How do you find sinhx and coshx when tanhx = 12/13?

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SUMMARY

The discussion focuses on finding the hyperbolic functions sinh(x) and cosh(x) given that tanh(x) = 12/13. Participants confirm that tanh(x) is defined as sinh(x)/cosh(x) and derive coth(x) = 13/12. The correct approach involves using the identity cosh²(x) - sinh²(x) = 1 to establish a system of equations. The solution requires calculating x using the inverse hyperbolic tangent function, tanh⁻¹(12/13), and then substituting back to find sinh(x) and cosh(x).

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  • Understanding of hyperbolic functions, specifically tanh, sinh, and cosh.
  • Familiarity with inverse hyperbolic functions, particularly tanh⁻¹.
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  • Basic algebra skills for solving equations involving hyperbolic functions.
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Homework Statement


find all other hyperbolic function at x for tanhx=12/13


Homework Equations


tanhx = sinhx/coshx
cothx=1/tanhx
etc...


The Attempt at a Solution


the only thing i got is cothx=13/12
all i need to know is how to find sinhx and i will be fine.
 
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Why not solve for x and then evaluate sinh(x)?
 
Bohrok said:
Why not solve for x and then evaluate sinh(x)?

so x= tanh-1(12/13) ?
then plug it in sinh(x)?

i am not sure if i get it..
 
tanh(x)=sinh(x)/cosh(x)=12/13. cosh(x)^2-sinh(x)^2=1. That's two equations in two unknowns. Can you find the solution?
 
\tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{e^{2x} - 1}{e^{2x} + 1}} = \frac{12}{13}
You can solve for x then plug it into the other hyperbolic trig functions. Or try what Dick suggested, looks like it should be easier.
 
yeah i asked a teacher today too and he explained the same thing.
that makes sense.

even though, tanhx = sinhx/coshx = 12/13 then from here i could use sinhx =12 and coshx = 13 right away..
but that's "the cheating way" i guess..

thanks for help!
 
sinh(x)=12 and cosh(x)=13 isn't only the 'cheating way', it's the wrong way. 13^2-12^2 isn't equal to 1.
 

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