vereinfachen (e^{ia}-e^{-ia})(e^{ib}+e^{-ib})

Lösung

vereinfachen

(e^{ia} - e^{- ia}) (e^{ib} + e^{- ib})

4 i sin (a) cos (b)

Schritte zur Lösung

(e^{ia} - e^{- ia}) (e^{ib} + e^{- ib})

e^{ia} = cos (a) + i sin (a)

e^{- ia} = cos (a) - i sin (a)

= (cos (a) + i sin (a) - (cos (a) - i sin (a))) (e^{ib} + e^{- ib})

e^{ib} = cos (b) + i sin (b)

e^{- ib} = cos (b) - i sin (b)

= (cos (a) + i sin (a) - (cos (a) - i sin (a))) (cos (b) + cos (b) + i sin (b) - i sin (b))

Vereinfache

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

= 2 cos (b) (cos (a) + i sin (a) - (cos (a) - i sin (a)))

Multipliziere aus

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

= 2 \cdot 2 i sin (a) cos (b)

Multipliziere die Zahlen:

2 \cdot 2 = 4

= 4 i sin (a) cos (b)

s im pl i f y (5 - x^{3}) \cdot (2 - \frac{3}{x + 1})

e x p an d \frac{3 x ^{2} - 5}{2 x ^{4} ( x ^{3} + 1 ) ^{7}}

e x p an d \frac{1}{a ( x - b ) ( x - c )}

e x p an d \frac{1 - e ^{- x}}{x ( x ^{2} + 2 x + 5 )}

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