Finding Inverses of Trig Functions Without Prior Knowledge

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SUMMARY

This discussion focuses on determining the inverses of trigonometric functions without prior knowledge of their properties. The key formula presented is \(\sec^2\theta = 1 + \tan^2\theta\), leading to the derivation of \(\sec\theta = \sqrt{1 + \tan^2\theta}\). The process for finding the inverse is outlined as follows: starting with \(x = 1 / \cos^2(y)\), it simplifies to \(y = \arccos(1/\sqrt{x})\). Additionally, understanding the domain and range of these inverses is emphasized as crucial for proper application.

PREREQUISITES
  • Understanding of basic trigonometric identities, specifically \(\sec\) and \(\tan\).
  • Familiarity with inverse trigonometric functions, particularly \(\arccos\).
  • Knowledge of domain and range concepts in mathematics.
  • Basic algebraic manipulation skills for solving equations.
NEXT STEPS
  • Study the properties of trigonometric identities, focusing on \(\sec\) and \(\tan\).
  • Learn about the domain and range of inverse trigonometric functions.
  • Explore the derivation and applications of other inverse trigonometric functions such as \(\arcsin\) and \(\arctan\).
  • Practice solving equations involving trigonometric functions and their inverses.
USEFUL FOR

Students and educators in mathematics, particularly those focusing on trigonometry and its applications, as well as anyone seeking to deepen their understanding of inverse trigonometric functions.

bill nye scienceguy!
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and how would i go about working out the inverses of trig functions if i didnt already know them?
 
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First off: [tex]\sec^2\theta=1+\tan^2\theta[/tex]
=

[tex]\sec\theta={\sqrt{1 +\tan^2\theta}}[/tex]
 
Last edited:
bill nye scienceguy! said:
and how would i go about working out the inverses of trig functions if i didnt already know them?

Vague question but I suspect you mean do the following.

x = 1 / cos^2(y)

sqrt(x) = 1/cos(y)

1/sqrt(x) = cos(y)

y = arccos(1/sqrt(x))

Now that's only half the solution, the most important part is working out the domain and range for which this inverse makes sense.
 

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