Post on 05-Jan-2022
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Calculus ATrigonometric Functions
Dr. Bisher M. Iqelanbiqelan@iugaza.edu.ps
Department of MathematicsThe Islamic University of Gaza
2019-2020, Semester 1
Dr. Bisher M. Iqelan (IUG) Sec1.3: Trigonometric Functions 1st Semester, 2019-2020 1 / 14
Trigonometric Functions
The number of radians in the centralangle A′CB ′ within a circle of radiusr is de�ned as the number of"radius units" contained in the arc ssubtended by the central angle.With the central angle measuring θradians, this meansθ = s/r or s = rθ.
De�nition
♣
One complete revolution of the unit circle is 360◦ or 2π radians.Therefore π radians = 180◦ and
1 radian =180
π≈ 57.3◦ or 1◦ =
π
180≈ 0.017 radians.
Note
Dr. Bisher M. Iqelan (IUG) Sec1.3: Trigonometric Functions 1st Semester, 2019-2020 2 / 14
Trigonometric Functions
An angle in the xy−plane is in standard position if its vertex lies at theorigin and its initial ray lies along the positive x−axis. Angles measured coun-terclockwise from the positive x−axis are assigned positive measures; anglesmeasured clockwise are assigned negative measures.
De�nition
A central angle in a circle of radius 8 is subtended by an arc of length10π. Find the angle's radian in degree measures.Solution:
θ = 225◦
Example
Dr. Bisher M. Iqelan (IUG) Sec1.3: Trigonometric Functions 1st Semester, 2019-2020 3 / 14
Trigonometric Functions
Convert the following radiansto degrees:
1.π
4. 2.
π
6. 3.
π
3. 4.
4π
3
Solution:
1π4= �π
4.180�π
= 45◦
2π6= �π
6.180�π
= 30◦
3π3= �π
3.180�π
= 60◦
44π3
= 4�π3.180�π
= 240◦
Example 1
Convert the following degreesto radians:
1. 40◦. 2. 120◦. 3. 15◦. 4. 270◦
Solution:
1 40◦ = 40. π180
= 2π9
2 120◦ = 120. π180
= 2π3
3 15◦ = 15. π180
= π12
4 270◦ = 270. π180
= 3π2
Example 2
Dr. Bisher M. Iqelan (IUG) Sec1.3: Trigonometric Functions 1st Semester, 2019-2020 4 / 14
Trigonometric Functions
Let 0 < θ < π2
using the right triangle we de�ne the sixtrigonometric functions as follows
De�nition: Right Triangle De�nition
Dr. Bisher M. Iqelan (IUG) Sec1.3: Trigonometric Functions 1st Semester, 2019-2020 5 / 14
Trigonometric Functions
We de�ne the six trigonometric functions for any angle θ by �rstplacing the angle in standard position in a circle of radius r . Thende�ne the trigonometric functions in terms of the coordinates of thepoint P(x , y) where the angle's terminal ray intersects the circle asfollows:
sin θ = yr cos θ = x
r
sec θ = rx csc θ = r
y
tan θ = yx cot θ = x
y
De�nition: Circular De�nition
The coordinates of the point P(x , y) on the circle x2 + y2 = r2 canbe expressed in term of θ and r as x = r cos θ and y = r sin θ. Now,if r = 1 we see that x = cos θ and y = sin θ.
Note 1
Dr. Bisher M. Iqelan (IUG) Sec1.3: Trigonometric Functions 1st Semester, 2019-2020 6 / 14
Trigonometric Functions
By de�nition above, we immediately have the trig identities:
tan θ = sin θcos θ cot θ = 1
tan θ
sec θ = 1
cos θ csc θ = 1
sin θ
Note 2
Based on the special 30◦ − 60◦ − 90◦ and 45◦ − 45◦ − 90◦ righttriangles, we can deduce the following trig functions for the "specialangles" 30◦ = π
6, 45◦ = π
4, and 60◦ = π
3:
cos π6=√3
2cos π
4=√2
2cos π
3= 1
2
sin π6= 1
2sin π
4=√2
2sin π
3=√3
2
tan π6= 1√
3tan π
4= 1 tan π
3=√3
Note 3
Dr. Bisher M. Iqelan (IUG) Sec1.3: Trigonometric Functions 1st Semester, 2019-2020 7 / 14
Trigonometric Functions
In Figure 1 we see in whichquadrants or on which axes theterminal side of an angle0◦ ≤ theta < 360◦ may fall.
