1. ## Simplify cosh(arsinh4)

Simplify cosh(arsinh4)

How would I got about simplifying this?

Thanks

2. 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

3. perhaps simpler:

$cosh^2(x)- sinh^2(x)= 1$ so $cosh(x)= \sqrt{1+ sinh^2(x)}$. With x= arcsinh(4), that is $cosh(arcsinh(4))= \sqrt{1+ sinh^2(arcsinh(4))}= \sqrt{1+ 16}= \sqrt{17}$

4. But my solution is 8

5. Originally Posted by Amer
But my solution is 8
Sorry, but $\sqrt{17}$ is the correct answer.

Originally Posted by Amer
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 ) Mr F says: That red minus should be a plus.

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

6. Hi, thanks for all the help everybody but I have a query with HallsOfIvy

$

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

$

cosh(arcsinh(4))= \sqrt{1+ sinh(4)}
$

Is that right?

Thanks

p.s. HallsOfIvy from physicsforums.com?

7. You're misunderstanding the statement:

$\sinh^2 ( \text{arcsinh} (4)) \equiv [\sinh(\text{arcsinh}(4))]^2 \equiv 4^2$.

8. Originally Posted by thomas49th
Hi, thanks for all the help everybody but I have a query with HallsOfIvy

$

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

$

cosh(arcsinh(4))= \sqrt{1+ sinh(4)}
$

Is that right?
No, it is not right. " $sinh^2(x)$ means sinh(x)*sinh(x), not sinh(sinh(x)).

Thanks

p.s. HallsOfIvy from physicsforums.com?
I know that guy- he's a troublemaker!

9. ahh yes. cool. I see

Thanks