2. HINT:
To
find
where
S A
Cos,
Sin,
Tan
is
posiVve.
You
can
use
This
Acronym:
ALL,
Students
Take
Calculus
T
C
Reference
Angles
A. What
is
a
reference
Angle:
-­‐A
reference
angle
is
the
angle
that
the
main
angle
makes
(
The
angle
is
measured
from
the
ini:al
side
to
the
terminal
side)
B.
How
to
find
the
reference
Angle
Step1:
Determine
The
Quadrants
in
which
the
angle
belongs
in
Step:2
Draw
the
angle
on
the
graph,
and
find
the
closet
x-­‐
axis.
-­‐If
you
get
a
reference
angle
that
is
less
then
90
degrees,
then
it
is
its
own
angle.
(
These
angles
would
fall
in
to
Quadrant
I)
-­‐In
quadrant
II
you
take
the
degrees
and
subtract
from
180
(180-­‐105=75)
-­‐Quadrant
III
subtract
180
from
the
degrees
-­‐Lastly
if
it
falls
onto
Quadrant
Iv
the
degrees
are
subtracted
by
360

3. TRIGONOMETRIC
EQUATIONS
Algebraic
Equa:on
What
is
a
trigonometric
equaVon?
-­‐ Trigonometric
EquaVon
is
an
equaVon
involving
the
Trigonometric
FuncVon
of
unknown
angles.
How
Do
You
Solve
A
Trig

If
you
can
EquaVon?
not
factor
use
the
-­‐To
solve
a
Trig
EquaVon,
We
use
the
same
Trig
Equa:on
QuadraVc
procedure
that
we
used
to
solve
algebraic
equaVons.
Formula
Step
1:
You
can
factor,
or
use
the
QuadraVc
Formula
to
solve
the
trig
equaVon.
Step2:
Aer
solving
trig
equaVon
Determine
which
Quadrant
they
are
in.

4. 
On
the
graph
STEP
2:
determine
where
Cos
is
posiVve
+
-­‐
Quadrants
I
and
II
Quadrants
III
and
SIN
IV
o
Cos
is
posiVve
so
it
Quadrants
I
and
Iv
Quadrants
II
and
III
is
in
TAN
Quadrant
I
Quadrants
I
and
Iv
Quadrants
II
and
III
and
IV
COS
2+√2=
3
o There
is
no
II
I
2
2
such
value
2-­‐√2=
.5857
=
sin-­‐1(.5857
)=
72
degrees
2
2
The
value
in
the
first
quadrant
is
72
degrees
and
is
also
III
the
reference
angle.
Therefore
in
the
fourth
Quadrant
IV
360-­‐73=287
Reference
angle
are
[73,287]

5. 3
different
methods
to
solve
this
trig
equaVon
Method
1:
Factor
Method
2:
square
root
Method
3:
QuadraVc
Formula

6. Periodic
FuncVon
IdenVfying
the
EquaVon
y=asine(x-­‐c)+d
A-­‐
is
the
amplitude
C-­‐
stands
for
phase
D-­‐
is
the
verVcal
shi
Example:
4sin(x-­‐2)+3

7. If
no
restricVons
are
set
(y
=
cos(x)
or
y
=
sin(x)
the
domain
is
(-­‐∞,
∞)
and
the
range
is
[1,
1].
In
360°
there
are
5
important
points
then
the
funcVon
repeats
and
conVnues
to
repeat
aer
1
wave
length
(λ),
this
is
called
a
period
(P)
and
is
equal
to
2π.
The
5
important
points
include
3
intersects
and
a
maximum
and
a
minimum
value
for
sinθ,
for
cosθ
there
are
2
maximums
2
intersects
and
1
minimum
over
the
course
of
1
period.
y
=
(A)sin(x
+
W)
+
k
periodic
funcVon
equaVon
where
A
is
the
amplitude,
W
is
the
phase
shi
of
the
funcVon
(
+
shis
the
graph
le
and
–
shis
the
graph
right),
K
shis
the
graph
verVcally
(
+
shis
the
graph
up
and
–
shis
the
graph
down).

