展开 (4e^tcos(t)-4e^tsin(t))^2

解答

展开

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

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

求解步骤

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

使用完全平方公式:

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

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

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

化简

(4 e^{t} cos (t))^{2} - 2 \cdot 4 e^{t} cos (t) \cdot 4 e^{t} sin (t) + (4 e^{t} sin (t))^{2} : 16 e^{2 t} cos^{2} (t) - 32 e^{2 t} cos (t) sin (t) + 16 e^{2 t} sin^{2} (t)

= 16 e^{2 t} cos^{2} (t) - 32 e^{2 t} cos (t) sin (t) + 16 e^{2 t} sin^{2} (t)

使用毕达哥拉斯恒等式:

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

= 16 e^{2 t} (- 2 cos (t) sin (t) + 1)

使用倍角公式:

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

= 16 e^{2 t} (1 - sin (2 t))

分解

16 e^{2 t} cos^{2} (t) - 32 e^{2 t} cos (t) sin (t) + 16 e^{2 t} sin^{2} (t) : 16 e^{2 t} (cos^{2} (t) - 2 cos (t) sin (t) + sin^{2} (t))

= 16 e^{2 t} (- sin (2 t) + 1)

e x p an d 5 (5 x^{2} + 3 x + 2)^{2}

e x p an d - e^{- x} (x + c) + e^{- x} + y

e x p an d x d y - y d x - (1 - x^{2}) d x

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

e x p an d (3 - 0, 1 x) x - (50 + 0.06 x)