Figure: 1
Figure 2 summarizes the signs (positiveor negative) for the trigonometricfunctions based on the angle's quadrant:
Figure: 2
Note 4
Dr. Bisher M. Iqelan (IUG) Sec1.3: Trigonometric Functions 1st Semester, 2019-2020 8 / 14
Trigonometric Functions
Find the exact values of all six trigonometric functions of 120◦.
Solution:
We know 120◦ = 180◦ − 60◦. We can use thepoint (−1,
√3) on the terminal side of the
angle 120◦ in QII , since a basic right trianglewith a 60◦ angle has adjacent side of length 1,opposite side of length
√3, and hypotenuse of
length 2, as in the �gure below. Drawing thattriangle in QII so that the hypotenuse is on theterminal side of 120◦ makes r = 2, x = −1 andy =√3. Hence:
sin 120◦ = yr =
√3
2cos 120◦ = x
r = −12
tan 120◦ = yx =
√3
−1
csc 120◦ = ry = 2√
3sec 120◦ = r
x = 2
−1 cot 120◦ = xy = −1√
3
Example 1
Dr. Bisher M. Iqelan (IUG) Sec1.3: Trigonometric Functions 1st Semester, 2019-2020 9 / 14
Trigonometric Functions
Suppose that cos θ = − 4
5. Find the exact values of sin θ and tan θ.
Solution: Since cos θ = − 4
5, we can use 4 as the length of the adjacent side and 5 as
the length of the hypotenuse. The length of the opposite side must then be 3. Sincecos θ is negative, so θ must be in either QII or QIII as shown in Figure 3 below.
When θ is in QII , we see from Figure 3(a) that the point (−4, 3) is on the terminal
side of θ, and so we have x = −4, y = 3, and r = 5. Thus, sin θ = yr = 3
5and
tan θ = yx = 3
−4 .When θ is in QIII , we see from Figure 3(b) that the point (−4,−3) is on the terminal
side of θ, and so we have x = −4, y = −3, and r = 5. Thus, sin θ = yr = −3
5and
tan θ = yx = 3
4. Thus, either sin θ = 3
5and tan θ = − 3
4or sin θ = − 3
5and tan θ = 3
4.
Example 2
Dr. Bisher M. Iqelan (IUG) Sec1.3: Trigonometric Functions 1st Semester, 2019-2020 10 / 14
Periodicity and Graphs of the Trigonometric Functions
A function f (x) is periodic if there is a positive number p such thatf (x + p) = f (x) for every value of x . The smallest value of p is theperiod of f .
De�nition
The periods of sine, cosine, secant, and cosecant are each 2π.The periods of tangent and cotangent are both π. This leads tothe trigonometric identities:
sin(x + 2π) = sin x sin(x + 2π) = sin x
sec(x + 2π) = sec x csc(x + 2π) = csc x
tan(x + π) = tan x cot(x + π) = cot x
Note
Dr. Bisher M. Iqelan (IUG) Sec1.3: Trigonometric Functions 1st Semester, 2019-2020 11 / 14
Periodicity and Graphs of the Trigonometric Functions
The graphs of the six trigonometric functions are as follows(the shading indicates a single period):
Note
Dr. Bisher M. Iqelan (IUG) Sec1.3: Trigonometric Functions 1st Semester, 2019-2020 12 / 14
Trigonometric Functions
Trigonometric Identities
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