Download 5.2 The Sine and Cosine Functions

5.2-The Sine and Cosine Functions
Objectives:
1.
2.
3.
4.
5.
6.
Evaluate Sine and Cosine Functions
Find the sign of a Sine or Cosine function.
Use reference angles to evaluate sine or cosine function.
Evaluate expressions containing sine or cosine functions.
Use the fundamental identity to find the value of sine or cosine function.
Use trig functions to model the motion of a spring and other applications.
Overview:
In this set of notes we will define two trigonometric functions whose domain is the set of all angles.
These two functions form the foundation of trigonometry.
Definition: Sine and Cosine Functions:
If α is an angle in standard position and (x, y) is the point of intersection of the terminal side and the unit
circle, then
1. sine of α—abbreviated sin(α) or sin α—is the y-coordinate of that point.
2. cosine of α—abbreviated cos(α) or cos α—is the x-coordinate of that point.
sin α = y , cos α = x.
Domain and Range:
The domain of the sine function and the cosine function is the set of angles in standard position, but
since each angle has a measure in degrees or radians, we generally use the set of degree measures or the
set of radian measures as the domain.
If (x, y) is on the unit circle, then –1 ≤ x ≤ 1 and –1 ≤ y ≤ 1, so the range of each of these functions is the
interval [–1, 1].
 1 3
 on the unit
Example: Find the sine and cosine of the angle that passes through the point  − ,

2
2


circle.
Solution: From the definition of sine and
cosine:
sin α = y
cos α = x
cos α = −
1
2
and
sin α =
3
2
Sine and Cosine of Quadrantal Angles:
Because the coordinates of the points where the unit circle crosses the x-axis and y-axis are easily
determined, the sine and cosine of the quadrantal angles are also easy to determine from the unit circle.
Example: Find the following trigonometric
functions:
1. sine 90 0
2. cosine 180 0
3π
3. cosine
2
4. sine 2π
Solution: From the unit circle:
1. sine 90 0 = 1
2. cosine 180 0 = -1
3π
=0
3. cosine
2
4. sine 2π = 0
Notice that the sine and cosine of the quadrantal angles will always be ± 1 or 0.
When determining the sine or cosine of a quadrantal angle, use the following procedure:
1. Sketch the unit circle.
2. Label the coordinates where the unit circle crosses the axes.
3. Determine which axis the angles terminal side lies on.
4. Assign the appropriate value.
Example: Find the following trigonometric functions:
1. sine − 450 0
2. sine 540 0
7π
3. cosine −
2
4. cosine 7π
Solution: Sketch a unit circle, label the
coordinates, and then determine which of the
axes the angle lies on. Assign the appropriate
value.
1. sine − 450 0 = −1
2. sine 540 0 = 0
7π
3. cosine −
=0
2
4. cosine 7π = -1
Signs of Sine and Cosine:
The signs of the sine and cosine functions depend on the quadrant in which the angle lies. The diagram
below summarizes these sign values.
Instead of memorizing the values in this table, it
may be easier to remember the definitions of the
functions.
Because sin α = y , the sine function will have the
same sign value as y in any quadrant.
Likewise, because cos α = x , the cosine function
will have the same sign value as x in any quadrant.
Sine and Cosine of 45 Degrees:
Since the terminal side of a 45 degree angle lies on the line y = x, the x and y coordinates at the point of
intersection with the unit circle are equal. The equation of the unit circle is x 2 + y 2 = 1 and because
y = x, we can use substitution to rewrite the equation as x 2 + x 2 = 1
2x 2 = 1
1
x2 =
2
1
x=±
2
x=±
2
2
 2 2
,


2
2 

Because y = x, the coordinates of the point
at the intersection of the unit circle and
y = x is:
 2 2


 2 , 2 


And therefore sin 45 =
2
2
and cos 45 =
2
2
Sine and Cosine of a Multiple of 45 Degrees:
There are four points where the lines y = x and y = -x intersect the unit circle. The coordinates of these
key points can be determined as above and are shown on the diagram below. This diagram shows the
coordinates of the key points for determining the exact value of the sine and cosine of any angle that is a
π
multiple of 45 degrees or radians.
4
Example: Find the exact value of
5π
1. sin
.
4
7π
2. cos
4
Solution: From the diagram we can
see that:
5π
2
1. sin
=−
4
2
7π
2
2. cos
=
4
2
Sine and Cosine of 30 and 60 Degrees:
Consider an equilateral triangle which has three sides of equal length (c = 1) and all three angles are 60
degrees. Now cut this triangle in half, creating two right triangles with angles of 30, 60, and 90 degrees.
It can be seen that the side opposite the 30 degree angle (side a) will be half the length of side c.
Using the Pythagorean Theorem, we can find the third side of the right triangle side b.
a2 + b2 = c2
2
1
2
  +b =1
2
 
