8-1 Simple Trigonometric Equations - · PDF file08.11.2014 · 8-1 Simple...
Transcript of 8-1 Simple Trigonometric Equations - · PDF file08.11.2014 · 8-1 Simple...
Warm Up
β’ Use your knowledge of UC to find at least one value for q.
1) sin π =1
2
2) cos π = β3
2
3) tan π = 1
β’ State as many angles as you can that are referenced by each:
1) 30Β°
2)π
3
3) 0.65 radians
Useful information to MEMORIZE: π
2β 1.57
3π
2β 4.71 2π β 6.28
8-1 Simple Trigonometric Equations
Objective: To solve simple Trigonometric Equations and apply
them
Copyright Β© by Houghton Mifflin Company, Inc.
All rights reserved. 3
-1
x
y
1 -19Ο
6 -11Ο
6 -7Ο
6 Ο
6 5Ο
6 13Ο
6 17Ο
6 25Ο
6
-Ο -2Ο -3Ο Ο 2Ο 3Ο 4Ο
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 π₯ =1
2
β’ We know that π₯ =π
6 and π₯ =
5π
6 are two
particular solutions.
β’ Since the period of sin π₯ is 2π, we can add integral multiples of 2π to get the other solutions:
β’ π₯ =π
6+ 2nπ and π₯ =
5π
6+ 2nπ When π 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 0Β° < π₯ < 360Β°for
which sinπ₯ = β0.35
Method 1: Algebraically: Step 1 Set the calculator in degree mode and use the inverse sine key
π₯ β β20.5Β°
Find the final answer(s) for the given range.
β’ 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 π₯ for which sin π₯ = β0.35 , then your answer would be:
π β πππ. π + ππππ AND
π β πππ. π + ππππ, for any integer π.
Example 2: Find the values of π₯ between
0 and 2π for which sin π₯ = 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 π₯ is positive, a Quadrant II angle also satisfies the equation.
π₯ = π β 0.6435 β 2.4981.
Final answers are: π β π. ππππ πππ π. ππππ.
If you had been asked to find ALL values of π₯ for which sin π₯ = 0.6 , then your answer would be:
π β π. ππππ + ππ π AND
π β π. ππππ + ππ π, for any integer π.
Example 1: Find the values of π₯ between
0 and 2π for which sin π₯ = 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.
β’ 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:
π cos π½ + π = π for ππ β€ π½ β€ ππππ
First apply the basic algebra rules and isolate the variable. 3 cos π + 9 = 7 3 cos π = β2
cos π = β2
3
Find the appropriate quadrant
Since cos π < 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: π β 131.8Β° or π β 228.2Β°
Another way; ignore the negative sign.
The reference angle is:
πππ β1(2
3) β 48.2π
The first solution is: π β 180π β 48.2π = 131.8π
The second solution is:
π β 180π + 48.2π = 228.2π
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. πππ π = 2 a) has 0 solution.
b) has 1 solution.
c) has 2 solutions.
d) has infinite number of solutions.
2. π‘πππ = 2 a) has 0 solution.
b) has 1 solution.
c) has 2 solutions.
d) has infinite number of solutions.
β’ Graph sine, cosine and tangent functions.
Inclination and Slope
β’ The inclination of a line is the angle π , where 0π β€ π < 180π, that is measured from the positive x-axis to the line.
Inclination and Slope
β’ The inclination of a line is the angle π , where 0π β€ π < 180π, that is measured from the positive x-axis to the line. The line at the left below has inclination 35π. The line at the right below has inclination 155π. The theorem that follows states that the slop of a nonvertical line is the tangent of its inclination.
Theorem
β’ For any line with slope π and inclination π π = tan π if π β 90π.
β’ If π = 90π, than the line has no slope. (The line is vertical.)
Example 3 to the nearest degree, find the
inclination of the line 2π₯ + 5π¦ = 15
Solution: rewrite the equation as π¦ = β2
5π₯ + 3
Slope = -2
5= tan π
π = π‘ππβ1 β2
5β β21.8π ( the reference
angle is 21.8π.)
β’ Since tan π is negative and π is positive angle, 90π < π < 180πthe inclination is 180π β21.8π β 158.2π.
β’ In section 6-7, you learned to graph conic sections whose equations have no π₯π¦ β π‘πππ. That is equation of the form.
π΄π₯2 + π΅π₯π¦ + πΆπ¦2 + π·π₯ + πΈπ¦ + πΉ = 0
β’ Where B=0. the graph at the right shows conic
section with center at the origin whose equation has an π₯π¦ β π‘πππ (π΅ β 0). Conics like this have one of their two axes inclined at an angle π to the x axis. To find this direction angle π , use the formula below.
π =π
4 if A=C
Tan 2π =π΅
π΄βπΆ if Aβ C , and 0< 2π < 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:
β’ Sec 8.1 Written exercises #1-21 odds
β’ Optional: Sec 8.2 written exercises 22-32 ALL