Question

cosA*cos2A*cos4A.cos(2的n-1次方)A 求化简,

Answer 1

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