化简 (e^{ipix}-e^{-ipix})^2

解答

化简

(e^{iπ x} - e^{- iπ x})^{2}

- 4 sin^{2} (π x)

求解步骤

(e^{iπ x} - e^{- iπ x})^{2}

e^{iπ x} = cos (π x) + i sin (π x)

e^{- iπ x} = cos (π x) - i sin (π x)

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

乘开

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

= (2 i sin (π x))^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} i^{2} sin^{2} (π x)

2^{2} = 4

= 4 i^{2} sin^{2} (π x)

i^{2} = - 1

= 4 (- 1) sin^{2} (π x)

整理后得

= - 4 sin^{2} (π x)

e x p an d N (- 5, - 1)

\frac{d ^{4}}{d x ^{4}} ((y) ((2 x + 1)^{4}))

e x p an d \frac{x ^{3} + 8}{( x ^{2} - 1 ) ( x - 2 )}

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

s im pl i f y (2 - 2 i 3)^{4}