cosA*cos2A*cos4A.cos(2的n-1次方)A 求化简,
问题描述:
cosA*cos2A*cos4A.cos(2的n-1次方)A 求化简,
答
等式乘以sinA
得1/2sin2A*cos2A*cos4A.cos(2^(n-1))A
=1/4sin4A*cos4A.cos(2^(n-1))A
=...=1/2^n*sin(2^n)*A
所以原式=1/2^n*sin(2^n*A)/sinA