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

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

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

expand

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

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

Solution steps

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

Apply the distributive law:

a (b + c) = ab + a c

a = \frac{1}{2}, b = \frac{a ^{2} - 1}{2}, c = \frac{1}{2} (e^{- 2 a} - \frac{1}{e ^{2}})

= \frac{1}{2} \cdot \frac{a ^{2} - 1}{2} + \frac{1}{2} \cdot \frac{1}{2} (e^{- 2 a} - \frac{1}{e ^{2}})

Simplify

\frac{1}{2} \cdot \frac{a ^{2} - 1}{2} + \frac{1}{2} \cdot \frac{1}{2} (e^{- 2 a} - \frac{1}{e ^{2}}) : \frac{1}{4} e^{- 2 a} - \frac{1}{4} - \frac{1}{4 e ^{2}} + \frac{a ^{2}}{4}

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

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(8 e^{t} cos (t) - 8 e^{t} sin (t))^{2}

(- 4, - 2) y (4, - 4)

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