8. The
funcVon
y=sinx+1
moves
up
one
unit
when
1
is
added
to
the
end.
-­‐so
when
we
change
the
value
of
d
we
slide
the
funcVon
up
or
down
-­‐when
the
amplitude
is
changed
the
graph
will
stretch
or
shrink
verVcally
This
graph
shows
the
funcVon
moving
to
the
le
2
units,
when
a
2
is
added
for
the
value
of
c

9. Ambiguous
Triangle
Case
Formula
Review:
• Pythagorean
Theorem=
a^2+b^2=c^2
(
a
and
b
are
the
legs
and
c
is
the
hypotenuse)
–
used
with
right
triangles
• SOH
CAH
TOA
SOH=
opposite/hypotenuse
CAH=
Adjacent/hypotenuse
TOA=opposite/adjacent
(this
is
also
used
with
right
triangles)
• Sine/Cosine
Law
Sine-­‐
we
use
this
method
to
find
the
angle
or
side
of
a
given
triangle.
Formula:
SinA/a=SinB/b=SinC/c
(used
for
oblique
triangles)
Cos-­‐
we
use
cosine
when
(sss)
is
given
or
(sas)
Formula:
a^2=b^2=c^2-­‐2ab
CosA

10. Ambiguous
triangles
come
from
a
limited
amount
of
informaVon
(having
only
2
sides
and
an
angle
opposite
angle).
There
are
4
possibiliVes
if
the
angle
is
acute
and
there
are
2
possibiliVes
if
the
angle
is
obtuse.
When
angle
A
is
acute
a
<
bsin(A):
There
is
no
triangle.
a
=
bsin(A):
Only
1
triangle
is
possible.
bsin(A)
<
a
<
b:
2
triangles
are
possible.
a
>
b:
1
triangle
is
possible.

11. Ambiguous
Triangles
(Cont.)
When
you
are
dealing
with
obtuse
angles
there
are
only
2
possible
outcomes
since
side
a
≠
b
and
a
cannot
be
less
than
b
or
there
will
be
no
triangle
possible.
If
a
<
b:
No
triangle
is
possible.
If
a
>
b:
1
triangle
is
possible.

12. How
to
use
the
Law
Sine
First
you
have
to
have
a
triangle
with
the
required
informaVon.
Then
you
have
to
set
up
the
equaVon,
for
this
parVcular
triangle
the
equaVon
will
look
like:
sin
36°
sin
76°
a
67
Then
you
cross
mulVply:
a(sin
76°)
=
67(sin
36°)
Isolate
a
then
solve:
A=67(sin
36°)
sin
76°
=
40.587

13. When
to
use
the
Law
of
Sines
The
law
of
Sines
is
used
when
at
least
2
Sides
are
given
with
one
having
their
respecVve
angle,
or
2
angles
are
given
with
one
having
their
respecVve
side.
Ex:
a
47
c
25
47°
64
43
54°

14. How
to
use
the
Law
of
Cosines
First
you
must
have
a
triangle
with
the
a
required
informaVon.
74
Then
you
have
to
set
up
the
equaVon,
for
this
parVcular
triangle
the
equaVon
will
look
like
57
41
this:
a²
=
74²
+
57²
-­‐
2
x
74
x
57
x
cos
41
You
then
isolate
the
unknown
then
solve:
a
=
√(74²
+
57²
-­‐
2
x
74
x
57
x
cos
41)
=
2358.270

15. When
to
use
the
Law
of
Cosines
The
Law
of
Cosines
can
only
be
used
when
you
have
3
sides,
or
2
sides
and
the
respecVve
angle
to
the
unknown
side
in
a
triangle.
Ex:
46
43
a
a
67
26
56
34