Proving Trig Problem: sec (2x) - 1 = sin^2 x / 2 sec (2x)

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Discussion Overview

The discussion revolves around proving the trigonometric identity sec(2x) - 1 = sin²(x) / (2 sec(2x)). Participants explore various approaches to manipulate the equation using trigonometric identities and relationships.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in proving the identity and requests assistance, noting a lack of familiarity with terms related to sec(2x).
  • Another participant provides a transformation for sec(2x) using the cosine double angle identity, suggesting that it can be expressed as 1/(cos²(x) - sin²(x)).
  • A different participant proposes starting with the left side of the equation and applying the double angle identity to express it in terms of sine and cosine.
  • A later reply indicates a struggle with the fractions involved in the problem, suggesting confusion in the manipulation of the terms.

Areas of Agreement / Disagreement

Participants do not reach a consensus on a specific method for proving the identity, and multiple approaches are suggested without resolution.

Contextual Notes

There are limitations regarding the assumptions made about the identities and transformations, as well as potential dependencies on specific definitions of trigonometric functions.

Sixlets
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I'm stuck :/ I have to prove the following:
sec (2x) - 1 = sin^2 x
____________
2 sec (2x)

Unfortunately, I don't know any terms that are equal to sec (2x). Any help would be awesome!
 
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[tex]sec(2x)=\frac{1}{cos(2x)}=\frac{1}{cos^2x-sin^2x}[/tex]

Also you can change the denominator by substituting in for [tex]cos^2x[/tex] or [tex]sin^2x[/tex].
 
I think I would start with the left side of the eq. and use the double angle identity to get it in terms of sin/cos.
 
I'll try that again. I kept getting way off track, but I probably just got confused with all the fractions haha. Thanks everyone!
 

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