Download 2. TRIGONOMETRIC FUNCTIONS and TABLES

SECTION TWO: TRIGONOMETRIC FUNCTIONS AND THEIR TABLES:
2.1 TRIGONOMETRIC FUNCTIONS AND THEIR TABLES:
Let P(x,y) be any point on the unit circle:
2
y
Unit Circle: x2 + y2 = 1
1
P(x,y)
x
α
-2
2
-1
-2
If P is the terminal point that corresponds to an angle
following functions:

, then we can define the
1. COSINE: The x-coordinate of the point P is called the “cosine of the angle  ”
and denoted by cos 
Cosine Function Domain:
∈0,
ℝ
, Range:


2
[−1,1]
∈
2
y

,
2
2
y
α is in the second quadrant
α is in the first quadrant
cos α < 0
cos α > 0
1
P(x,y)
P(x,y)
1
x
α
-1
Unit Circle: x2 + y2 = 1
x
α
2
2
-1
Unit Circle: x2 + y2 = 1
∈ ,
3

2
∈
2
y
3
, 2 
2
2
α is in the third quadrant
y
α is in the forth quadrant
cos α < 0
cos α > 0
1
1
x
α
x
α
2
2
P(x,y)
-1
P(x,y)
-1
Unit Circle: x2 + y2 = 1
Unit Circle: x2 + y2 = 1
2. SINE: The y-coordinate of the point P is called the “sine of the angle  ” and
denoted by sin 
Sine Function Domain:
∈0,
ℝ
, Range:


2
[−1,1]
∈
2
y

, 
2
2
α is in the second quadrant
y
α is in the first quadrant
sin α > 0
sin α > 0
1
1
P(x,y)
P(x,y)
x
α
-1
Unit Circle: x2 + y2 = 1
x
α
2
2
-1
Unit Circle: x2 + y2 = 1
∈ ,
3

2
∈
2
3
, 2
2
y
y
2
α is in the third quadrant
α is in the forth quadrant
sin α < 0
sin α < 0
1
1
x
α
x
α
2
P(x,y)
2
-1
P(x,y)
-1
Unit Circle: x2 + y2 = 1
Unit Circle: x2 + y2 = 1
SUMMARY: Any point P on the unit circle coterminal to an angle
cos  , sin 
2

has coordinates
y
Unit Circle: x2 + y2 = 1
P(cos α , sin α)
1
x
α
2
-1
An immediate result that follows summary:
cos2 sin 2 =1
Explanation: Since P is any point on the unit circle, its coordinates must satisfy the
2
2
equation of the unit circle. Therefore we get cos  sin  =1
Notation:
cos 
2
is denoted as
cos 2  ,and
sin 
2
2
is denoted as sin  , etc.
3. TANGENT: The y-coordinate of the point Q is called the “tangent of the angle 
” and denoted by tan 
2
y
Q (1 , tan α)
(0,1)
1
P (cos α , sin α)
α
(-1,0)
x
(1,0)
2
-1
(0,-1)
Unit Circle: x2 + y2 = 1
From the figure, it is understood that the y-coordinate of point Q is well-defined

3
except when  is different from 2 , 2
, etc. Therefore the domain of the

tangent function is ℝ−{ k } , where k is any integer. Since Q can be anywhere
2
on its axis, it takes any value in the interval −∞ , ∞ . Therefore the range of the
−∞ ,∞ , namely all real numbers.
tangent function is

ℝ−{ k } , k ∈ℤ
2
Tangent Function Domain:
∈0,


2
∈
2
(0,1)
1
2
α
(-1,0)
-1
Q (1 , tan α)
P (cos α , sin α)
(1,0)
(0,-1)
Unit Circle: x2 + y2 = 1
y
α is in the second quadrant
tan α < 0
(0,1)
1
P
x
2
−∞ ,∞

,
2
y
α is in the first quadrant
tan α > 0
, Range:
x
α
(1,0)
(-1,0)
-1
2
Q (1 , tan α)
(0,-1)
Unit Circle: x2 + y2 = 1
∈ ,
3

