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

#1**0 **

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**0 **

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