Question

sin(kπ/6)=1/2sin(π/12)[cos(2k-1)π/12-cos(2k+1)π/12] 这个式子为什么相

Answer 1

cos(2k-1)π/12 = cos(kπ/6-π/12) = cos(kπ/6)cos(π/12) + sin(kπ/6)sin(π/12)

cos(2k+1)π/12 = cos(kπ/6+π/12) = cos(kπ/6)cos(π/12) - sin(kπ/6)sin(π/12)

上述两式相减,得 2*sin(kπ/6)sin(π/12),

因此 1/[2sin(π/12)] * [cos(2k-1)π/12-cos(2k+1)π/12]

=1/[2sin(π/12)] * 2*sin(kπ/6)sin(π/12)

= sin(kπ/6)