Prove the given hyperbolic trigonometry equation

chwala
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Homework Statement
See attached.
Relevant Equations
trigonometry- Further Maths
1715772024268.png


I have,

Using ##\ cosh 2x = 2 \cosh^2 x - 1##

##\cosh x = 2 \cosh^2\dfrac{x}{2} -1##

Therefore,

##\cosh x -1 = 2 \cosh^2\dfrac{x}{2} -1 - 1##

##\cosh x -1 = 2 \cosh^2\dfrac{x}{2} -2##

##=2\left[ \cosh^2 \dfrac{x}{2} -1 \right]##

##= 2\left[\left(\dfrac{(e^\frac{x}{2} + e^\frac{-x}{2})}{2}\right)^2-1\right] ##

##= 2\left[\dfrac{(e^\frac{x}{2} + e^\frac{-x}{2})^2}{4}-1\right] ##

##= 2\left[\dfrac{(e^\frac{x}{2} + e^\frac{-x}{2})^2-4}{4}\right] ##

##= 2\left[\dfrac{(e^\frac{x}{2} + e^\frac{-x}{2})^2-4}{4}\right] ##

##= \dfrac{1}{2}\left[ e^x +e^{-x}-2 \right]##

##= \dfrac{1}{2}\left[ e^{0.5x}e^{0.5x} +e^{-0.5x}e^{-0.5x}-e^{0.5x}e^{-0.5x}-e^{0.5x}e^{-0.5x} \right]##


##= \dfrac{1}{2}\left[ e^{0.5x}(e^{0.5x}- e^{-0.5x})+ e^{-0.5x}(e^{-0.5x}-e^{0.5x}) \right]##

##= \dfrac{1}{2}\left[ e^{0.5x}(e^{0.5x}- e^{-0.5x})- e^{-0.5x}(e^{0.5x}-e^{-0.5x}) \right]##

##= \dfrac{1}{2}\left[ e^{0.5x}-e^{-0.5x}\right]^2##


I know that there could be a better straightforward approach...your insight is welcome.
 
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Typos in your exponents. Right now you have zero at the very end. Explain the 5th equality from the bottom.
 
nuuskur said:
Typos in your exponents. Right now you have zero at the very end. Explain the 5th equality from the bottom.
I'll amend the last line...the other steps are straightforward.
 
chwala said:
Homework Statement: See attached.
Relevant Equations: trigonometry- Further Maths

View attachment 345288

I have,

Using ##\ cosh 2x = 2 \cosh^2 x - 1##

You need to end up with \sinh^2 \frac x2, so start from \cosh 2x - 1 = 2\sinh^2 x. But using these identities comes very close to assuming exactly what the question is asking you to show.

In any event, the simplest approach is the straight forward \begin{split}<br /> (e^{x/2} - e^{-x/2})^2 &amp;= (e^{x/2})^2 - 2(e^{x/2})(e^{-x/2}) + (e^{-x/2})^2 \\<br /> &amp;= e^{x} - 2 + e^{-x} \\<br /> &amp;= 2\cosh x - 2.\end{split}
 
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pasmith said:
You need to end up with \sinh^2 \frac x2, so start from \cosh 2x - 1 = 2\sinh^2 x. But using these identities comes very close to assuming exactly what the question is asking you to show.

In any event, the simplest approach is the straight forward \begin{split}<br /> (e^{x/2} - e^{-x/2})^2 &amp;= (e^{x/2})^2 - 2(e^{x/2})(e^{-x/2}) + (e^{-x/2})^2 \\<br /> &amp;= e^{x} - 2 + e^{-x} \\<br /> &amp;= 2\cosh x - 2.\end{split}
Yeah using ##\sinh^2 \dfrac{x}{2}## is easier...and the appropriate approach...has only 2 lines...will post that later...

Using,

##\cosh x = 2\left(1 + \sinh^2 \left( \dfrac{x}{2}\right) \right)-1##

##\cosh x = 2+2 \sinh^2 \left(\dfrac{x}{2}\right) -1= 2 \sinh^2 \left(\dfrac{x}{2} \right)+1##

Therefore

##\cosh x - 1 = 2 \sinh^2 \left(\dfrac{x}{2}\right)##

##\cosh x - 1 = 2\left[\dfrac{(e^{0.5x} - e^{-0.5x})}{2}\right]^2 = \left[\dfrac{(e^{0.5x} - e^{-0.5x})^2}{2}\right]##
 
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