新高考三角裂项公式

luyuanhong

luyuanhong

luyuanhong

王守恩

\(\frac{2\sin(a)\sin(b)}{\cos(a - b) - \cos(a + b)}≡1\)

\(\frac{2\sin(a)\cos(b)}{\sin(a - b) + \sin(a + b)}≡1\)

\(\frac{2\cos(a)\sin(b)}{\sin(a + b) - \sin(a - b)}≡1\)

\(\frac{2\cos(a)\cos(b)}{\cos(a + b) + \cos(a - b)}≡1\)

王守恩

短的可以有4个？长的只有2个？

\(\displaystyle\frac{2\sin(a)\sin(b)}{\cos(a - b) - \cos(a + b)}≡1\)

\(\displaystyle\frac{2\sin(a)\cos(b)}{\sin(a - b) + \sin(a + b)}≡1\)

\(\displaystyle\frac{2\cos(a)\sin(b)}{\sin(a + b) - \sin(a - b)}≡1\)

\(\displaystyle\frac{2\cos(a)\cos(b)}{\cos(a + b) + \cos(a - b)}≡1\)

\(\displaystyle\frac{2\sin(a)\sum_{k=0}^9\sin(2 k*a + b)}{\cos(a - b) - \cos(19 a + b)}≡1\)

\(\displaystyle\frac{2\sin(a)\sum_{k=0}^9\cos(2 k*a + b)}{\sin(a - b) + \sin(19 a + b)}≡1\)

luyuanhong

luyuanhong