Simplify cosh(arsinh4)
How would I got about simplifying this?
Thanks
Ok
arsinh4 = ln(4+ 17^1/2 )
cosh (arsinh4)
cosh (ln (4 + 17^1/2 ) = 1/2(e^ln(4 + 17^1/2 ) + e^-ln(4 + 17^1/2 ))
cosh (ln (4 + 17^1/2 ) = 1/2( 4+17^1/2 - 1/(4 + 17^1/2 )
cosh (ln (4 + 17^1/2 ) = 1/2 ( ((4+17^1/2)^(2) -1))/(4 + 17^1/2)
= ( 16 + 8(17)^1/2 + 17 -1 )/(4 + 17^1/2 )
= ( 32 + 8(17)^1/2 )/(4+17^1/2)
multiply with (-4+17^1/2)/(-4+17^1/2)
= ( 32(17)^1/2 + 8(17) - 4(32) - 4*8(17)^1/2 )/(17 - 16 )
= (136 - 128 )/1 =
you continue
perhaps simpler:
$\displaystyle cosh^2(x)- sinh^2(x)= 1$ so $\displaystyle cosh(x)= \sqrt{1+ sinh^2(x)}$. With x= arcsinh(4), that is $\displaystyle cosh(arcsinh(4))= \sqrt{1+ sinh^2(arcsinh(4))}= \sqrt{1+ 16}= \sqrt{17}$
Hi, thanks for all the help everybody but I have a query with HallsOfIvy
$\displaystyle
cosh(arcsinh(4))= \sqrt{1+ sinh^2(arcsinh(4))}= \sqrt{1+ 16}= \sqrt{17}
$
you have sinh^2 and arcsinh, but how does 1 arcsinh cancel with sinh^2, I thought It would only cancel with one of the sinh giving you
$\displaystyle
cosh(arcsinh(4))= \sqrt{1+ sinh(4)}
$
Is that right?
Thanks
p.s. HallsOfIvy from physicsforums.com?