8-1 simple trigonometric equations - · pdf file08.11.2014 · 8-1 simple...

Warm Up

• Use your knowledge of UC to find at least one value for q.

1) sin 𝜃 =1

2

2) cos 𝜃 = −3

2

3) tan 𝜃 = 1

• State as many angles as you can that are referenced by each:

1) 30°

2)𝜋

3

3) 0.65 radians

Useful information to MEMORIZE: 𝜋

2≈ 1.57

3𝜋

2≈ 4.71 2𝜋 ≈ 6.28

8-1 Simple Trigonometric Equations

Objective: To solve simple Trigonometric Equations and apply

them

Copyright © by Houghton Mifflin Company, Inc.

All rights reserved. 3

-1

x

y

1 -19π

6 -11π

6 -7π

6 π

6 5π

6 13π

6 17π

6 25π

6

-π -2π -3π π 2π 3π 4π

All the solutions for x can be expressed in the form of

a general solution.

y = 2

1

y=sin x

There are many solutions to the

trigonometric equation sin 𝑥 =1

2

• We know that 𝑥 =𝜋

6 and 𝑥 =

5𝜋

6 are two

particular solutions.

• Since the period of sin 𝑥 is 2𝜋, we can add integral multiples of 2𝜋 to get the other solutions:

• 𝑥 =𝜋

6+ 2n𝜋 and 𝑥 =

5𝜋

6+ 2n𝜋 When 𝑛 is any

integer

Solving for angles that are not on UC

We will work through solutions algebraically and graphically.

Learning both methods will enhance your understanding of the work.

90 180 270 360

-1

1

x

y

degrees

Example 1: Find the values of 0° < 𝑥 < 360°for

which sin𝑥 = −0.35

Method 1: Algebraically: Step 1 Set the calculator in degree mode and use the inverse sine key

𝑥 ≈ −20.5°

Find the final answer(s) for the given range.

• Since the answer given by your calculator is NOT between 0 and 360 degrees, find the proper answers by using RA.

90 180 270 360

-1

1

x

y

degrees

Solving Graphically

If you had been asked to find ALL values of 𝑥 for which sin 𝑥 = −0.35 , then your answer would be:

𝒙 ≈ 𝟐𝟎𝟎. 𝟓 + 𝟑𝟔𝟎𝒏 AND

𝒙 ≈ 𝟑𝟑𝟗. 𝟓 + 𝟑𝟔𝟎𝒏, for any integer 𝑛.

Example 2: Find the values of 𝑥 between

0 and 2𝜋 for which sin 𝑥 = 0.6

Method 1: Algebraically: Step 1 Set the calculator in radian mode and use the inverse sine key

Step 2: Determine the proper quadrant

0.6435 is the reference angle for other solutions.

Since sin 𝑥 is positive, a Quadrant II angle also satisfies the equation.

𝑥 = 𝜋 − 0.6435 ≈ 2.4981.

Final answers are: 𝒙 ≈ 𝟎. 𝟔𝟒𝟑𝟓 𝒂𝒏𝒅 𝟐. 𝟒𝟗𝟖𝟏.

If you had been asked to find ALL values of 𝑥 for which sin 𝑥 = 0.6 , then your answer would be:

𝒙 ≈ 𝟎. 𝟔𝟒𝟑𝟓 + 𝟐𝝅𝒏 AND

𝒙 ≈ 𝟐. 𝟒𝟗𝟖𝟏 + 𝟐𝝅𝒏, for any integer 𝑛.

Example 1: Find the values of 𝑥 between

0 and 2𝜋 for which sin 𝑥 = 0.6

Method 2: Graphically: Step 1 Set the calculator in radian mode.

Use your Knowledge of trig functions to choose an appropriate window

Use the intersect Key once more for the second point of intersection. i.e solution.

When you use the graphing method, you can easily see there is more than one solution.

When using the graphing method, it might take a while to set the window properly.

The algebraic method is quicker, however, you have make sure to look for a possible second answer.

