展开 (e^tcos(t)-e^tsin(t))^2

解答

展开

(e^{t} \cdot cos (t) - e^{t} \cdot sin (t))^{2}

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

求解步骤

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

使用完全平方公式:

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

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

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

化简

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

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

使用倍角公式:

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

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

使用毕达哥拉斯恒等式:

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

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

e x p an d \frac{2 ( x - y )}{3}

e x p an d \frac{500}{( s + 1 ) ( s + 3 ) ( s + 10 )}

e x p an d d y (3 x - \frac{2}{x} + 1)

e x p an d \frac{9}{1 - x ^{2}}

e x p an d (t + Δ t)^{4}