help me please
if t=tanh(1/2 x),
prove sinhx=2t/(1-t^2)

Question

help me please
if t=tanh(1/2 x), 
prove sinhx=2t/(1-t^2)

anonymous · Answer

thank you, so here's the question, isnt tanhx=$$\frac{ e^x - e^{-x} }{ e^x + e ^{-x} }$$

solomonzelman · Answer

$\color{#000000 }{ \displaystyle t=\frac{e^{(x/2)}-e^{-(x/2)}}{e^{(x/2)}+e^{-(x/2)}}  }$

$\color{#000000 }{ \displaystyle t=\frac{e^{x}-1}{e^{x}+1}  }$

We can solve for $e^x$ in terms of $t$.

$\color{#000000 }{ \displaystyle t=\frac{e^{x}-1}{e^{x}+1}  }$

$\color{#000000 }{ \displaystyle e^x=\frac{1+t}{1-t}  }$

$\color{#000000 }{ \displaystyle t=\frac{e^{x}-1}{e^{x}+1} \quad \Longrightarrow \quad  \frac{1+t}{1-t}=e^{x}\quad  \Longleftrightarrow   \quad  \frac{1-t}{1+t}=e^{-x} }$

then,

$\color{#000000 }{ \displaystyle \sinh(x)=\frac{e^{x}-e^{-x}}{2} =\frac{2t}{1-t^2} }$

$\color{#000000 }{ \displaystyle \frac{\dfrac{1+t}{1-t}-\dfrac{1-t}{1+t}}{2} =\frac{2t}{1-t^2} }$

solomonzelman · Answer

Sorry for making that erro, but nice that you caught it.

solomonzelman · Answer

and yes, it becomes indeed equivalent when algerba correctly applied.

anonymous · Answer

Ok, tq, I'll try first