Question

Need help with proofing. Question and work attached, I got up to a certain point, but got confused. Someone please explain how to do this problem.

Answer 1

no there are quite a few mistakes

Answer 2

(a + b)^2 = a^2 + 2ab + b^2

Answer 3

But the easiest way to do this problem is to recognize:

p^2 - q^2 = (p + q)(p - q)

Answer 4

firstly, the second term of the first line should have a -ve sign between the e's

Answer 5

and as ranga has stated your expansion is incorrect, try again

Answer 6

Ah ok. So something like this?

Answer 7

no, with the top line of the first term we have (e^x + e^-x)^2
if you make e^x = a and e^-x =b then use ranga's expansion above
for the bottom line it's just 2^2

Answer 8

Ok, I rewrote the equation and it came out to be something like this. Is it right?

Answer 9

It looks like you are now using the second method I gave you:
p^2 - q^2 = (p + q)(p - q)
It should be:
{ (e^x + e^-x) / 2 + (e^x - e^-x) / 2 } * { (e^x + e^-x) / 2 - (e^x - e^-x) / 2 } =
e^x * e^-x = e^0 = 1

Answer 10

looks good ranga

Answer 11

well done Az

Answer 12

now just continue form there to get the answer

Answer 13

First Method:
{ (e^x + e^-x)/2 }^2 - { (e^x - e^-x)/2 }^2 =
1/4{ e^2x + e^-2x + 2e^x * e^-x } - 1/4{ e^2x + e^-2x - 2e^x * e^-x } =
1/4 { e^2x + e^-2x + 2 } - 1/4 { e^2x + e^-2x - 2 } =
1/4 { e^2x + e^-2x + 2 - e^2x - e^-2x + 2 } =
1/4 { 4 } = 1

Thanks alekos.

Answer 14

@ranga Yeah, I tried fitting the first one in, but it got really complicated, so I just went with the second one. Thanks for helping @ranga @alekos . I was able to arrive at the final step after some simplification once I got the correctly expanded equation done.

Answer 15

that's good. i hope you understand everything we showed you