Question

Hello, I need to find \[A ^{2}-B ^{2}\] if
\[A=\frac{ 1 }{ 2}(2^{x}+2^{-x)}\]
\[B=\frac{ 1 }{ 2 }(2^{x}-2^{-x})\]
Any tips?

Answer 1

@Mertsj

Answer 2

\(A^2-B^2= (A+B)(A-B)\)
first, find A+B= ... ?

Answer 3

do I need, firstly, "unbracket" it? Like \[2^{-1}(2^{x}+2^{-x})=2^{x-1}+2^{-x-1}\] or not?

Answer 4

no need,
\((1/2) [2^x+2^{-x}] +(1/2) [2^x-2^{-x}] = (1/2) [2^x+2^{-x} +2^x-2^{-x} ]=...? \)

Answer 5

is it? \[ 2*2^{x}=2^{x+1}\]

Answer 6

i have forgotten 1/2

Answer 7

then it is 2^x

Answer 8

rigth or not?

Answer 9

yes, A+B = 2^x is correct.
what about A-B , similarly ?

Answer 10

2^-x?

Answer 11

good! absolutely correct :)

Answer 12

so, what about (A+B)(A-B) = 2^x * 2^(-x) =... ?

Answer 13

2^x+(-x)=2^x-x=2^0=1?

Answer 14

tank you a lot, hartnn! You have helped me a lot!!

Answer 15

you are \(\huge \color{red}{\checkmark}\)
welcome ^_^