2
∈
2
3
, 2 
2
y
2
α is in the third quadrant
tan α > 0
(0,1)
1
(0,1)
1
Q (1 , tan α)
(1,0)
x
α
x
α
(-1,0)
y
α is in the forth quadrant
tan α < 0
(1,0)
(-1,0)
2
P
P
2
Q (1 , tan α)
-1
(0,-1)
Unit Circle: x2 + y2 = 1
-1
(0,-1)
Unit Circle: x2 + y2 = 1
4. COTANGENT: The x-coordinate of the point R is called the “cotangent of the
angle  ” and denoted by cot 
2
y
(0,1)
1
R(cot α , 1)
P(cos α , sin α)
α
(-1,0)
x
(1,0)
2
-1
(0,-1)
Unit Circle: x2 + y2 = 1
Once again, from the figure, it is understood that the x-coordinate of point R is welldefined except when  is different from 0 ,  , etc. Therefore the domain of
the cotangent function is ℝ−{k } , where k is any integer. Since R can be
anywhere on its axis, it takes any value in the interval −∞ , ∞ . Therefore the
−∞ ,∞ , namely all real numbers.
range of the cotangent function is
Cotangent Function Domain:
ℝ−{k } , k ∈ℤ
, Range:
−∞ ,∞
∈0,


2
∈
2

,
2
y
α is in the first quadrant
cot α > 0
(0,1)
1
α
(-1,0)
α is in the second quadrant
cot α < 0
R(cot α , 1)
R(cot α , 1) (0,1)
1
P(cos α , sin α)
P
(1,0)
x
-1
(-1,0)
3
, 2 
2
y
2
α is in the third quadrant
cot α > 0
α
y
α is in the forth quadrant
cot α < 0
R (cot α , 1)
R(cot α , 1) (0,1)
1
x
α
(1,0)
x
2
-1
∈
(0,1)
1
(1,0)
(0,-1)
Unit Circle: x2 + y2 = 1
3

2
2
α
(-1,0)
2
(0,-1)
Unit Circle: x2 + y2 = 1
∈ ,
y
2
2
(-1,0)
(1,0)
x
2
P
P
-1
(0,-1)
Unit Circle: x2 + y2 = 1
-1
(0,-1)
Unit Circle: x2 + y2 = 1
CONCLUSION: In the unit circle, the x-axis may be called the cosine axis, the y-axis
may be called the sine axis. The axis on which point Q travels is called the tangent
axis, and the axis on which point R travels is called the cotangent axis. (See figure
below)
y axis (or sine axis)
(0,1)
R
cotangent axis
Q
P
(-1,0)
x axis (or cosine axis)
θ
(1,0)
P ( cos α , sin α )
Q ( 1 , tan α )
R ( cot α , 1 )
(0,-1)
tangent axis
5. SECANT and COSECANT FUNCTIONS: These are little less important
trigonometric functions. I will just give their expressions and move on.
 ” : sec
Notation for “secant of the angle
1
Definition of secant: sec = cos 
Notation for “cosecant of the angle
1
Definition of cosecant: csc = sin 

 ” : csc

2.2 TRIGONOMETRIC RATIOS OF QUADRANTAL ANGLES:


(DEG)
(RAD)
0o
0
90o

2
1
0

0
-1
180
o
270o
3
2
sin 
cos 
tan 
cot 
sec 
csc 
0
1
0
Undefined
1
Undefined
Undefined
Undefined
0
0
Undefined
undefined
-1
0
1
-1
Undefined
Undefined
0
-1
2.3 TRIGONOMETRIC FUNCTIONS ASSOCIATED WITH ACUTE ANGLES AND
SOLVING RIGHT TRIANGLE:
If

