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Trigonometry: Angles in Standard Position and Trigonometric Functions - Prof. Daniel E. Sm, Assignments of Mathematics

The concept of angles in standard position in trigonometry, the definition of central angles and subtended angles, and the introduction of trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. It also includes examples and homework problems.

Typology: Assignments

2009/2010

Uploaded on 02/25/2010

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Chapter 7
Section 1
We will now look at trigonometry. We are going to put our angles in a coordinate system such that the vertex of the angle
is at the origin and the initial side of the angle lies on the positive x-axis. This is called standard position.
θ
x
y
A counter-clockwise rotation is positive and clockwise is negative.
45
45
Recall that a full revolution is 360. So 1is 1
360 of a revolution. A degree is divided into minutes. There are 60 minutes
in a degree. We denote minutes by the symbol (0). So
1= 600.
A minute is divided into seconds. There are 60 seconds in a minute. We denote seconds by ( 00). So
10= 6000.
Example 1. Covert 21503000 to a decimal degree.
Example 2. Covert 5.7623to degree-minute-second format.
One way of measuring angles is in degrees. Another is with radians. Suppose you have a circle of radius r. Divide the
circle into two arcs such that one of them is of length r. Now draw 2 line segments starting from the center of the circle and
intersecting the circle at each endpoint of the arc of length r.
1
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pf4
pf5
pf8
pf9
pfa

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Chapter 7

Section 1

We will now look at trigonometry. We are going to put our angles in a coordinate system such that the vertex of the angle is at the origin and the initial side of the angle lies on the positive x-axis. This is called standard position.

x

y

A counter-clockwise rotation is positive and clockwise is negative.

45 ◦

− 45 ◦

Recall that a full revolution is 360◦. So 1◦^ is 3601 of a revolution. A degree is divided into minutes. There are 60 minutes in a degree. We denote minutes by the symbol (′). So

1 ◦^ = 60′.

A minute is divided into seconds. There are 60 seconds in a minute. We denote seconds by ( ′′). So

1 ′^ = 60′′.

Example 1. Covert 2 ◦ 15 ′ 30 ′′^ to a decimal degree.

Example 2. Covert 5. 7623 ◦^ to degree-minute-second format.

One way of measuring angles is in degrees. Another is with radians. Suppose you have a circle of radius r. Divide the circle into two arcs such that one of them is of length r. Now draw 2 line segments starting from the center of the circle and intersecting the circle at each endpoint of the arc of length r.

θ

r

r

The angle θ has measure 1 radian.

Definition (Central Angle). A central angle in a circle is an angle whose vertex is at the center of the circle.

Definition (Subtended). If an angle intersects a curve at the points P and Q, then the segment of the curve P Q is said to subtend the angle.

So we say the measure of a central angle that is subtended by an arc whose length is the radius of the circle is one radian.

Property (Angle Conversions).

  1. 360 ◦^ ≡ 2 π.
  2. 1 ◦^ ≡ 60 ′.
  3. 1 ′^ ≡ 60”

Example 3. Convert

  1. 50 ◦^ to radians
  2. 1 to degrees.

HW 7.1: 1-23 odd

Example 4. Evaluate the trig functions at θ if

θ 3

4

5

Example 5. Let the point (2, 3) be on the terminal side of the angle θ. Find sin θ and tan θ.

θ 2

3

r

When using the calculator to evaluate a trig function at an angle, be sure that your calculator is set to degrees or radians as appropriate.

Example 6.

  1. sin 30◦^ = 2. sin π 4 =

Example 7. (HW 10) If (3, 5) is a point on the terminal side of an angle find sin, cos and tan of the angle.

Example 8. (HW 26) Find sin, cos and tan of the angle 19. 2 ◦

Example 9. (HW 30) Find sin, cos and tan of the angle 0. 385

Example 10. (HW 32) Evaluate the expression

2 .84 cos 73. 4 ◦^ − 3 .83 tan 36. 2 ◦

Example 11. (HW 34) Evaluate the expression

11 .2 tan 5 + 15.3 cos 3

HW 7.2: 11,15,17-35 odd

Section 3

Sometimes we know the ratio of two sides of a right triangle and want to find the corresponding angle. To do this we use the inverse trig functions. Note: All angles will be in the first quadrant for this chapter. The inverse function of sin is arcsin or sin−^1. So if sin θ = A then arcsin A = θ or sin−^1 A = θ. Note:

(sin θ)−^1 =

sin θ

= sin−^1 θ.

Example 12. Given the triangle below find θ.

θ √ 3

1 2

Example 13. (HW 2) Find the acute angle (in degrees) if tan D = 1. 53.

Example 14. (HW 8) Without using the calculator find sin, cos and tan of angle A if

cot A =

Example 15. (HW 10) Evaluate arcsin 0. 635.

Example 16. (HW 14) Evaluate cos−^1 0. 229.

HW 7.3: 1-15 odd

Example 21. Express the following as a function of the complementary angle:

HW 22 cos 73◦^ =

HW 24 sec 85. 6 ◦^ =

HW 7.4: 1,5,9,11,15,19,21-29 odd

Section 5

In this section we will look at some applications of right triangles. First let us note that given:

x

r y

sin θ =

y r cos θ =

x r

y = r sin θ x = r cos θ

We need another definition.

Definition (Angle of Elevation(Depression)). The angle of elevation(depression) is the angle between the line of sight and the horizontal.

observer

horizontal

line of sight

angle of elevation

observer

horizontal

line of sight

angle of depression

Example 22. HW 2,8,

HW 7.5: 1,5,9,11,13,19,21,25,27,29,

So in 2-dimensional space a vector is an ordered pair. In 3-dimensional space vectors are represented as ordered triples: (2, 3 , 4). In general a vector in n-dimensional space is represented by an ordered n-tuple: (v 1 , v 2 ,... , vn). To describe a moving rotating sphere you need 9-dimensional space:

3 dimensions for position 3 dimensions for velocity 3 dimensions for rotation

Can we add vectors? Yes. So how do we do it? Let us start with the easiest (non-trivial) case. Let vectors ~u and ~v be perpendicular. Let ~r be their sum: ~r = ~u + ~v. We call ~r the resultant. If ~u = (x, 0) and ~v = (0, y), then ~r is given by

|~r| =

x^2 + y^2

θ = tan−^1 y x

x

y (^) (x, y) ~r

~u

~v θ

Example 24. Let ~u = (3, 0) and ~v = (0, 4). Find ~u + ~v.

What do we do if the vectors are not perpendicular? To do this we first need to learn how to break a vector into components. Suppose we have a vector ~v. It can be viewed as consisting of two components: an x-component, vx, and a y-component, vy. So ~v = (vx, vy ).

~v

vx

vy

So if we are given a vector ~v whose magnitude is v = |~v| and whose angle is θ, then

cos θ =

vx v

sin θ =

vy v

vx = v cos θ vy = v sin θ

Example 25. Let ~v = 2∠ 30 ◦. Break ~v into components.

Now to add ~u and ~v we break ~u and ~v into components. Add the x components together and add the y components together. The resultant vector ~r = ~u + ~v will be ~r = (ux + vx, uy + vy ).

Example 26. Find ~r = ~u + ~v, if ~u = 2∠ 40 ◦^ and ~v = 3∠ 20 ◦.

HW 7.6: 7, 9, 11, 15, 17, 19, 21, 23, 25

Section 7

Example 27. HW 2,4,6,12,

HW 7.7: 1,3,5,11,