You will always do this, this is part of
your ‘work’ on a test and is required
You want to study the sine and
Starting point is phase shift. Ending
cosine graphs. Remember:
point is Phase shift + Period
Sine  0, 1, 0, -1, 0
You will take the starting and ending
Cosine
 1,find
0, -1,
1
points and
the0,average,
then find
the average
againperforming
to break it up into
You
are basically
four equal regions
transformations
on those key points
5
45 9

2
2
y  2 cot(2 x   )  1
A) Factor out the coefficient of x, and use
even-odd properties to simplify
   
y  2 cot 2 x     1
2 
 
1) Find Vertical Stretch and period
2) Find Phase Shift, and vertical shift
3) Find starting and ending x-coordinates cot, asymp 1
0
-1 asymp
Find the average, then find the
4) Divide into 4 equal parts
averages again.
5) Label key points

6) Connect
Vertical =
Stretch
Period =
|


Phase Shift =
|2
+

END 2


2

3
2

3
 3
2
4


7
5

4
2 4 

8
2
2
8

1
START

4

2

Vertical Shift =
START
END


y  2 tan x    1
4
2
A) Factor out the coefficient of x, and use
even-odd properties to simplify
1) Find Vertical Stretch and period
1 
 
y  2 tan  x     1
2 
2
2) Find Phase Shift, and vertical shift
3) Shift zero (the middle)
-1, 1 between
4) Divide the period in half, add and subtract
from the middle, sketch asymptotes.
Go between the asymptotes and
the middle and put -1, 1, then
transform.
5) Perform transformations.
6) Connect
Vertical =
Stretch
T=
|

2

P.S. =
V.S. =
| 2
Remember,
1
it’s T
π\2
2
 1
2 2
1
2
1

2
1

2
1

2
3

2
A) First, sketch the cos graph, stating all
the information as your for cosine.
1) At the ‘zeros’ of cosine (or the middle),
sketch asymptotes.
y  3 sec(x  3 )  2
y  3 cos( x  3 )  2
2) At the maxes and mins (tops and
bottoms), make your U’s
1
You aren’t going to need three
cycles. Probably just one
cycle.
3
5
Write the equation of the sin and cos graph.

Remember, for sin and cos, the amplitude,
period, and vertical shift are all the same,
only the phase shift is different.
4
Cosine
Vertical
Amplitude.
The period
starts
Shift,
You
ofhow
at
sin
can
the
and
much
use
topcos
(1
common
did
0
are
the
-1the
0middle
1). middle
move
So
sense,
same,
findfrom
so
how
a max,
it’s
the
far
easiest
x-axis?
the
is the
x-coordinate
(IMHO)
max
Or from
youtocan
the
is
find
ause
the
possible
middle,
the formula.
period
orphase
you
using
shift.
cosine.
use the
Toformula.
do
that,
Sin
starts
in
thecan
middle,
then
goes
up
Concept,
Amplitude
is the
thein
find
far apart
maxes
(0
1 out
0 are
-1how
0).
Find
a point
the are.
There
many
possible
solutions.
Concept,
V.S.
is average
of
distance
between
max
and
5could
 use
middle
where
the
graph
goes
upFor
this
problem,
you
max
and
min.
So
you
add
Period

T


min,
sopi\4,
you5pi\4,
subtract
afterwards.
is one possible
3pi\4,
and
divide
byThat
2. etc.
(distance)
then divide4by 2.4
phase shift.

max
min
max
min
VAmp
.S .0,and
bApi
-pi,
viable2
options.
 are all
22 
2

T
0 0( 4()4)
Amp
b 
 2 2
2 2
1
Max
middle
Then up

middle
2
Then up
5
4
Max
Max
Then up
Min
y  2A sin(
0 ))  b2
2 (x  

y  2A cos(
 b2
2 ( x  )) 
4
Vertical Shift
Amplitude
Period
Clear
Clear
Clear
cos Phase Shift sin Phase Shift
Clear
Clear
You could try to graph tangent using old
style transformations if you wanted to.
Factor out the coefficient of x.
Key Points
  