is an acute angle, then
∈0o , 90 o 
∈0 rad ,
or

Consider this right triangle. For an acute angle

rad 
2
in a right triangle,
B
β
c
A
* cos 
Hence
* sin 
Hence
α
a
C
b
= (length of side adjacent to
b
a
cos = and cos =
c
c
 )
= (length of side opposite to
a
b
sin = and sin =
c
c
 )
* tan  = (length of side opposite to  )
a
b
Hence tan = and tan =
b
a
(length of hypotenuse)
÷
(length of hypotenuse)
÷
÷
(length of side adjacent to  )
÷
* cot  = (length of side adjacent to  )
(length of side opposite to  )
b
a
Hence cot = and cos =
a
b
1
c
c
* sec  = cos  Hence sec = and sec =
b
a
1
c
c
* csc  =
Hence csc = and csc =
sin 
a
b
Example: When the angle of elevation of the sun is
10 m long. How tall is the tree?
Ans: With the help of the figure, we can write
From this we get
a=10 tan 30=5.77 m
B
60
c
A
30
b=10 m
a=?
C
30o
tan 30=
a
b
, a tree casts a shadow of
. Equivalently,
a=b tan 30
2.4 RELATION BETWEEN TRIGONOMETRIC RATIOS OF TWO
COMPLEMENTARY ANGLES:
If two angles are complementary, that is, their sum is equal to
90o

2
(or
radians) then sine, tangent, and secant functions of one angle is equal to the COfunction of the other one.
B
α+β=90°, therefore,
sin α = cos β = a / c
tan α = cot β = a / b
sec α = csc β = c / b
A
α
β
c
a
C
b
In other words,
y
α+β=90°, therefore, β = 90 - α
sin β = sin (90 - α) = cos α
tan β = tan (90 - α) = cot α
sec β = sec (90 - α) = csc α
β
α
x
Examples:
sin 40o=cos90 o−40o =cos 50o
sec 89=csc90o−89o =csc 1o
cot
3
 3

=tan  −
=tan
8
2
8
8
tan 10=cot 90o−10o =cot 80o
cos 2=sin 90 o−2o =sin 88o
cos
5
 5

=sin  −
=cos
12
2 12
12
We can obtain the values of trigonometric functions for
trigonometric ratios in a right triangle.
30o , 45o and 60o
by using
2.5 TRIGONOMETRIC RATIOS OF
30 o , 45o , 60 o
The values of these functions for 30o and 60o can be evaluated by drawing an
equilateral triangle with side length 1, and for 45o by drawing a square with side
length 1.
First, let us use the equilateral triangle of side length 1 below to determine the
trigonometric ratios of 30o and 60o
C
1
1
1
A
B
Let us draw the altitude to one of the sides to bisect it:
C
30
1
1
A
D
60
1/2
1/2
B
Namely we obtain a 30 o−60o −90 o
right triangle ADC. It remains to determine its
sides. The hypotenuse and one of the right sides is known. Therefore we can use the
pythagorean theorem to determine the unknown side:
Using
Namely
∣AC 2∣=∣AD 2∣∣CD 2∣
3
∣CD∣= 
2
.
, we can write
1 2
1 3
12=  ∣CD 2∣⇒∣CD 2∣=1 – =
2
4 4
.
C
30
1
1
3/2
A
sin 30=
cos 30=
D
60
B
1/2
1/2
1/2
=1/2
1
sin 60=
  3/ 2
=  3/ 2
1
tan 30=
cot 30=
  3/ 2
= 3/ 2
1
cos 60=
1/ 2
=1/  3
3
 
