Help with Proof: sin4x/(1-cos4x) * (1-cos2x)/cos2x = tan x

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Homework Help Overview

The discussion revolves around proving the identity sin4x/(1-cos4x) * (1-cos2x)/cos2x = tan x, which involves trigonometric identities and simplifications.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore various trigonometric identities, such as sin(2x+2x) and the relationships between sin and cos functions. There are attempts to simplify the expression by substituting known identities and questioning the validity of certain steps.

Discussion Status

Some participants have provided guidance by suggesting substitutions and identities that could simplify the problem. There is an ongoing exploration of the expressions involved, with no explicit consensus reached yet.

Contextual Notes

There are indications of missing information and assumptions that may affect the approach to the problem, as noted by participants questioning the completeness of the original post's attempt.

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Homework Statement



sin4x/(1-cos4x) * (1-cos2x)/cos2x = tan x

Homework Equations





The Attempt at a Solution


 
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Well, gosh, there seems to be some things missing! Do you really think saying "I don't feel like making any attempt at all" is a good way to convince people to help you?
 
HallsofIvy said:
Well, gosh, there seems to be some things missing! Do you really think saying "I don't feel like making any attempt at all" is a good way to convince people to help you?

I have tried sin4x=sin(2x+2x)=2sin2xcos2x

(2sin2x/(1-cos4x))*1-cos2x

(2sin2x-2sin2xcos2x)/(1-cos4x)

2sin2x(1-cos2X)/1-cos4x

cos2x=1-2sin^2x

cos4x= cos^2 2x + sin^2 2x

can I do sin2x=sin(x+x)=sinxcosx+cosxsinx=sinx(2cosx)

I end up with 4sinxcosx=tanx?
 
kathyjoan said:
I have tried sin4x=sin(2x+2x)=2sin2xcos2x

(2sin2x/(1-cos4x))*1-cos2x

(2sin2x-2sin2xcos2x)/(1-cos4x)

2sin2x(1-cos2X)/1-cos4x

cos2x=1-2sin^2x

cos4x= cos^2 2x + sin^2 2x

can I do sin2x=sin(x+x)=sinxcosx+cosxsinx=sinx(2cosx)

I end up with 4sinxcosx=tanx? not great
 
Let me start you off on an easier path
[tex]\frac{sin4x}{1-cos4x} * \frac{1-cos2x}{cos2x}[/tex]

remember that [itex]sin4x=2sin2xcos2x[/itex], you replace sin4x by that identity...will anything there cancel out and make the expression simpler to prove?

EDIT:2sin2x(1-cos2X)/1-cos4x

you're nearly there actually...remember [itex]cos2A=1-2sin^2A[/itex] if A=2x then you'll have an identity for cos4x...use it and you'll get it out
 
Okay

I am left with 2sin2x(1-cos2x)/-2sin^2 2x
wow! okay then cos 2x=1-2sin^2x
2sin^2x/-sin2X
2sin^2x/-2sinxcosx=sin/cos wha la Thanks so very much!
 

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