Trig Toolkit
Name ________________________
Trigonometry Toolkit
90° = π/2
30-60-90 Triangles:
120° = 2π/3
45-45-90 Triangles
60° = π/3
135° = 3π/4
2
60°
30°
90°
45° = π/4
150° = 5π/6
30° = π/6
2 45°
1
45° 90°
2
3
180° = π
or any multiples of.
0 or 360° =2π
Conversions:
210° = 7π/6
330° = 11π/6
225° = 5π/4
240° = 4π/3
315° = 7π/4
300° = 5π/3
or any multiples of.
Acute angle α is the
reference angle for
a (180 – α)˚ angle QII,
a (180 + α)˚ angle QIII,
a (360 – α)˚ angle QIV
 π 
degrees 
 = radians
 180 
 180 
radians 
 = degrees
 π 
as well as all of the coterminal
angles, (α ±360n)˚
270° = 3π/2
y
P(x,y)
(
P cos
,sin
P(cos
0,θsin
0)θ
Note: In the table below, r can be
r
y
2
replaced with
x
r= x +y
3 basic
Sine (opp/hyp)
y
sin θ = , r ≠ 0
r
Cosine (adj/hyp)
x
cos θ = , r ≠ 0
r
Tangent (opp/adj)
y
tan θ = ,x ≠ 0
x
3 other
Cosecant
r
csc θ = , y ≠ 0
y
Secant
r
sec θ = , x ≠ 0
x
Cotangent
x
cot θ = , y ≠ 0
y
2
x 2 + y 2 , since
1
)
y
2
reciprocals
1
r
cscθ =
=
sin θ y
Rad.
0
π
6
sec θ =
1
r
=
cos θ x
1
x
cot θ =
=
tan θ y
π
4
π
3
π 2
π
3π 2
2π
x
x
cos θ
1
3
2
2
2
1
2
0
-1
0
1
sin θ
0
1
2
tan θ
0
3
3
1
2
2
3
2
3
Undef.
1
0
-1
0
0
Undef.
0
Now just use the reference angles!
( 0,1)
Positive Values?
Sin All
π/2
1 3
 ,

2 2 
 2 2
,


 2 2 
π/3
π/4
Tan Cos
 3 1
, 

 2 2
π/6
0
S. Stirling
(1, 0 )
Page 1 of 8
Trig Toolkit
Relationships Among the Functions
Reciprocal
Relationships:
1
sin θ
1
sec θ =
cos θ
1
cot θ =
tan θ
cscθ =
Relationships with Negatives:
Pythagorean Relationships:
sin ( −θ ) = − sin θ and cos ( −θ ) = cos θ
sin 2 θ + cos2 θ = 1
1 + tan 2 θ = sec2 θ
1 + cot 2 θ = csc2 θ
csc ( −θ ) = − cscθ and sec ( −θ ) = secθ
tan ( −θ ) = − tan θ and cot ( −θ ) = − cot θ
Quotient Identities:
sin θ
tan θ =
, cos θ ≠ 0
cos θ
cos θ
cot θ =
, sin θ ≠ 0
sin θ
Periodicity Relationships:
sin(θ + 2π ) = sin θ (same for csc.)
cos(θ + 2π ) = cos θ (same for sec.)
(same for cot.)
tan(θ + π ) = tan θ
S. Stirling
Name ________________________
Double-Angle Formulas
(These will be given to you by
AP; however, your book won’t
do this.)
sin( 2α) = 2sin α cos α
cos( 2α) = cos2 α – sin2 α
cos( 2α) = 1 – 2 sin2 α
cos( 2α) = 2 cos2 α – 1
2 tan α
tan(2α ) =
1 − tan 2 α
Definition Periodic Function:
f(x) is periodic if there is a positive
number p such that f(x + p) = f(x)
for every value of x. The smallest
such value of p is the fundamental
period of f.
Page 2 of 8
Trig Toolkit
Name ________________________
Graphs of the Trigonometric Functions
For all functions, n = an integer.
y
Sine
y = sin θ .
y
Cosine
y = cos θ .
1
1
Period: 2π
x
−3π/2 −π −π/2
π/2
π
3π/2
2π
Domain: ℜ
Period: 2π
x
−3π/2 −π −π/2
cscθ =
Range:
{ f ( x) | −1 ≤ f ( x) ≤ 1}
y
1
.
sin θ
sec θ =
x
Period: 2π
−3π/2 −π −π/2
Tangent
tan θ =
π/2
π
3π/2
−2
1
x
Domain:
π


 x | x ≠ + nπ 
2


Range: ℜ
−3π/2 −π −π/2
1
.
cos θ
1
x
Period: 2π
−3π/2 −π −π/2
Domain:
π


 x | x ≠ + nπ 
2


Range:
{ f ( x) | f ( x) ≤ −1 or f ( x) ≥ 1}
y
Period: π
y
2π
−1
sin θ
.
cos θ
2π
−2
Secant
1
Domain:
{ x | x ≠ 0 + nπ }
Range:
{ f ( x) | f ( x) ≤ −1 or f ( x) ≥ 1}
3π/2
−1
−2
Cosecant
π
Domain: ℜ
−1
Range:
{ f ( x) | −1 ≤ f ( x) ≤ 1}
π/2
π/2
π
3π/2
2π
−1
Cotangent
1
cot θ =
tan θ
cos θ
cot θ =
sin θ
Period: π
π/2
π
3π/2
2π
−1
−2
y
1
x
−3π/2 −π −π/2
π/2
π
3π/2
2π
−1
Domain:
−2
{ x | x ≠ 0 + nπ }
−2
Range: ℜ
Even and Odd Functions:
cos(x) and sec(x) are even functions.
sin(x), csc(x), tan(x) and cot(x) are odd functions.
S. Stirling
Page 3 of 8
Trig Toolkit
Sine and Cosine Functions and Modeling
Domain: ℜ Range: ℜ , where -a ≤ f(x) ≤ a Period: 2π
Name ________________________
How to write models: y = a cos [b(x – c)] + d
Determine the amplitude, a, vertical stretch or compression:
a = (maximum – minimum) ÷ 2
Distance from midline to the maximum or minimum.
Determine the horizontal stretch or compression, b:
b = 2π / period.
Period = the horizontal distance between successive maximums (or
minimums).
Determine the horizontal shift, c:
Use the x-value of a maximum value (or minimum value) and figure
the number of x-units needed to map the parent function onto the
transformed function.
Determine the vertical shift, d, location of the midline:
d = average of the maximum and minimum value, midpoint.
The midline is also called “the axis of the wave”.
Graph transformations
Example:
y = 3cos(4 x − 2π ) − 7
Must get it in the correct form!
y = a cos [b(x – c)] + d
 
π 
y = 3cos  4  x −   − 7
2 
 
Stretch & Compress
Translations
y = cos( x)
y = 3cos( x)
vertical stretch (factor of 3) or amplitude = 3
y = 3cos(4 x) − 7
translation = shift down 7 or midline at y = −7
y = 3cos(4 x)
horizontal compress (factor of 4) or
2π π
period =
=
4
2
Stretch
factor 3
 
