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}

(3 (x - 73.75))^{3} - 10

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

\frac{3 x ^{4} + x ^{6}}{96} + (\frac{( x + 2 ) ( x ^{2} + x + 1 )}{24})

\frac{2 ( 45 - 15 )}{15}

sin (2 (t + π))

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