If sinhx =3/4, why is cosh x not = 4/5 by using triplet triangle

I appreciate that it is incorrect but can anyone tell me why it does not work with the triangle.

Original post by fnatic NateDestiel

I appreciate that it is incorrect but can anyone tell me why it does not work with the triangle.

What triangle?

\cosh(x)^2 - \sinh(x)^2=1

Original post by fnatic NateDestiel

I appreciate that it is incorrect but can anyone tell me why it does not work with the triangle.

Hyperbolic functions do not have the same properties of trigonometric functions

Original post by BuryMathsTutor

What triangle?

\cosh(x)^2 - \sinh(x)^2=1

yep i did it that way after!

Also inv sinh gives values of x to be 0.693 iirc and then it yields 5/4 as the answer.

Thanks though

Hyperbolic functions do not have the same properties of trigonometric functions

aha thanks.

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