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
Hi, thanks for all the help everybody but I have a query with HallsOfIvy
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
Is that right?
Thanks
p.s. HallsOfIvy from physicsforums.com?