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# Inverse of hyperbolic function when the power isn't one.

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trying to find t where 6=sinh^2/3(4t), i'm struggling to understand the inverse of sinh when the power isn't one

Guest Apr 5, 2017
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#1
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Solve for t:
6 = 4 t (sinh(x))^(1/3)^2

4 t (sinh(x))^(1/3)^2 = 4 t sinh^(2/3)(x):
6 = 4 t sinh^(2/3)(x)

6 = 4 t sinh^(2/3)(x) is equivalent to 4 t sinh^(2/3)(x) = 6:
4 t sinh^(2/3)(x) = 6

Divide both sides by 4 sinh^(2/3)(x):
Answer: | t = 3/(2 sinh^(2/3)(x))

Guest Apr 5, 2017
#2
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Can you first raise both sides to the 3/2 power...then take the sinh inverse of both sides?

Like this:

$$6=\sinh^{\frac23}(4t) \\~\\ \\~\\ 6^{\frac32}=\sinh^{\frac23*\frac32}(4t) \\~\\ 6\sqrt{6}=\sinh(4t) \\~\\ \sinh^{-1}(6\sqrt{6})=4t \\~\\ \frac{\sinh^{-1}(6\sqrt{6})}{4}=t \\~\\ t \approx 0.845 \text{ radians} \\ t \approx 48.443 \text{ degrees}$$

Assuming that was the question.

It took me awhile to check this because in most calculators you need to input it like this:

[sinh(4(0.845))]^(2/3)

*edited to clarify that it is 0.845 radians, and to add degrees.*

hectictar  Apr 5, 2017
edited by hectictar  Apr 5, 2017

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