 
Horizontal
x
,1 0 , 0  , 1

2
2
 4

4 
Stretch:

Reflect:
Shift:
Vertical
1
2
Stretch:
2
Reflect:
None
Shift:
y  2 tan  x     1
2 
2
2
 1 
1 


,

1
0
,
0


 , 1 x  1
I recommend the
2
2




other method for
3
1
 1 
faster graphing.
 1,1  , 0  0 , 1 x 
x
2
2
 2 
This is different
  1than the other tangent

 1,2  , 0  0 , 2
graph because
 2there
 is an addition sign
inside, not a subtraction
sign.
 1 
 1,1  ,1 0 , 3
 2 
x  1
None
Left
x



y  2 tan x    1
4
2
 
1 
Up 1
1

2
Asymptote changes are only affected by
horizontal transformations.
1
2

2
Composite trig functions.
1) Set up triangle in the correct quadrant.
1) Pythagorean Theorem may be
necessary. (r always positive)
2) Find solution using correct sides.
 1   7   2
sin  sec 
  
 3  7


 3  4
cos tan1    
 4  5

 1   13   12
cos csc 
  
 5   13

Remember, r is positive.
Note, secant negative means it’s in
quadrant II.
Quadrants I , IV


x
2
2


tan :
x
2
2


csc :
x
2
2
sin :
Quadrants I , II
cos : 0  x  
cot : 0  x  
sec : 0  x  
x0
x
xy
cos
tan 
rx
2
7
 3

2
5
3
412
13
ry
sec


rx
sin 

csc 
rx cos
ry
5
sin( A  B)  sin A cos B  cos A sin B
sin( A  B)  sin A cos B  cos A sin B
Basic sum\difference formula usages.
Exam
ple
Break
it upBinto
common
sin(
A

B
)

sin
A cos
 cos
A sin B
 5 
radian
sin   5 
1 values. A chart may Note, onlyMatch
2up
 examples
2 for8
showing
the expression
   ) be
sin(csc
sinhelpful.
110  cos(Note,
50  cos
 12
use110  sin 50


5

sin.
These
problems
may
show
12


with
the
correct
formula.
  sin
6
12
3
12
same
 denominators)

up
in
cos
and
tan
format.
If
you
60)
     sin(
 12
 cot, 3sec,
 or csc,
3 use 9
sin   
need to do
These
 problems are

13 appropriate formula
 6 4   Apply
the formula
of
their
reciprocal
4designed
12 to give4 you a12

counterparts,
thenunit
take
the value.
2 2 6
familiar
circle

4

5You 10
reciprocal
of
the
solution.




 4  
   
12 though,6 you 12
 to be3 careful
sin    sin  cos   cos  sin need
the

   4  2 6    will have to rationalize
6
12





denominator
at
times.
The


 12  6  22 6  3   2 
2 will
12be very helpful.
12
CONJUGATE


  

4(2 2   62) 
 2   2 


 
4
2 6
  2  6
4
Application
Reverse
Note
Clear
Clear
Clear
Many combinations are
possible. These add up to
In this example,
find12.
csc, I use
equal 5pito
over
sin to find a solution, and since
csc is the reciprocal of sin, I take
the reciprocal of my answer. The
work is shown on how to break it
down. Watch for the conjugate.
Basic Half-Angle formula Usage
sin A 
5
; A is in QIII
13
A
Find cos   
2
  A
1  cos A
2
1
  13
2

1
26

26
26
sin15
30 
 30
1 cosA
sin
 
 2 
2

1
2
3
2 
What quadrant is A\2
in?
3
1  cos A
 A
sin    
2
2
2
1  cos A
 A
2 2 2
cos   
2
2

A
3

Set up triangle,
  use QII
1  cos A
original
 A
2 A when
2
4
t
an


 
plugging into formula.
1  cos A
2
Your solution will be plus
OR minus, not plus and
1  cos A

minus. Use the quadrant
sin A
of A\2 to12determine SIGN!
sin A
A
5
Cosine
is negative in  12

