展开-pisin(1/(2pi)(t-1))

解答

展开

- π sin (\frac{1}{2 π} (t - 1))

- π sin (\frac{t - 1}{2 π})

求解步骤

- π sin (\frac{1}{2 π} (t - 1))

乘开

\frac{1}{2 π} (t - 1) : \frac{1}{2 π} t - 1 \cdot \frac{1}{2 π}

= - π sin (\frac{1}{2 π} t - 1 \cdot \frac{1}{2 π})

使用三角恒等式改写

= - π (cos (\frac{1}{2 π}) sin (\frac{t}{2 π}) - sin (\frac{1}{2 π}) cos (\frac{t}{2 π}))

使用角差恒等式:

sin (s) cos (t) - cos (s) sin (t) = sin (s - t)

= - π sin (\frac{t}{2 π} - \frac{1}{2 π})

化简

\frac{t}{2 π} - \frac{1}{2 π} : \frac{t - 1}{2 π}

= - π sin (\frac{t - 1}{2 π})

e x p an d \frac{2 ( - 4 π - 9 )}{3 π}

e x p an d ab - 9 a c + 3 b c + 9 b c - 2 a c + 3 ab

e x p an d (1 + y + y^{2}) d x

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

e x p an d (4 - 2 ln (t))^{4}