Can someone kindly explain? Thanks

  • Thread starter Thread starter whatisphysics
  • Start date Start date
  • Tags Tags
    Explain Thanks
Click For Summary
SUMMARY

The discussion centers on the mathematical relationship between the arcsecant and arctangent functions, specifically how arcsec(α) can be expressed in terms of arctan(β). The user illustrates this by constructing a right triangle with a hypotenuse labeled as x and a base of 4, leading to the equations α = sec-1(x/4) and β = tan-1(4/sqrt(x2 - 16)). The conclusion drawn is that the angles α and β are related through the properties of right triangles and trigonometric identities.

PREREQUISITES
  • Understanding of trigonometric functions, specifically arcsecant and arctangent.
  • Knowledge of right triangle properties and the Pythagorean theorem.
  • Familiarity with inverse trigonometric functions and their relationships.
  • Basic algebra skills for manipulating equations involving square roots.
NEXT STEPS
  • Study the relationship between inverse trigonometric functions, focusing on arcsec and arctan.
  • Learn how to derive trigonometric identities using right triangles.
  • Explore the applications of inverse trigonometric functions in solving real-world problems.
  • Practice problems involving the conversion between arcsec and arctan using various triangle configurations.
USEFUL FOR

Students studying trigonometry, mathematics educators, and anyone seeking to deepen their understanding of inverse trigonometric functions and their applications in geometry.

whatisphysics
Messages
30
Reaction score
0

Homework Statement


I have attached my question. I don't understand how A can turn into B? (the highlighted parts)


Homework Equations





The Attempt at a Solution


I finished the whole question, but I don't understand why arcsec turns into arctan...
 

Attachments

  • calculus.PNG
    calculus.PNG
    4.1 KB · Views: 576
Physics news on Phys.org
Draw a right triangle with acute angle α at the left and another acute angle β at the upper right. Label the hypotenuse as x and the base as 4. Then the altitude is sqrt(x2 - 16).

Then x/4 = sec(α), so α = sec-1(x/4).

And 4/sqrt(x2 - 16) = tan(β), so β = tan-1( 4/sqrt(x2 - 16)).

What can you say about the angles α and β?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K