化简三角函数:sin[(2n-1/2)π-α]+cos[(2n+1/2)π+α],n∈N
问题描述：
化简三角函数:sin[(2n-1/2)π-α]+cos[(2n+1/2)π+α],n∈N
答
sin[(2n-1/2)π-α]+cos[(2n+1/2)π+α]
=sin[2nπ-π/2-α]+cos[2nπ+π/2+α]
=sin[-π/2-α]+cos[π/2+α]
=-sin[π/2+α]+cos[π/2+α]
=-cosα-sinα
答
sin[(2n-1/2)π-α]+cos[(2n+1/2)π+α],
=sin(-π/2-a)+cos(π/2+a)
=-sin(π/2+a)+cos(π/2+a)
=-sin(π/2-a)-sina
=-cosa-sina
=-√2(sina+π/4)