化简 e^{-ix}(6+e^{ix}+e^{-ix})

解答

化简

e^{- i \cdot x} \cdot (6 + e^{i \cdot x} + e^{- i \cdot x})

(2 cos^{2} (x) + 6 cos (x)) + i (- 6 sin (x) - 2 cos (x) sin (x))

求解步骤

e^{- i x} (6 + e^{i x} + e^{- i x})

e^{- i x} = cos (x) - i sin (x)

= (cos (x) - i sin (x)) (e^{i x} + e^{- i x} + 6)

e^{i x} = cos (x) + i sin (x)

e^{- i x} = cos (x) - i sin (x)

= (cos (x) - i sin (x)) (cos (x) + i sin (x) + cos (x) - i sin (x) + 6)

化简

6 + cos (x) + sin (x) i + cos (x) - i sin (x) : 2 cos (x) + 6

= (cos (x) - i sin (x)) (2 cos (x) + 6)

将

(cos (x) - i sin (x)) (2 cos (x) + 6)

改写成标准复数形式:

(2 cos^{2} (x) + 6 cos (x)) + (- 2 sin (x) cos (x) - 6 sin (x)) i

= (2 cos^{2} (x) + 6 cos (x)) + (- 2 sin (x) cos (x) - 6 sin (x)) i

e x p an d d (cos (x) + 2)

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

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

e x p an d (- \frac{1}{4} x - \frac{3}{2}) - (\frac{3}{2} x + \frac{1}{3})

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