展开 (4(sin(2pin)-2pincos(2pin)))/(pi^2n^2)

解答

展开

\frac{4 ( sin ( 2 πn ) - 2 πn cos ( 2 πn ) )}{π ^{2} n ^{2}}

\frac{4 sin ( 2 πn )}{π ^{2} n ^{2}} - \frac{8 cos ( 2 πn )}{πn}

求解步骤

\frac{4 ( sin ( 2 πn ) - 2 πn cos ( 2 πn ) )}{π ^{2} n ^{2}}

乘开

4 (sin (2 πn) - 2 πn cos (2 πn)) : 4 sin (2 πn) - 8 πn cos (2 πn)

= \frac{4 sin ( 2 πn ) - 8 πn cos ( 2 πn )}{π ^{2} n ^{2}}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{4 sin ( 2 πn ) - 8 πn cos ( 2 πn )}{π ^{2} n ^{2}} = \frac{4 sin ( 2 πn )}{π ^{2} n ^{2}} - \frac{8 πn cos ( 2 πn )}{π ^{2} n ^{2}}

= \frac{4 sin ( 2 πn )}{π ^{2} n ^{2}} - \frac{8 πn cos ( 2 πn )}{π ^{2} n ^{2}}

消掉

\frac{8 πn cos ( 2 πn )}{π ^{2} n ^{2}} : \frac{8 cos ( 2 πn )}{πn}

= \frac{4 sin ( 2 πn )}{π ^{2} n ^{2}} - \frac{8 cos ( 2 πn )}{πn}

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