2
tan 60=
  3/ 2
= 3
1
 
2
cot 60=
1/ 2
=1/ 2
1
  3/ 2
= 3
1
 
2
1/ 2
=1/  3
3
 
2
Remark: For any two complementary angles, the values of the functions sine and
tangent of one angle is equal to the co-function values of the other angle. Namely:
sin 30=cos 60 ,sin 60=cos 30 , tan 30=cot 60 , tan 60=cot 30
.
Next, let us use the square of side length 1 below to determine the trigonometric
ratios of 45o
D
1
1
A
C
1
1
B
Let us draw one of the diagonals to divide the square into two right triangles:
1
D
C
45
1
A
1
45
B
1
Namely we obtain a 45o −45o −90 o right triangle ABC. It remains to determine its
sides. The right sides are known. Therefore we can use the pythagorean theorem to
determine the hypotenuse:
Using
∣AC 2∣=∣AB 2∣∣CB 2∣
∣AC 2∣=1212=2⇒∣AC∣= 2
, we can write
1
D
.
C
45
2
1
A
1
45
B
1
sin 45=
cos 45=
1
2
1
2
1
tan 45= =1
1
1
cot 45= =1
1
Let's repeat the remark about complementary angles: For any two complementary
angles, the values of the functions sine and tangent of one angle is equal to the cofunction values of the other angle. Namely sin 45=cos 45 , tan 45=cot 45 .
Now, let us summarize these results together in a single table:
C
1
D
C
45
30
1
1
2
1
1
3/2
A
D
60
1/2
cos 30=
tan 30=
cot 30=
1/2
cos 45=
cos 60=
1
2
1
2
  3/ 2
= 3
1
 
2
1
tan 45= =1
1
1/ 2
=1/  3
3

 
2
1
cot 45= =1
1
tan 60=
cot 60=
B
1
1/ 2
=1/ 2
1
  3/ 2
=  3/2
1
  3/ 2
= 3
1
 
2
45
sin 45=
sin 60=
1/2
=1/  3
3

 
2
A
  3/ 2
= 3/ 2
1
1/ 2
=1/ 2
1
sin 30=
B
2.6 FUNDAMENTAL TRIGONOMETRIC IDENTITIES:
IDENTITY ONE: For all
∈ℝ
,
sin 2 cos2 =1
We proved this identity using a unit circle. Now let us prove it using a right triangle:
sin α = a / c ⇒ sin2α = a2 / c2
cos α = b / c ⇒ cos2α = b2 / c2
⇒ sin2α + cos2α = (a2 / c2) + (b2 / c2)
c
⇒ sin2α + cos2α = (a2 + b2) / c2
By pythagorean identity, a2 + b2 = c2
⇒ sin2α + cos2α = c2 / c2 = 1
α
A
b
B
β
a
C
Conclusion:
sin 2 =1 – cos 2  ⇒ sin =± 1−cos2 
Example: Calculate
sin  cos 
if
2
2
2
, cos =1 – sin  ⇒ cos =± 1−sin 
sin  – cos =
1
3
1
.
9
Using the identity one, the first two terms simplify to 1. Hence the equation becomes
1
1
8
4
1 – 2sin  cos =
from which we get – 2sin  cos = −1=− ⇒ sin  cos =
9
9
9
9
2
2
Answer: Squaring the difference, we get an equation sin cos  – 2 sin  cos =
IDENTITY TWO:
tan =
sin 
cos 
, cot =
cos 
sin 
It it very easy to derive this using a right triangle. Let's do it with unit circle. Let's
work with the figure below: Just use the definition of tangent and cotangent.
y axis (or sine axis)
C(0,1)
cot α
Q
cotangent axis
R
P
tan α
1
1
sin α
(-1,0)
A
x axis (or cosine axis)
α
cos α
(0,-1)
M
B(1,0)
P ( cos α , sin α )
Q ( 1 , tan α )
R ( cot α , 1 )
tangent axis
We will use similarity of triangles AMP and ABQ first:
Namely
tan =
sin 
cos 
. Next we use similarity of triangles AMP and ACR:
∣CR∣ ∣AM ∣ cot  cos 
=
⇒
=
. Namely
∣CA∣ ∣MP∣
1
sin 
Conclusion:
∣BQ∣ ∣MP∣ tan  sin 
=
⇒
=
∣BA∣ ∣MA∣
1
cos 
tan  cot =1
cot =
cos 
.
sin 
tan cot 
Example: Calculate
tan  – cot =4
if
 AB2= A−B24 AB
Answer: Here we can use a shortcut:
Namely
.
tan cot 2=tan −cot 24 tan  cot 
Using the given information and the identity two, we get
2
2
tan cot  =4 4⋅1=164=20
Taking the square root, we get
IDENTITY THREE:
sec =
Example: Prove that
tan cot =± 20
1
1
, csc =
cos 
sin 
1tan 2 =sec 2 
Here use the identity two to write
1tan 2 =1
sin 2  cos2  sin 2  cos 2 sin 2 
=

