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 January 1st, 2017, 05:49 AM #1 Newbie   Joined: Oct 2015 From: huddersfield Posts: 16 Thanks: 1 trigonometric identities given 5e^x - 4e^-x ≡A sinh x + B cosh x Find A and B
January 1st, 2017, 06:16 AM   #2
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Quote:
 Originally Posted by greeny93 given 5e^x - 4e^-x ≡A sinh x + B cosh x Find A and B
$5e^x-4e^{-x} = \dfrac{A(e^x - e^{-x})}{2} + \dfrac{B(e^x + e^{-x})}{2}$

$5e^x-4e^{-x} = \dfrac{(A+B)e^x + (B-A)e^{-x}}{2}$

$10e^x-8e^{-x} = (A+B)e^x + (B-A)e^{-x}$

$A+B = 10$

$B-A = -8$

can you finish?

 January 1st, 2017, 04:42 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,576 Thanks: 931 Math Focus: Elementary mathematics and beyond $$5e^x-4e^{-x}=8\sinh x+e^x=8\sinh x+ \sinh x+\cosh x=9\sinh x+\cosh x$$
January 4th, 2017, 01:56 PM   #4
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 Originally Posted by skeeter $5e^x-4e^{-x} = \dfrac{A(e^x - e^{-x})}{2} + \dfrac{B(e^x + e^{-x})}{2}$ $5e^x-4e^{-x} = \dfrac{(A+B)e^x + (B-A)e^{-x}}{2}$ $10e^x-8e^{-x} = (A+B)e^x + (B-A)e^{-x}$ $A+B = 10$ $B-A = -8$ can you finish?
unfortunately not, more studying required on this topic i think

January 4th, 2017, 02:00 PM   #5
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 Originally Posted by greeny93 unfortunately not, more studying required on this topic i think
come on ... it's basic algebra for solving a system of equations.

adding the last two equations term for term eliminates $A$ and yields the equation

$2B=2$

find $B$ and $A$ now?

January 4th, 2017, 02:44 PM   #6
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Quote:
 Originally Posted by skeeter come on ... it's basic algebra for solving a system of equations. adding the last two equations term for term eliminates $A$ and yields the equation $2B=2$ find $B$ and $A$ now?

of course, i couldnt see the wood for the trees in that instance i had got myself so confused in the other questions i am doing, i had the answer for this one a long time ago, your method and stages really helped me so thank you

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