3
b2 =
4
3
b=
2
From this work, we can determine the coordinates of the point where a 60 degree angle intersects the
1 3

unit circle to be P =  ,

2 2 
Therefore,
sin 60 =
cos 60 =
3
and
2
1
2
We can use the same 30-60-90 degree triangle to determine the point of intersection of the terminal side
of a 30 degree angle with the unit circle.
From the diagram, the coordinates of the point where a 30 degree angle intersects the unit circle are
 3 1
P = 
, 
2
2

Therefore, cos 30 =
and sin 30 =
3
2
1
2
Notice the relationship between the sine and cosine of 30 and 60 degree angles. This is a significant
relationship which should prove helpful for remembering these values.
sin30 = cos60
and
sin60 = cos30
Sine and Cosine of a Multiple of 30 and 60 Degrees:
The sine and cosine values of multiples of 30 degree and 60 degree angles can be determined from the
diagrams below. Please do not try to memorize these values. Once you understand the relationship
demonstrated in these diagrams it will be easy to determine the function values of multiples of 30 and 60
degree angles by simply sketching the angle.
Multiples of 30 Degrees
Multiples of 60 Degrees
Example: Find the exact value of the following angles.
5π
4π
.
sin
cos
6
3
Solution: From the diagram we determine that:
sin
5π 1
=
6
2
cos
4π
1
=−
3
2
Strategy for Finding Exact Values of Certain Angles:
When determining the exact sine or cosine values of a 30, 45, or 60 degree angle, use the following
procedure:
1. Determine if the angle is a multiple of 30, 45, or 60 degrees.
2. Sketch the angle in the appropriate quadrant on the unit circle.
3. Assign the appropriate value.
Example: Find the sine and cosine of
5π
3
Solution: Because the denominator is a 3, this angle is a multiple of 60 degrees or
π
3
. The angle lies in
Quadrant IV.
Therefore sin
5π
3
5π 1
and cos
=−
=
3
2
3
2
Example: Find the sine and cosine of −
5π
4
Solution: Because the denominator is a 4, this angle is a multiple of 45 degrees or
Quadrant II.
Therefore sin −
5π
2
5π
2
and cos −
=
=−
4
2
4
2
π
4
. The angle lies in
Reference Angles:
Another way to find the sine and cosine of an angle is to find the sine and cosine of the corresponding
reference angle. A reference angle is the positive acute angle formed by the terminal side of an angle
and the closest x-axis
Definition: Reference Angle
If θ is a nonquandrantal angle in standard position, then the reference angle for θ is the positive acute
angle θ' (read “theta prime”) formed by the terminal side of θ and the positive or negative x-axis.
Theorem: Evaluating Trigonometric Functions Using Reference Angles
For an angle θ in standard position that is not a quadrantal angle, the value of a trigonometric function of
θ can be found by finding the value of the function for its reference angle θ' and prefixing the
appropriate sign.
Procedure for Finding Sine/Cosine Using Reference Angles:
1. Sketch the terminal side of the given angle in the appropriate quadrant.
2. Determine the angle measure between the terminal side and the nearest x-axis.
3. The sine/cosine of the given angle is the same as that of the reference angle after adjusting the
sign value for that quadrant.
Example: For each of the following, identify the quadrant the terminal side is in, find the reference
angle and then determine the sine and cosine of the angle.
120 0
210 0
315 0
5π
4
−
7π
3
11π
6
Solution: Sketch each angle in the appropriate quadrant than determine the positive angle measure
between the angle and the nearest x-axis. This angle measure is the reference angle.
3
2
120 0
II
180 – 120 = 60 0
sin 120 0 = sin 60 0 =
210 0
III
210 – 180 = 30 0
sin 210 0 = − sin 30 0 = −
315 0
IV
360 – 315 = 45 0
sin 315 0 = − sin 45 0 = −
5π
4
III
5π 4π π
−
=
4
4
4
sin
IV
−
6π 7π π
+
=
3
3
3
sin −
IV
12π 11π π
−
=
6
6
6
sin −
−
7π
3
11π
6
cos 120 0 = − cos 60 0 = −
1
2
2
2
5π
π
2
= − sin = −
4
4
2
7π
7π
3
= − sin
=−
3
3
2
11π
11π
1
= − sin
=−
6
6
2
1
2
cos 210 0 = − cos 30 0 = −
cos 315 0 = cos 45 0 =
cos
2
2
5π
π
2
= − cos = −
4
4
2
cos −
cos−
7π
7π 1
= cos
=
3
3
2
11π
11π
3
= cos
=
6
6
2
3
2
Approximate Values for Sine and Cosine:
The sine and cosine for any angle that is a multiple of 30, 45 or 60 degrees can be found exactly. These
