Cosh2x+2sinh^2x, not sure if this goes in this forum

  • Thread starter Thread starter pat666
  • Start date Start date
  • Tags Tags
    Forum
AI Thread Summary
The function cosh(2x) + 2sinh^2(x) needs to be expressed in terms of exponential functions. Initial attempts to simplify the expression included incorrect algebra, particularly in handling the term 2sinh^2(x). The correct expression for sinh^2(x) is 1/2(e^x - e^(-x))^2, which should be expanded using the FOIL method rather than simply squared. A consensus emerged that the factor of 1/2 cancels out with the 2 in front of sinh^2(x), leading to confusion about the final simplification. Clarification on these algebraic steps is essential for accurate results.
pat666
Messages
703
Reaction score
0

Homework Statement



Express the function cosh(2x)+2 sinh^2(x)= in terms of exponential functions.

Homework Equations


The Attempt at a Solution


i have an answer but I am not sure if its right can someone check please.
=1/2 (e^2x+e^(-2x) )+1/2(e^2x-e^(-2x)) if this is right i need some help simplifying.
 
Physics news on Phys.org
No, it's not right. Check your algebra.
 
almost certain that the problem is the 2sinh^2(x), i got 1/2(e^x-e^(-x))^2 which my 89 says is right, then i try to simplify to 1/2(e^(2x)-e^-(2x)) which it says is wrong but i can't see why?
 
If you're trying to say (a+b)2 = a2+b2, that's the reason why. It's not true. You need to FOIL it out. Also, the 1/2 shouldn't be there because it cancels with the 2 in front.
 
thanks i agree with one problem you found but i think that the 1/2 should still be there because its squared in sinh^2 (the 2 isnt) giving 1/4 which then becomes 1/2?? not positive though??
 
Oops, you're right. Sorry.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Back
Top