π 
y = 3cos  4  x −   − 7
2 
 
translation = shift right
Compress
factor 4
y
4
down 7
−π/2
−1
2
3
2
1
−π
π
y
2
−3π/2
2
or x =
4
3
−2π
π
1
x
π/2
π
3π/2
2π
5π/2
−2π
−3π/2
−π
−π/2
−1
−2
−2
−3
−3
−4
−4
−5
−5
−6
−6
−7
−7
−8
−8
−9
−9
−10
−10
−11
−12
right
π/2
x
π/2
π
x=
3π/2
2π
5π/2
π
2
y = −7
−11
−12
This process works for any transformations on any function.
S. Stirling
Page 4 of 8
Trig Toolkit
Name ________________________
Graphs of the Inverse Trigonometric Functions
With inverses, the roles of the x and y are interchanged. You must limit the domains of the original (to
get a function) and then reflect over the line y = x in order to get the inverse function. Note: the ranges of
the inverse functions are the limited domain of the original!
Inverse
Trigonometric
Functions
graph splits the x-axis
(works around y = 0)
graph above the x-axis
(works around y = π/2)
sin −1 ( x ) or arcsin ( x )
cos −1 ( x ) or arccos ( x )
π
π/2
x
−1
1
−1 ≤ x ≤ 1
π/2
2
−π/2
x
−1
( x)
csc−1 ( x ) = sin −1 1
y≠0
Domain x-values
IN = value, v
(are ranges of
trig. functions)
1
2
( x)
sec−1 ( x ) = cos −1 1
y ≠π
2
π/2
x ≤ −1 or x ≥ 1
or
x ≥1
π
x
−1
1
π/2
2
x
−π/2
−1
tan −1 ( x )
y ≠ ±π
1
2
cot −1 ( x ) = π − tan −1 ( x )
2
y ≠ 0 or y ≠ π
2
−∞ ≤ x ≤ ∞
π
π/2
x
−1
1
−π/2
Range y-values
OUT = angle = θ
(Limited domains
of trig. functions)
−
π
π/2
2
x
−1
≤ y≤
π
2
2
θ in Quadrant I or IV
1
2
0≤ y ≤π
θ in Quadrant I or II
The graph reflects over the y-axis.
Helpful website: http://www.math.uakron.edu
S. Stirling
Lesson 10 is good for Trig Functions!
Page 5 of 8
Trig Toolkit
Name ________________________
Moving From Sine to Arcsine
Concept of inverse: v = sin (θ ) so sin −1 ( v ) = θ or arcsin ( v ) = θ , think “the angle whose sine is value”
Inverse Sine: Graphically
To make the reflection over y = x a function, an inverse, you must limit the domain of sine. (Which
limits the range of arcsine.)
y = sin ( x ) D:
( −∞, ∞ )
Negative θ
R: [ −1,1]
Positive θ
y
QI
Q II
−
1
x
−π
−π/2
π/2
π
3π/2
π
2
≤θ ≤
π
2
Interchange
x and y &
solve for y.
y = sin ( x )
sin
−1
 π π
y = sin −1 ( x ) D: [ −1,1] R:  − , 
 2 2
Limit the
domain, the
angle values
−1
( x) = y
y
.
π
Pos.
θ
π/2
Q xI
−1
Neg.
θ
1
Q IV
−π/2
−π
Q III Q IV
−3π/2
Inverse Sine: Numerically
Interchanging the x and y values.
Inverse Sine: Unit Circle
Arcsin only in Quadrant I and Quadrant IV because of the
limited domain.
 π
sin  −  = −1
 2
sin −1 ( −1) = −
3
 π
sin  −  = −
2
 3

3
π
sin −1  −
 = −
3
 2 
sin −1 ( 0 ) = 0
sin ( 0 ) = 0
π  1
sin   =
6 2
2
π 
sin   =
4 2
3
 2π 
sin 
=
 3  2
2
1 π
sin −1   =
2 6
 2 π
sin −1 
 =
 2  4
 3 π
sin −1 
 =
 2  3
sin −1 ( 2 ) = ?
2 is out of
the domain.
S. Stirling
π
2π
is out
3
of the range.
1 3
 ,

2 2 
 3
3
sin 
 = θ ask sin (θ ) =
2
 2 
−1
2
Sine positive in Quad I, so
 3 π
3
π 
sin   =
and sin −1 
 =
3 2
 2  3

