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

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

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

expand

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

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

Solution steps

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

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

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

2^{2} = 4

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

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

Cancel

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

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

\frac{s ^{3} + 8 s ^{2} + 4 s + 3}{s ^{2} ( s ^{2} + 4 s + 3 )}

(- ux) d x + (x (u d x + x d u) + x y (u d x + x d u)) d y

\frac{1}{3} (2 + \frac{1}{5} (4 z - 1) = \frac{3}{4} z - \frac{1}{3})

2 e^{- x^{2} - y^{2}} y - 2 e^{- x^{2} - y^{2}} y (y^{2} - x^{2})

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

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