A
13 II, so cos
1  cos A
quadrant
we will use
13
the negative sign.
Common
denominator
maywill
alsoalso
be involve
done with
TheseThese
problems
radians.
A
30





o – 30o),
o, it’s formulas
sum\difference
(45I,
Lookin
at a15
quadrant
so sin
They work

similar
 infashion.

butPOSITIVE
see
2 it may I.give you a
is
in
 2 if doubling
 quadrant
common unit circle measure.
So A  30
2 3
2 3
2 3
2


2
4
2
Give the
formula
all the solutions.
Find
all general
solutions
for forbetween
[0,2 )
 1
 2
x  
cos  
2 2
 3
sin, cos,csc,sec  2
tan,cot  
1
Where does cos(x)  ?
2 Be careful, sometimes you can
combine general formulas.
5

cos repeats every 2
 2k; k  
 2k

3
3

2
It'
s
not
really
x,
but




2
7
2

2
3

 2k
    2k
3
6
3
6

2
You can think of it like x     
7


19
5


11
13


2

3



3

k
    3k
44
44
k

k  -1
k0
7

4
k 1
Plug in values for k, remember, k
is an integer.
Cross out the ones not between
[0,2 )
Equations (Basic)
Give the general formula for all the solutions.This does NOT cover all possible types of
equations. Some common things to
sin A  cos 2 A  1  0
watch out for regarding equations:
sin A  (1  sin 2 A)  1  0
sin A  1  sin 2 A  1  0
sin A  sin 2 A  0
sin A(1  sin A)  0
sin A  0 1  sin A  0
sin A  1
3
A  2k
A
 2k
2
A    2k k  
3
A  k
A
 2k
2
k 
1) Move everything to one side.
2) Use properties when possible (Sum to
Product, pythagorean, double angle)
3) Many times, changing things into the
same trig function may be helpful.
4) Factoring may occur many times.
5) Remember to plus\minus when square
rooting both sides.
6) You can combine general formulas
sometimes.
Identities: Pg 513: 45
RULE: WORK ON ONE SIDE ONLY!
?
2 cot A cot 2 A  cot A  1
2
 1 ?
2
2 cot A
  cot A  1
 tan2 A 
?
1
2 cot A
 cot2 A  1
 2 tan A 


2
 1  tan A 
 1  tan2 A  ?
  cot2 A  1
2 cot A
 2 tan A 
?
cot A(1  tan A)  cot2 A  1
2
2
?
cot A  cot A tan A  cot 2 A  1
2
2
2
cot2 A  1  cot2 A  1
Remember the question mark.
Helpful Items, in no particular order.
1)
Changing to sin\cos helps.
2)
Look for Pythagorean, double angle,
product-to-sum, reciprocal, even\odd
identities.
3)
Look at the other side.
4)
Conjugates and multilying by one
helps.
5)
Combining or splitting up fractions is
also helpful.
6)
Factoring may be helpful.
7)
Work on more complicated side.
I noticed there was ‘cot’ on the
other side, that’s why I didn’t
change to sin\cos in this case.
Identities: Pg 512: 30
RULE: WORK ON ONE SIDE ONLY!
cos2 A ?
1
 sin A
1  sin A
2
1  sin A1 cos
sin A ?
 sin A
1
1  sin
1 A
sin A
 sin
sin2 A) ?
1 (1sin
A A()(11sin
1
 sin A
sinAA
11sin
2
??
sin1 
A1sin
A
sin A sin
sin AA
1  sin A
sinAA) ? sin A
sin A(1  sin
 sin A
1  sin A
Remember the question mark.
Helpful Items, in no particular order.
1)
Changing to sin\cos helps.
2)
Look for Pythagorean, double angle,
product-to-sum, reciprocal, even\odd
identities.
3)
Look at the other side.
4)
Conjugates and multilying by one
helps.
5)
Combining or splitting up fractions is
also helpful.
6)
Factoring may be helpful.
7)
Work on more complicated side.
sin A  sin A
Here is another method
14