展开 i/8*(cos(2pix)+isin(2pix))^3

解答

展开

\frac{i}{8} \cdot (cos (2 π x) + i sin (2 π x))^{3}

\frac{sin ^{3} ( 2 π x ) - 3 cos ^{2} ( 2 π x ) sin ( 2 π x )}{8} + i \frac{cos ^{3} ( 2 π x ) - 3 sin ^{2} ( 2 π x ) cos ( 2 π x )}{8}

求解步骤

\frac{i}{8} (cos (2 π x) + i sin (2 π x))^{3}

(cos (2 π x) + i sin (2 π x))^{3} = cos^{3} (2 π x) + 3 i cos^{2} (2 π x) sin (2 π x) - 3 sin^{2} (2 π x) cos (2 π x) - i sin^{3} (2 π x)

= \frac{i}{8} (cos^{3} (2 π x) + 3 i cos^{2} (2 π x) sin (2 π x) - i sin^{3} (2 π x) - 3 sin^{2} (2 π x) cos (2 π x))

使用复数算数法则:

(ai) (b + c i) = - a c + abi

a = \frac{1}{8}, b = cos^{3} (2 π x) - 3 sin^{2} (2 π x) cos (2 π x), c = 3 cos^{2} (2 π x) sin (2 π x) - sin^{3} (2 π x)

= - \frac{1}{8} (3 cos^{2} (2 π x) sin (2 π x) - sin^{3} (2 π x)) + \frac{1}{8} (cos^{3} (2 π x) - 3 sin^{2} (2 π x) cos (2 π x)) i

\frac{1}{8} (cos^{3} (2 π x) - 3 sin^{2} (2 π x) cos (2 π x)) i = \frac{i}{8} (cos^{3} (2 π x) - 3 sin^{2} (2 π x) cos (2 π x))

= - \frac{1}{8} (3 cos^{2} (2 π x) sin (2 π x) - sin^{3} (2 π x)) + \frac{i}{8} (cos^{3} (2 π x) - 3 sin^{2} (2 π x) cos (2 π x))

= - \frac{1}{8} (3 cos^{2} (2 π x) sin (2 π x) - sin^{3} (2 π x)) + \frac{i}{8} (cos^{3} (2 π x) - 3 sin^{2} (2 π x) cos (2 π x))

乘开

- \frac{1}{8} (3 cos^{2} (2 π x) sin (2 π x) - sin^{3} (2 π x)) : - \frac{3}{8} cos^{2} (2 π x) sin (2 π x) + \frac{1}{8} sin^{3} (2 π x)

= - \frac{3}{8} cos^{2} (2 π x) sin (2 π x) + \frac{1}{8} sin^{3} (2 π x) + (cos^{3} (2 π x) - 3 sin^{2} (2 π x) cos (2 π x)) \frac{i}{8}

乘开

\frac{i}{8} (cos^{3} (2 π x) - 3 sin^{2} (2 π x) cos (2 π x)) : \frac{i}{8} cos^{3} (2 π x) - \frac{3 i}{8} sin^{2} (2 π x) cos (2 π x)

= - \frac{3}{8} cos^{2} (2 π x) sin (2 π x) + \frac{1}{8} sin^{3} (2 π x) + \frac{i}{8} cos^{3} (2 π x) - \frac{3 i}{8} sin^{2} (2 π x) cos (2 π x)

将

- \frac{3}{8} cos^{2} (2 π x) sin (2 π x) + \frac{1}{8} sin^{3} (2 π x) + \frac{i}{8} cos^{3} (2 π x) - \frac{3 i}{8} sin^{2} (2 π x) cos (2 π x)

改写成标准复数形式:

\frac{- 3 cos ^{2} ( 2 π x ) sin ( 2 π x ) + sin ^{3} ( 2 π x )}{8} + \frac{cos ^{3} ( 2 π x ) - 3 sin ^{2} ( 2 π x ) cos ( 2 π x )}{8} i

= \frac{- 3 cos ^{2} ( 2 π x ) sin ( 2 π x ) + sin ^{3} ( 2 π x )}{8} + \frac{cos ^{3} ( 2 π x ) - 3 sin ^{2} ( 2 π x ) cos ( 2 π x )}{8} i

e x p an d \frac{( n + 1 ) ( n + 3 ) ( 2 n + 9 )}{6}

e x p an d \frac{4 x ^{2}}{( 2 x ^{3} - 11 ) ^{\frac{1}{3}} \cdot ( 2 x - 1 )}

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

e x p an d (e^{5 x} + 4 e^{- 3 x})^{2}

e x p an d \frac{( 13 - 11 ) 13 + 11}{79}