angles are so common that it is important to know these exact values. However, for most other angles
we use approximate values for sine and cosine, found with the help of a scientific calculator.
Example: Find each function value rounded to four decimal places.
a.
b.
c.
d.
sin ( 438°)
cos ( -433°)
sin (3.13)
cos ( 32.5°)
Solution: Using a scientific calculator, we obtain:
a. sin ( 438°) = 0.9781
b. cos ( -433°) = 0.2924
c. sin (3.13) = 0.0116
d. cos ( 32.5°) = 0.8434
The Fundamental Identity of Trigonometry:
An identity is an equation that is satisfied for all values of the variable for which both sides are defined.
The most fundamental identity in trigonometry involves the squares of the sine and cosine functions.
The equation of the unit circle is x 2 + y 2 = 1 . Because sin α = y and cosα = x we can write the
following identity called the fundamental identity of trigonometry.
sin 2 α + cos 2 α = 1
This identity is mathematically useful in many ways. One such use is to find the value of sine or cosine
if the value of the other function is known.
Example: Find sin (α), given that cos (α) =
4
and α in quadrant IV.
7
Solution: Use the fundamental identity of trigonometry to solve for sin α .
sin 2 α + cos 2 α = 1
2
4
sin 2 α +   = 1
7
16
sin 2 α = 1 −
49
33
sin 2 α =
49
33
sin α = ±
7
Because the angle is in Quadrant IV, the sine is negative. Therefore, sin α = −
33
7
Modeling the Motion of a Spring:
The sine and cosine functions are used in modeling the motion of a spring. If
the weight is at rest while hanging from a spring then it is at an equilibrium
position, or 0 on a vertical number line. If the weight is set in motion with
an initial velocity v0 from location x0 then the location at time t is given by
x=
v0
ω
sin(ω ⋅ t ) + x0 cos(ω ⋅ t )
where ω is the spring constant, and t is time. A downward initial velocity
is considered positive and an upward initial velocity is negative. Positive
values of x are below the equilibrium position and negative values are
above.
Example: A weight on a vertical spring is given an initial downward velocity of 90 cm/sec from a point
10 cm above equilibrium. Assuming that the spring constant has a value of ω = 4.5, write the formula
for the location of the weight at time t. Find its location at 10 seconds after it is set in motion. Write your
answer using only exact values.
Solution: Use the formula:
x=
v0
x=
90
sin( 4.5t ) − 10 cos( 4.5t )
4.5
ω
sin(ω ⋅ t ) + x0 cos(ω ⋅ t )
Using v0 = 90 and x0 = −10 .
Solving the above equation for x at t=10.
x=
v0
ω
sin(ω ⋅ t ) + x0 cos(ω ⋅ t )
90
sin(4.5 ⋅ 10) − 10 cos(4.5 ⋅ 10)
4.5
x = 20 sin 45 − 10 cos 45
x=
2
2
− 10
2
2
x = 10 2 − 5 2
x = 20
x=5 2
Therefore the spring will be located 5 2 cm. below the equilibrium point 10 seconds after the spring is
released.
Problem: A weight on a vertical spring is given an initial downward velocity of 72 cm/sec from a point
2 3 cm below equilibrium. If the spring constant has a value of ω = 12, write the formula for the
location of the weight at time t. Then, find its location 5 seconds after it is set in motion. Write your
answer using only exact values.
Solution: Use the formula:
x=
v0
ω
sin(ω ⋅ t ) + x0 cos(ω ⋅ t )
where v0 is an initial velocity from location x 0 , ω is the spring constant, and t is time. A downward
initial velocity is considered positive and an upward initial velocity is negative. Positive values of x are
below the equilibrium position and negative values are above.
Using v0 = 72 and x0 = 2 3 the formula is:
x=
72
sin(12t ) + 2 3 cos(12t )
12
Solving the above equation for t=5 seconds..
72
sin(12 ⋅ 5) + 2 3 cos(12 ⋅ 5)
12
x = 6 sin 60 + 2 3 cos 60
x=
3
1
+2 3⋅
2
2
x=3 3+ 3
x = 6⋅
x=4 3
Therefore the spring will be located 4 3 cm. below the equilibrium point 10 seconds after the spring is
released.
1 2
v 0 sin(2θ ) gives the distance (d) in feet that a projectile will travel when
32
its launch angle is θ and its initial velocity is v0 feet per second. Approximately what initial velocity in
feet per second does it take to throw a javelin 256 feet with launch angle 15 degrees?
Example: The formula d =
Solution: Let d = 256 and θ = 15 in the formula.
1 2
v0 sin(2θ )
32
1 2
256 =
v 0 sin(2 ⋅ 15)
32
1 2
256 =
v 0 sin(30)
32
1 2 1 
256 =
v0  
32  2 
16,384 = v02
d=
v 0 = 128
The initial velocity is 128 feet per second.