What is expand 1+(((e^x-e^{(-x)}))/((2)))^2 ?

The solution to expand 1+(((e^x-e^{(-x)}))/((2)))^2 is 1+(e^{2x})/4-1/2+(e^{-2x})/4

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expand 1+(((e^x-e^{(-x)}))/((2)))^2

expand

1 + (\frac{( e ^{x} - e ^{(- x)} )}{( 2 )})^{2}

1 + \frac{e ^{2 x}}{4} - \frac{1}{2} + \frac{e ^{- 2 x}}{4}

Solution steps

1 + (\frac{e ^{x} - e ^{- x}}{2})^{2}

(\frac{e ^{x} - e ^{- x}}{2})^{2} = \frac{e ^{2 x} - 2 + e ^{- 2 x}}{2 ^{2}}

= 1 + \frac{e ^{2 x} + e ^{- 2 x} - 2}{2 ^{2}}

2^{2} = 4

= 1 + \frac{e ^{2 x} + e ^{- 2 x} - 2}{4}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{e ^{2 x} - 2 + e ^{- 2 x}}{4} = \frac{e ^{2 x}}{4} - \frac{2}{4} + \frac{e ^{- 2 x}}{4}

= 1 + \frac{e ^{2 x}}{4} - \frac{2}{4} + \frac{e ^{- 2 x}}{4}

Cancel

\frac{2}{4} : \frac{1}{2}

= 1 + \frac{e ^{2 x}}{4} - \frac{1}{2} + \frac{e ^{- 2 x}}{4}

\frac{x + 5}{( x + 2 ) ^{2}}

4 (\frac{x}{3} + 6 = 40)

\frac{1}{2 ( ( 2 x - 3 ) ^{3} )}

(m_{1} \cdot m_{2}) x^{''} + k (x - l) (m_{1} - m_{2})

\frac{( 5 + i ) ^{4}}{( 4 - 3 i ) ^{4}}

What is expand 1+(((e^x-e^{(-x)}))/((2)))^2 ?
The solution to expand 1+(((e^x-e^{(-x)}))/((2)))^2 is 1+(e^{2x})/4-1/2+(e^{-2x})/4