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

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The discussion revolves around proving the equation sin4x/(1-cos4x) * (1-cos2x)/cos2x = tan x. Participants share various trigonometric identities and simplifications to manipulate the left side of the equation. Key steps include using the identity sin4x = 2sin2xcos2x and recognizing that cos4x can be expressed in terms of sin2x. The conversation emphasizes the importance of substituting identities to simplify the expression effectively. Ultimately, the group arrives at a solution that confirms the original equation.
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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
\frac{sin4x}{1-cos4x} * \frac{1-cos2x}{cos2x}

remember that sin4x=2sin2xcos2x, 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 cos2A=1-2sin^2A 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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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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