2
sin −1  −
 = θ
2


− 2
ask sin (θ ) =
2
π
3
3
1
2
−π
4
2
Sine negative in Quad IV, so

2
π
2
 π
and sin −1  −
sin  −  = −
 = −
2
4
 4
 2 
− 2
 2 2
,


2
2 

Page 6 of 8
Trig Toolkit
Name ________________________
Inverse Trigonometric Functions (continued)
Inverses: sin (θ ) = value so sin −1 ( value ) = θ when −
π
2
≤θ ≤
π
2
and −1 ≤ value ≤ 1 .
Composites: sin ( sin −1 ( value ) ) = value as long as −1 ≤ value ≤ 1
and sin −1 ( sin(θ ) ) = θ as long as −
π
≤θ ≤
π
Limit domain of sine to get a function.
2
2
The “undoing” works so long as the domain and range of the inverse trig functions are met (see page 5).
So when simplifying trig. expressions or solving trig. equations, you must be aware of the restrictions on
the input variable.
1
1
π
 1
Ex 1. Evaluate sin −1  −  , “the angle whose sign is − ” sin (θ ) = − Only in Q IV, so − .
2
2
6
 2
It works the same way for the other inverse trig functions. See the inverse trig. chart.
π
π 
Ex 2. y = cos −1 ( 0 ) implies cos y = 0 when 0 ≤ y ≤ π . So, cos   = 0 and y = .
2
2
Ex 3. y = tan −1 3 implies tan y = 3 when −
π
2
≤ y≤
π
2
. So, y = tan −1 3 =
π
3
.
The easiest way to do Ex 4 – Ex 6 is to draw right triangles. Keep in mind what quadrant you are in
(determined by the range of the inverse trig function).
Ex 4.
Ex 5.

 1 
Evaluate cot sin −1  −  
 2 

 −1 
sin −1   = θ
 2
Quad. IV, sine negative.
Set up triangle using arcsin:
Solve sin −1
(
)
2 x = cos −1
( x)
both positive in Quad. I, so take sine both sides
2x = sin cos −1 x  and set up the triangle using arccos:


( )
Use Pythagorean thm.
2
12 = x + b 2
b = 1− x
Cotangent is adj/opp
3
cot [θ ] =
=− 3
−1
OR sin −1 − 1 = − π
2
6
cot  − π  = − 3 = − 3
1
 6
(
S. Stirling
)
Sine is opp/hyp, so sin θ =
2x = 1 − x , 0 ≤ x ≤ 1
Now solve for x.
2x = 1 − x
1
x=
3
1− x
now substitute
1
watch the domains:
radicals 2 x ≥ 0 , x ≥ 0
inverse cosine between –1 and 1.
x ≤1, x ≤1
Page 7 of 8
Trig Toolkit
Name ________________________
Ex 6. Write y = sin sec−1 ( x )  in algebraic form.
y = sin sec−1 ( x )  in Quad. I or II because of restrictions
on secant 0 ≤ θ ≤ π , θ ≠
π
2
 x
sec−1   = θ , Set up triangle, secant is hyp/adj.
1
Use Pythagorean thm.
x 2 = 12 + b 2
b = x2 − 1
Sine is opp/hyp and sine would be positive …
So y = sin θ =
x2 −1
, the x guarantees sin θ positive.
x
To help you remember your relationships.
Cosecant is defined as csc x =
1
. They are reciprocals. (The inverses are not!)
sin ( x )
1
Why is csc−1 x = sin −1   ?
 x
Start with
y = csc−1 x , where x ≥ 1 and −
So
csc y = x and we know cscθ =
π
2
≤ y≤
π
2
, y≠0
1
, replace csc y
sin θ
1
= x now solve for y
sin y
1
= sin y and now undo sine
x
1
sin −1   = y and y = csc−1 x
 x
1
Which means sin −1   = csc−1 x .
 x
1
= csc−1 x !!! Careful!
This is NOT the same as
sin −1 x
S. Stirling
Page 8 of 8