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

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

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

expand

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

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

Solution steps

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

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

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

Apply the fraction rule:

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

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

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

Apply rule

\frac{a}{a} = 1

\frac{2}{2} = 1

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

Group like terms

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

1 - 1 = 0

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

\frac{1}{2} \cdot 0.5 \cdot x^{2} + 0.5 \cdot 30 \cdot 10 - 400

x + 2^{2}

\frac{2 5}{3 - 5}

\frac{10 ( Ks + X )}{s ^{3} + 11 s ^{2} + 10 s + 10 Ks + 10 X}

(f \circ g) (3 x + 7)

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