Please see the attached file.When does the graph of

  • Context: Undergrad 
  • Thread starter Thread starter jlawson
  • Start date Start date
  • Tags Tags
    Graph
Click For Summary
SUMMARY

The discussion focuses on solving the equation 5/12 = 5/6*sin(10t - 0.927) to find the values of t where the graph intersects specific points. The user identifies one solution using arcsin(y) within the range (-π/2, +π/2] and acknowledges a second solution, π - arcsin(y), which applies to the range (+π/2, +π]. The red points on the graph correspond to these additional solutions. The conversation emphasizes the importance of understanding the properties of the sine function and its inverse for accurate calculations.

PREREQUISITES
  • Understanding of trigonometric functions, specifically sine and its properties.
  • Knowledge of inverse trigonometric functions, particularly arcsin.
  • Familiarity with graphing techniques for periodic functions.
  • Basic algebra skills for solving equations involving trigonometric identities.
NEXT STEPS
  • Study the properties of the sine function and its periodicity.
  • Learn how to apply inverse trigonometric functions in solving equations.
  • Explore graphing techniques for trigonometric functions to visualize intersections.
  • Investigate the implications of multiple solutions in trigonometric equations.
USEFUL FOR

Students and educators in mathematics, particularly those studying trigonometry, as well as anyone involved in solving equations involving periodic functions.

jlawson
Messages
5
Reaction score
0
Please see the attached file.

When does the graph of x(t)=5/6*sin⁡(10t-0.927) equal 5/12

5/12=5/6*sin⁡(10t-0.927)

One answer, attached, is when the graph goes through the green points but I can't seem to figure out how to calculate the values of the red points.

Any help would be greatly appreciated.
 

Attachments

  • Untitled.jpg
    Untitled.jpg
    32.5 KB · Views: 491
Physics news on Phys.org


The value of the function arcsin(y) is defined to be the solution of sin(x) = y that lies in the range (-π/2, +π/2]. There is a second solution π-arcsin(y) in the range (+π/2, +π]. This will form the basis of the 'red' points.
 


Oh, yep, duh.

Thank you very much for your help.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
10
Views
3K
Replies
7
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 115 ·
4
Replies
115
Views
9K
  • · Replies 9 ·
Replies
9
Views
3K