vereinfachen (e^{ti}-e^{-ti})^3

Lösung

vereinfachen

(e^{t i} - e^{- t i})^{3}

- 8 i sin^{3} (t)

Schritte zur Lösung

(e^{t i} - e^{- t i})^{3}

e^{t i} = cos (t) + i sin (t)

e^{- t i} = cos (t) - i sin (t)

= (cos (t) + i sin (t) - (cos (t) - i sin (t)))^{3}

Multipliziere aus

cos (t) + sin (t) i - (cos (t) - i sin (t)) : 2 i sin (t)

= (2 i sin (t))^{3}

Wende Exponentenregel an:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{3} i^{3} sin^{3} (t)

2^{3} = 8

= 8 i^{3} sin^{3} (t)

i^{3} = - i

= 8 (- i) sin^{3} (t)

Entferne die Klammern:

(- a) = - a

= - 8 i sin^{3} (t)

e x p an d (\frac{1 - x}{2})^{8}

e x p an d (e^{x} + y) d x + (2 + x + y e^{x}) d y

e x p an d \frac{f ( c + h ) - f c ^{^{'}}}{h}

e x p an d - 14 (- 2 e^{- x^{2}} x^{2} + e^{- x^{2}})

e x p an d \frac{( x + 1 ) x ( x - 1 )}{6}