化简sin[(4n-1)π/2-a]+cos[(4n+1)π/2-a]
问题描述:
化简sin[(4n-1)π/2-a]+cos[(4n+1)π/2-a]
答
sin[(4n-1)π/2-a]+cos[(4n+1)π/2-a]
=sin(2nπ-π/2-a)+cossin(2nπ+π/2-a)
=sin(-π/2-a)+cos(π/2-a)
=-sin(π/2+a)+sina
=-cosa+sina