• To solve an equation involving a single trigonometric function, we first transform the equation so that the function is alone on one side of the equals sign. Then we follow the same procedure used in Example 1.

Example 2

To the nearest tenth degree, solve:

𝟑 cos 𝜽 + 𝟗 = 𝟕 for 𝟎𝒐 ≤ 𝜽 ≤ 𝟑𝟔𝟎𝒐

First apply the basic algebra rules and isolate the variable. 3 cos 𝜃 + 9 = 7 3 cos 𝜃 = −2

cos 𝜃 = −2

3

Find the appropriate quadrant

Since cos 𝜃 < 0 , the final answers are in the QII and QIII.

Use your knowledge of reference angle to find the second answer:

The final answers are: 𝜃 ≈ 131.8° or 𝜃 ≈ 228.2°

Another way; ignore the negative sign.

The reference angle is:

𝑐𝑜𝑠−1(2

3) ≈ 48.2𝑜

The first solution is: 𝜃 ≈ 180𝑜 − 48.2𝑜 = 131.8𝑜

The second solution is:

𝜃 ≈ 180𝑜 + 48.2𝑜 = 228.2𝑜

Graphing Calculator:

Although this is a reasonable window to start with, it does not capture the graph. So change Ymin and Ymax.

Warm Up Day 2;

1. 𝑐𝑜𝑠𝜃 = 2 a) has 0 solution.

b) has 1 solution.

c) has 2 solutions.

d) has infinite number of solutions.

2. 𝑡𝑎𝑛𝜃 = 2 a) has 0 solution.

b) has 1 solution.

c) has 2 solutions.

d) has infinite number of solutions.

• Graph sine, cosine and tangent functions.

Inclination and Slope

• The inclination of a line is the angle 𝑎 , where 0𝑜 ≤ 𝑎 < 180𝑜, that is measured from the positive x-axis to the line.

Inclination and Slope

• The inclination of a line is the angle 𝑎 , where 0𝑜 ≤ 𝑎 < 180𝑜, that is measured from the positive x-axis to the line. The line at the left below has inclination 35𝑜. The line at the right below has inclination 155𝑜. The theorem that follows states that the slop of a nonvertical line is the tangent of its inclination.

Theorem

• For any line with slope 𝑚 and inclination 𝑎 𝑚 = tan 𝑎 if 𝑎 ≠ 90𝑜.

• If 𝑎 = 90𝑜, than the line has no slope. (The line is vertical.)

Example 3 to the nearest degree, find the

inclination of the line 2𝑥 + 5𝑦 = 15

Solution: rewrite the equation as 𝑦 = −2

5𝑥 + 3

Slope = -2

5= tan 𝑎

𝑎 = 𝑡𝑎𝑛−1 −2

5≈ −21.8𝑜 ( the reference

angle is 21.8𝑜.)

• Since tan 𝑎 is negative and 𝑎 is positive angle, 90𝑜 < 𝑎 < 180𝑜the inclination is 180𝑜 −21.8𝑜 ≈ 158.2𝑜.

• In section 6-7, you learned to graph conic sections whose equations have no 𝑥𝑦 − 𝑡𝑒𝑟𝑚. That is equation of the form.

𝐴𝑥2 + 𝐵𝑥𝑦 + 𝐶𝑦2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0

• Where B=0. the graph at the right shows conic

section with center at the origin whose equation has an 𝑥𝑦 − 𝑡𝑒𝑟𝑚 (𝐵 ≠ 0). Conics like this have one of their two axes inclined at an angle 𝑎 to the x axis. To find this direction angle 𝑎 , use the formula below.

𝑎 =𝜋

4 if A=C

Tan 2𝑎 =𝐵

𝐴−𝐶 if A≠C , and 0< 2𝑎 < 𝜋

The direction angle a is useful in finding the equation of the axes of these conic sections. This is shown in method 1 of example 4 on the next page

Homework:

• Sec 8.1 Written exercises #1-21 odds

• Optional: Sec 8.2 written exercises 22-32 ALL