展開する (2e^{2t}cos(t)+4e^{2t}sin(t))^2

解

展開する

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

4 e^{4 t} cos^{2} (t) + 16 e^{4 t} cos (t) sin (t) + 16 e^{4 t} sin^{2} (t)

解答ステップ

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

完全平方式を適用する:

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

a = 2 e^{2 t} cos (t), b = 4 e^{2 t} sin (t)

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

簡素化

(2 e^{2 t} cos (t))^{2} + 2 \cdot 2 e^{2 t} cos (t) \cdot 4 e^{2 t} sin (t) + (4 e^{2 t} sin (t))^{2} : 4 e^{4 t} cos^{2} (t) + 16 e^{4 t} cos (t) sin (t) + 16 e^{4 t} sin^{2} (t)

= 4 e^{4 t} cos^{2} (t) + 16 e^{4 t} cos (t) sin (t) + 16 e^{4 t} sin^{2} (t)