化简 (e^{ipix}+e^{-ipix})^2

解答

化简

(e^{iπ x} + e^{- iπ x})^{2}

4 cos^{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}

化简

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

= (2 cos (π x))^{2}

使用指数法则:

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

= 2^{2} cos^{2} (π x)

2^{2} = 4

= 4 cos^{2} (π x)

e x p an d \frac{x}{x - x - 6} - \frac{1}{x + 2} - \frac{2}{x - 3}

e x p an d (- 3 a cos^{2} (t) + sin (t))^{2}

e x p an d (a + 4 b) (a - 2 b)

e x p an d (y cos (x) + 2 x e^{y}) + (5 m x + x^{2} e^{y} - 1) y^{'} (x)

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