展开 (sin((2k+1)2pi))/(2k+1)

解答

展开

\frac{sin ( ( 2 k + 1 ) 2 π )}{2 k + 1}

\frac{sin ( 4 πk )}{2 k + 1}

求解步骤

\frac{sin ( ( 2 k + 1 ) \cdot 2 π )}{2 k + 1}

乘开

(2 k + 1) \cdot 2 π : 4 πk + 2 π

= \frac{sin ( 4 πk + 2 π )}{2 k + 1}

使用三角恒等式改写

= \frac{sin ( 4 πk )}{2 k + 1}

e x p an d \frac{- 2 s ^{3} + 6 s ^{2} - 17 s + 54}{( s ^{2} + 9 ) ^{2}}

e x p an d \frac{2 ( 5 + 2 \cdot 5 ^{\frac{1}{2}} )}{5}

e x p an d (\frac{1}{3} e^{x} + \frac{5}{3} e^{- 2 x}) \cdot (e^{x} - 2 y)

e x p an d a (- 2, 4) b (1, 2) c (0, - 3)

e x p an d 5^{(n - 1)} + 2^{(n - 1)}