CalculusMan
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#1
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#1
Prove that cosh 2x = 2 cosh2 x - 1

My working:

cosh x = 1/2(ex + e-x)
cosh2 x = 1/4(e2x + 2e0 + e-2x)
2cosh2 x = 1/2(e2x + e-2x + 2)
2cosh2 x = 1/2(e2x + e-2x) + 1
2cosh2 x - 1 = 1/2(e2x + e-2x)
= cosh 2x

Is this enough? Are there any other ways to prove it?
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generalebriety
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#2
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You could note that coshx = cos(ix), and thus the identity cos2a = 2cos2a - 1 holds.

Note though that sinhx = -isin(ix), and so cosh2x = cos2(ix) = 1 - sin2(ix) = 1 + (-i)2sin2(ix) = 1 + sinh2x. So it follows that cosh2x - sinh2x = 1, and also that cosh2x = cos2(ix) = cos2(ix) - sin2(ix) = cosh2x + sinh2x.

This can be verified using your method, of course:

We know that coshx = 1/2(ex + e-x), so cosh2x = 1/2(e2x + e-2x).
cosh2x = 1/4(e2x + e-2x + 2)
sinh2x = 1/4(e2x + e-2x - 2)
Adding,
cosh2x + sinh2x = 1/2(e2x + e-2x) = cosh2x.

Bit irrelevant, but meh, useful.
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Juwel
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#3
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Looks like you did alright mate. I might have started with cosh(2x) = [e^(2x) + e^(-2x)]/2 and kept working.
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CalculusMan
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#4
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(Original post by Juwel)
Looks like you did alright mate. I might have started with cosh(2x) = [e^(2x) + e^(-2x)]/2 and kept working.
I tried this way but couldn't get actually get anywhere. Any ideas on how to start it off?

Cheers guys for your help.
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generalebriety
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(Original post by CalculusMan)
I tried this way but couldn't get actually get anywhere. Any ideas on how to start it off?

Cheers guys for your help.
cosh(2x) = [e^(2x) + e^(-2x)]/2
= [(e^x)^2 + (e^-x)^2]/2
= [(e^x)^2 + (e^-x)^2 + 2 - 2]/2
= [(e^x + e^-x)^2 - 2]/2
= [(e^x + e^-x)^2]/2 - 1
= cosh^2 x - 1.
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CalculusMan
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#6
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You've just proved something which isn't true.

I've followed your working but can't seem to spot how it should be (2cosh^x - 1)
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Jan
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#7
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cosh(2x) = [e^(2x) + e^(-2x)]/2
= [(e^x)^2 + (e^-x)^2]/2
= [(e^x)^2 + (e^-x)^2 + 2 - 2]/2
= [(e^x + e^-x)^2 - 2]/2
= [(e^x + e^-x)^2]/2 - 1
= cosh^2 x - 1.
cosh(2x) = [e^(2x) + e^(-2x)]/2
[e^(2x) + e^(-2x)]/2
= [(e^x)^2 + (e^-x)^2]/2
= [(e^x)^2 + (e^-x)^2 + 2 - 2]/2
= [(e^x + e^-x)^2 - 2]/2
= 2[(e^x + e^-x)^2]/4 - 1
= 2[(e^x + e^-x)/2]^2 - 1
= 2cosh^2 x - 1
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generalebriety
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#8
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(Original post by Jan (Honza))
cosh(2x) = [e^(2x) + e^(-2x)]/2
[e^(2x) + e^(-2x)]/2
= [(e^x)^2 + (e^-x)^2]/2
= [(e^x)^2 + (e^-x)^2 + 2 - 2]/2
= [(e^x + e^-x)^2 - 2]/2
= 2[(e^x + e^-x)^2]/4 - 1
= 2[(e^x + e^-x)/2]^2 - 1
= 2cosh^2 x - 1
Oops. Looks like I've forgotten how to think. Thanks.
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Noumbi
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#9
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Where is that -2 comming from the second = signe
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1753739109
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#10
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Senh^2 (x) -2cosh^2(2)=1
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