Proving Identities: cotx-tanx=2cot 2x

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Homework Help Overview

The discussion revolves around proving the trigonometric identity cot(x) - tan(x) = 2cot(2x). Participants are exploring the relationships between cotangent and tangent functions, particularly in the context of manipulating trigonometric identities.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to express 2cot(2x) in terms of sine and cosine, questioning whether it can be represented as 2(cos(2x)/sin(2x). There are discussions about simplifying fractions and the implications of separating terms in the context of the identity.

Discussion Status

There is an ongoing exploration of how to manipulate the fractions involved in the identity. Some participants are providing guidance on how to approach the simplification, while others express confusion about the mechanics of fraction manipulation and the implications of the denominators.

Contextual Notes

Some participants indicate a need for a review of basic fraction addition and subtraction, suggesting that foundational understanding may be affecting their ability to engage with the proof effectively.

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Homework Statement


Prove that cotx-tanx=2cot 2x

Homework Equations



Is 2cot 2x the same as 2(cos2x/sin2x)

The Attempt at a Solution

 
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keishaap said:

Homework Statement


Prove that cotx-tanx=2cot 2x

Homework Equations



Is 2cot 2x the same as 2(cos2x/sin2x)

The Attempt at a Solution


Yes.
 
LCKurtz said:
Yes.

Okay so i get 2(cos^2-sin^2)/(2sinxcosx) when i divide my sins what do i do with the 2 in the denominator?
 
keishaap said:
Okay so i get 2(cos^2-sin^2)/(2sinxcosx) when i divide my sins what do i do with the 2 in the denominator?

Just write that as two fractions and compare it to the left side.
 
LCKurtz said:
Just write that as two fractions and compare it to the left side.

So its 2(cosx/2sinx)-(sinx/cosx) do the 2 go on both fractions? And i thought you couldn't separate the fractions unless tgey had the same denominator?
 
keishaap said:
Okay so i get 2(cos^2-sin^2)/(2sinxcosx) when i divide my sins what do i do with the 2 in the denominator?
You have a 2 in the numerator and a 2 in the denominator. You can cancel the 2's. Then split what you have into two fractions.
 
you would want to write (2(cos^2x-sin^2x))/(2sinxcosx) as 2 fractions. the two cancels out.
 
462chevelle said:
you would want to write (2(cos^2x-sin^2x))/(2sinxcosx) as 2 fractions. the two cancels out.

So would the fractions look like cos^2 x/sinx and sin^2x/cosx ?
 
your whole denominator needs to carry into both fractions
 
  • #10
462chevelle said:
your whole denominator needs to carry into both fractions

So what would it look like?
 
  • #11
keishaap said:
So would the fractions look like cos^2 x/sinx and sin^2x/cosx ?
You need to review basic fraction addition and subtraction. One place to start would be khanacademy.org.
 
  • #12
Mark44 said:
You need to review basic fraction addition and subtraction. One place to start would be khanacademy.org.
this.
 
  • #13
Lol ever have one of those brain farts where you think things are harder than the seem? Yeah give me a break.
 
  • #14
keishaap said:
Okay so i get 2(cos^2-sin^2)/(2sinxcosx) when i divide my sins what do i do with the 2 in the denominator?

keishaap said:
So its 2(cosx/2sinx)-(sinx/cosx) do the 2 go on both fractions? And i thought you couldn't separate the fractions unless tgey had the same denominator?

keishaap said:
Lol ever have one of those brain farts where you think things are harder than the seem? Yeah give me a break.
The questions you asked in the first two quotes above suggest that you're having trouble with the basics, rather than just a momentary lapse. Some review of how to add and multiply fractions would go a long way. If you don't have the mechanics down cold, you absolutely will not be able to complete proofs like the one you posted in this thread.

This isn't meant to be a personal criticism of you - I'm just pointing out an area with some weakness that will keep you from being successful.
 

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