erweitern (e^{2t}sin(t)+e^{2t}cos(t))^2

Lösung

erweitern

(e^{2 t} sin (t) + e^{2 t} cos (t))^{2}

e^{4 t} (sin (2 t) + 1)

Schritte zur Lösung

(e^{2 t} sin (t) + e^{2 t} cos (t))^{2}

Wende Formel für perfekte quadratische Gleichungen an:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = e^{2 t} sin (t), b = e^{2 t} cos (t)

= (e^{2 t} sin (t))^{2} + 2 e^{2 t} sin (t) e^{2 t} cos (t) + (e^{2 t} cos (t))^{2}

Vereinfache

(e^{2 t} sin (t))^{2} + 2 e^{2 t} sin (t) e^{2 t} cos (t) + (e^{2 t} cos (t))^{2} : e^{4 t} sin^{2} (t) + 2 e^{4 t} sin (t) cos (t) + e^{4 t} cos^{2} (t)

= e^{4 t} sin^{2} (t) + 2 e^{4 t} sin (t) cos (t) + e^{4 t} cos^{2} (t)

Verwende die Doppelwinkelidentität:

2 sin (x) cos (x) = sin (2 x)

= cos^{2} (t) e^{4 t} + sin^{2} (t) e^{4 t} + sin (2 t) e^{4 t}

Verwende die Pythagoreische Identität:

cos^{2} (x) + sin^{2} (x) = 1

= e^{4 t} (sin (2 t) + 1)

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

e x p an d \frac{2 x ^{4} - 6 x ^{2} + 12}{( x + 1 ) ( x - 1 )}

e x p an d \frac{d}{d x} (3 x cos (2 x)) + (3 x^{2} - 5) sin (2 x)

e x p an d A (- 4, - 2) y B (5, 3)

e x p an d \frac{4 m n ^{2} + p n ^{2}}{32 km + 8 k p}