=
cos 2  cos2  cos2 
cos2 
The numerator equals 1 (by identity one) hence the equality simplifies to
1
=sec2 
2
cos 
2.7 TRIGONOMETRIC VALUES OF
−
Let %alpha be an acute angle. Since the angle − is in Quadrant IV, we can write
the following rules (just by using the definition of sin, cos, tan, and cot)
1
sin (- α ) = - sin (α )
P (sin(α ) , cos(α ))
cos (- α ) = cos (α )
α
-α
-2
tan (- α ) = - tan (α )
cot (- α ) = - cot (α )
2
P' (sin(- α ) , cos(-α ))
-1
Remark: Sine, tangent, and cotangent are odd functions whereas cosine is an even
function.
Example: sin (-60) + cos (-45) + tan (-30) + cot (-45)
= −sin 60cos45−tan 30−cot 45=−
 3  1 – 1 −1
2  2 3
2.8 FINDING THE REMAINING TRIGONOMETRIC RATIOS PROVIDED THAT
ONE OF THEM IS KNOWN
Example: Let
∈0,


2
and
sin =
3
5
cos 
. Find
and
tan 
.
B
Solution:
1. Draw a suitable right triangle such tha sin α = 3 / 5
2. Use pythagorean theore m to de te rmine the unknown side:
⇒ c2 = a2 + b 2⇒ b 2 = c2 - a2 = 25 - 9 = 16 ⇒ b = 4
c=5
a=3
3. Use de finition of cosine, tange nt to find cos α and tan α :
cos α = b / c = 4 / 5 and tan α = a / b = 3 / 4
α
A
Example: Let
∈


2,
and
cos =−
5
13
. Find
C
b=unknown
sin  , tan  and cot 
.
B
Remark: α is not an acute angle . Le t's assume it is an acute angle and we
will change the answe r late r accordingly.
1. Le t us assume α is an acute angle whence cos α = 5 / 13
2. Draw a suitable right triangle such that cos α = 5 / 13
3. Use pythagorean theore m to de te rmine the unknown side:
⇒ c2 = a2 + b 2⇒ a2 = c2 - b 2 = 169 - 25 = 144 ⇒ a = 12
c=13
a=?
4. Use de finition of sine , tange nt, cotange nt:
sin α = a / c = 12 / 13 , tan α = a / b = 12 / 5 and cot α = b / a = 5 / 12
5. In Quadrant II, sine is positiv e while cosine, tange nt and cotange nt
ar e negative (se e section 2.1) There fore, the final answe r is:
sin α = 12 / 13 , tan α = - 12 / 5 and cot α = - 5 / 12
A α
b=5
C
2.9 SUMMARY
Let's summarize section two with a big unit circle that includes all the multiples of
30o and 45o angles. Analyze the figure below carefully:
A2 (0,1)
1
1
3
D2 (- ,
)
2 2
120°
2
2
C2 (,
)
2
2
135°
3 1
B2 (, )
2 2
150°
90°
0.8
0.6
0.4
1 3
D1 ( ,
)
2 2
60°
2 2
C1 (
,
)
2
2
45°
3 1
B1 (
, )
2 2
30°
0.2
A3 (-1,0) 180°
-1.5
-1
-0.5
0° A1 (1,0)
360° 1
0.5
-0.2
B3 (-
-0.4
3
1 210°
,- )
2 2
C3 (-
2
2
,-
2
2
)
-0.6
225°
1
3
D3 (- ,)
2 2
330°
240°
-0.8
270°
-1
A4 (0,-1)
315°
300°
C4 (
1
3
D4 ( ,)
2 2
B4 (
2
2
,-
3
2
2
2
1
,- )
2
)
1.5