Question

prove arcsinh 3/4 + arcsinh 5/12 =arcsinh 4/3

Answer 1

Hey there :)
Hmm this problem looks kinda fun.

Answer 2

hello, its fun for me once i get the answer

Answer 3

Are you familiar with this definition of arcsinh?\[\large\rm arcsinh(x)=\ln\left(x+\sqrt{x^2+1}\right)\]If not, we can derive that equation from sinh(x) and take inverse.
Otherwise we can start from this point.

Answer 4

yeah, i do learnt about it

Answer 5

\[\rm arcsinh\left(\frac{3}{4}\right)+arcsinh\left(\frac{5}{12}\right)\]So converting to log stuff,\[\rm \ln\left(\frac{3}{4}+\sqrt{\left(\frac{3}{4}\right)^2+1}\right)+\ln\left(\frac{5}{12}+\sqrt{\left(\frac{5}{12}\right)^2+1}\right)\]First step is simply getting everything set up correctly.
Ok with that? :o

Answer 6

yup

Answer 7

From there it should be easy peasy :)
Just a bunch of algebra (leave everything as fractions, no decimals),
then one application of log rule: \(\rm log(a)+log(b)=log(a\cdot b)\)

Answer 8

Square your things,\[\rm \ln\left(\frac{3}{4}+\sqrt{\frac{9}{16}+1}\right)+\ln\left(\frac{5}{12}+\sqrt{\frac{25}{144}+1}\right)\]and then get common denominators, ya? :)

Answer 9

just a question, when should we used the triangle and when we should apply the ln form

Answer 10

Ooo I don't really remember triangle method for hyperbolic,
I'm not sure I would be able to answer that :(

Answer 11

alright, thanks for helping :)

Answer 12

c: