Prove the given hyperbolic trigonometry equation

[chwala]

Homework Statement

See attached.

Relevant Equations

trigonometry- Further Maths

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.

 

[nuuskur]

Typos in your exponents. Right now you have zero at the very end. Explain the 5th equality from the bottom.

 

[chwala]

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.

 

[pasmith]

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 [itex]\sinh^2 \frac x2[/itex], so start from [itex]\cosh 2x - 1 = 2\sinh^2 x[/itex]. 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 [tex]\begin{split}
(e^{x/2} - e^{-x/2})^2 &= (e^{x/2})^2 - 2(e^{x/2})(e^{-x/2}) + (e^{-x/2})^2 \\
&= e^{x} - 2 + e^{-x} \\
&= 2\cosh x - 2.\end{split}[/tex]

 

[chwala]

You need to end up with [itex]\sinh^2 \frac x2[/itex], so start from [itex]\cosh 2x - 1 = 2\sinh^2 x[/itex]. 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 [tex]\begin{split}
(e^{x/2} - e^{-x/2})^2 &= (e^{x/2})^2 - 2(e^{x/2})(e^{-x/2}) + (e^{-x/2})^2 \\
&= e^{x} - 2 + e^{-x} \\
&= 2\cosh x - 2.\end{split}[/tex]

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]##