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

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SUMMARY

The discussion focuses on proving the reciprocal identity (secx + 1)/(sin2x) = (tanx)/(2cosx - 2cos2x). Participants analyze both sides of the equation, simplifying expressions involving trigonometric identities such as secant, tangent, and sine. Key steps include manipulating the left side to (1/2sinxcosx) and the right side to (2sin2xcosx), ultimately confirming the identity through algebraic simplification. The consensus emphasizes the importance of reversing steps for clarity and rigor in proofs.

PREREQUISITES
  • Understanding of trigonometric identities, including secant, tangent, and sine functions.
  • Familiarity with algebraic manipulation of fractions and expressions.
  • Knowledge of the double angle formulas for sine and cosine.
  • Ability to perform simplifications and factorizations in trigonometric equations.
NEXT STEPS
  • Study the derivation and applications of double angle formulas for sine and cosine.
  • Learn more about algebraic techniques for simplifying trigonometric expressions.
  • Practice proving trigonometric identities using various methods, including working backwards.
  • Explore the implications of tautologies in mathematical proofs and their role in establishing identities.
USEFUL FOR

Students studying trigonometry, mathematics educators, and anyone looking to enhance their skills in proving trigonometric identities.

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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