Proving Reciprocal Identities: (secx+1)/(sin2x) = (tanx)/2cosx-2cos2x

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Homework Help Overview

The discussion revolves around proving the reciprocal identities involving trigonometric functions, specifically the equation (secx+1)/(sin2x) = (tanx)/(2cosx-2cos2x). Participants are exploring various approaches to manipulate both sides of the equation to demonstrate their equality.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants attempt to simplify both sides of the equation separately, with some focusing on manipulating the left side and others on the right side. Questions arise about the validity of canceling terms and the steps taken to eliminate denominators. There is also discussion about the implications of starting with the equation to be proved.

Discussion Status

The discussion is active, with participants providing feedback on each other's attempts and suggesting alternative methods for simplification. Some guidance has been offered regarding the need for reversible steps in the proof process, and there is acknowledgment of the need to work backwards from a tautology.

Contextual Notes

Participants are navigating the complexities of trigonometric identities and the rules of algebraic manipulation. There is an emphasis on ensuring that all steps taken in the proof are valid and reversible, reflecting the constraints of the homework context.

Schaus
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Homework Statement


(secx+1)/(sin2x) = (tanx)/2cosx-2cos2x)

Homework Equations

The Attempt at a Solution


Left Side
((1+cosx)/cosx)/2sinxcosx

((1+cosx)/cosx) x (1/2sinxcosx)
cancel the a cosx from both to get
(1/2sinxcosx)
This is all I could manage with left side so I tried right side
Right Side
(sinx/cosx)/2cosx-2cos2x)
I'm stuck here. I've been trying to find something to change the denominator of the Right Side but I can't think of anything that will work. If someone could let me know where I am going wrong it would be greatly appreciated!
 
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Schaus said:
((1+cosx)/cosx) x (1/2sinxcosx)
cancel the a cosx from both to get
(1/2sinxcosx)
What happened to the 1+cos?

Rather than working each side separately, multiply out to get rid of all the denominators.
 
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I thought I could cancel a cosx, maybe I cannot. I tried to eliminate the denominator like you said. Here's what I got.
(secx+1)(2cosx-2cos2x) = (Sin2x)(tanx)

((1+cosx)/cosx)(2cosx-2cos2x)=(2sinxcosx)(sinx/cosx)
Expanding
((2cosx-2cos2x+2cos2x-2cos3x)/cosx) = 2
Moving the cosx under left side to the right side and simplifying
2cosx-2cos3x = 2cosx
Does this look right? And if so, where do I go from here?
 
Schaus said:
((2cosx-2cos2x+2cos2x-2cos3x)/cosx) = 2
Check the right hand side.
 
Woops. I think I should have gotten 2sin2xcosx
2cosx-2cos3x = 2sin2xcosx
Does this look right?
 
Schaus said:
Woops. I think I should have gotten 2sin2xcosx
2cosx-2cos3x = 2sin2xcosx
Does this look right?
Yes. Keep simplifying.
 
2cosx-2cos3x = 2sin2xcosx
2cosx(1-cos2x) = 2sin2xcosx
2cosx(sin2x) = 2sin2xcosx
2cosxsin2x = 2sin2xcosx
Does this work?
 
Schaus said:
2cosx-2cos3x = 2sin2xcosx
2cosx(1-cos2x) = 2sin2xcosx
2cosx(sin2x) = 2sin2xcosx
2cosxsin2x = 2sin2xcosx
Does this work?
Yes.
Of course, it is not strictly kosher to start with the thing to be proved and deduce a tautology. You need all the steps to be reversible. They are in this case, but it is cleaner to rewrite it in the more persuasive sequence: start with the tautology and deduce the thing to be proved.
 
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So I should start with 2sin2xcosx and work backwards? Thank you for all the help by the way.
 
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Schaus said:
So I should start with 2sin2xcosx and work backwards?
Ideally, yes.
 
  • #11
Ok, I'll try it. I'm going to have to practice these quite a bit more I think.
 

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