展开 (3e^tcos(t)-3e^tsin(t))^2

解答

展开

(3 e^{t} cos (t) - 3 e^{t} sin (t))^{2}

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

求解步骤

(3 e^{t} cos (t) - 3 e^{t} sin (t))^{2}

使用完全平方公式:

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

a = 3 e^{t} cos (t), b = 3 e^{t} sin (t)

= (3 e^{t} cos (t))^{2} - 2 \cdot 3 e^{t} cos (t) \cdot 3 e^{t} sin (t) + (3 e^{t} sin (t))^{2}

化简

(3 e^{t} cos (t))^{2} - 2 \cdot 3 e^{t} cos (t) \cdot 3 e^{t} sin (t) + (3 e^{t} sin (t))^{2} : 9 e^{2 t} cos^{2} (t) - 18 e^{2 t} cos (t) sin (t) + 9 e^{2 t} sin^{2} (t)

= 9 e^{2 t} cos^{2} (t) - 18 e^{2 t} cos (t) sin (t) + 9 e^{2 t} sin^{2} (t)

使用毕达哥拉斯恒等式:

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

= 9 e^{2 t} (- 2 cos (t) sin (t) + 1)

使用倍角公式:

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

= 9 e^{2 t} (1 - sin (2 t))

分解

9 e^{2 t} cos^{2} (t) - 18 e^{2 t} cos (t) sin (t) + 9 e^{2 t} sin^{2} (t) : 9 e^{2 t} (cos^{2} (t) - 2 cos (t) sin (t) + sin^{2} (t))

= 9 e^{2 t} (- sin (2 t) + 1)

e x p an d \frac{( s + 13 )}{s ^{2} ( s ^{2} + 6 s + 10 )}

e x p an d d x (x + 1) (x - 1) (x - 2)

e x p an d \frac{867 ( 2 + 3 ) 2 - 3}{1960}

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

e x p an d \frac{x ^{3} - 6 x ^{2} + 12 x - 8}{x - 2}