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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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}

d v (2 x + 1)

\frac{\partial}{\partial x} ((z) (e^{x yz} + x y))

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

\frac{1}{( 1 + k cos ( x ) ) ^{2}}

e^{2 t} (2 - t - 2 e^{- t})

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