展开 (4sin(x)-4cos(x))^2

解答

展开

(4 sin (x) - 4 cos (x))^{2}

- 16 sin (2 x) + 16

求解步骤

(4 sin (x) - 4 cos (x))^{2}

使用完全平方公式:

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

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

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

化简

(4 sin (x))^{2} - 2 \cdot 4 sin (x) \cdot 4 cos (x) + (4 cos (x))^{2} : 16 (- sin (2 x) + 1)

= 16 (- sin (2 x) + 1)

使用分配律:

a (b + c) = ab + a c

16 (- sin (2 x) + 1) = 16 (- sin (2 x)) + 16 \cdot 1

= 16 (- sin (2 x)) + 16 \cdot 1

化简

16 (- sin (2 x)) + 16 \cdot 1 : - 16 sin (2 x) + 16

= - 16 sin (2 x) + 16

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

e x p an d \frac{6 ( x + y ) + 5 - 6 x + 5}{y}

s im pl i f y 3 (x + 1) (x - \frac{1}{2}) (x - 1)

e x p an d (32 + 1) (2 - 3)

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