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}

(7 - 2 x - x + 2)^{2}

(tan (x) + 1)^{2} cos (x)

sec (a) csc (a) (tan (a) + cot (a))

(1 + cos (a)) (1 - cos (a))

(x - \frac{1}{x})^{4}

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