Rahul's question at Yahoo Answers regarding hyperbolic and circular trigonometry

[MarkFL]

Here is the question:

If coshx=sec (theta) then prove that tanh^2(x/2)= tan^2 (theta/2)?


tanh^2 (x/2)=coshx-1/.coshx+1
this is how the problem starts i can't understan how did they get this.. can anybody help me out...

I have posted a link there to this thread so the OP can see my work.

 

[MarkFL]

Hello Rahul,

Let's begin with:

\(\displaystyle \tanh\left(\frac{x}{2} \right)=\frac{\sinh\left(\frac{x}{2} \right)}{\cosh\left(\frac{x}{2} \right)}\)

Next, we can employ the definitions:

\(\displaystyle \sinh(u)\equiv\frac{e^u-e^{-u}}{2}\)

\(\displaystyle \cosh(u)\equiv\frac{e^u+e^{-u}}{2}\)

To obtain:

\(\displaystyle \tanh\left(\frac{x}{2} \right)=\frac{\dfrac{e^{\frac{x}{2}}-e^{-\frac{x}{2}}}{2}}{\dfrac{e^{\frac{x}{2}}+e^{-\frac{x}{2}}}{2}}\)

Multiplying the right side by \(\displaystyle 1=\frac{2}{2}\) we obtain:

\(\displaystyle \tanh\left(\frac{x}{2} \right)=\frac{e^{\frac{x}{2}}-e^{-\frac{x}{2}}}{e^{\frac{x}{2}}+e^{-\frac{x}{2}}}\)

Squaring both sides, we obtain:

\(\displaystyle \tanh^2\left(\frac{x}{2} \right)=\frac{e^{x}+e^{-x}-2}{e^{x}+e^{-x}+2}\)

Dividing each term in the numerator and denominator on the right by $2$ we may write:

\(\displaystyle \tanh^2\left(\frac{x}{2} \right)=\frac{\dfrac{e^{x}+e^{-x}}{2}-1}{\dfrac{e^{x}+e^{-x}}{2}+1}\)

Using the definition for the hyperbolic cosine function, we obtain:

\(\displaystyle \tanh^2\left(\frac{x}{2} \right)=\frac{\cosh(x)-1}{\cosh(x)+1}\)

Now, we are given \(\displaystyle \cosh(x)=\sec(\theta)\), and so we may write:

\(\displaystyle \tanh^2\left(\frac{x}{2} \right)=\frac{\sec(\theta)-1}{\sec(\theta)+1}\)

Multiply the right side by \(\displaystyle 1=\frac{\cos(\theta)}{\cos(\theta)}\) to get:

\(\displaystyle \tanh^2\left(\frac{x}{2} \right)=\frac{1-\cos(\theta)}{1+\cos(\theta)}\)

In the numerator use the double-angle identity for cosine \(\displaystyle \cos(2u)=1-2\sin^2(u)\) and in the denominator use the double-angle identity for cosine \(\displaystyle \cos(2u)=2\cos^2(u)-1\) to get:

\(\displaystyle \tanh^2\left(\frac{x}{2} \right)=\frac{1-\left(1-2\sin^2\left(\dfrac{\theta}{2} \right) \right)}{1+\left(2\cos^2\left(\dfrac{\theta}{2} \right)-1 \right)}=\frac{2\sin^2\left(\dfrac{\theta}{2} \right)}{2\cos^2\left(\dfrac{\theta}{2} \right)}\)

Hence, we have:

\(\displaystyle \tanh^2\left(\frac{x}{2} \right)=\tan^2\left(\dfrac{\theta}{2} \right)\)

And this is what we were asked to show.