已知cos(π/2-α)-sin(π/2+α)=1/5,且0<α<π,求下列各式的值:
问题描述:
已知cos(π/2-α)-sin(π/2+α)=1/5,且0<α<π,求下列各式的值:
(1)sin(π+α)cos(π+α);;(2)sin(3π/2-α)+cos(3π/2-α);;
(3)cos3^(3π/2+α)-sin^3(3π/2+α)
答
cos(π/2-α)-sin(π/2+α)=sina-cosa=1/5 ①两边平方得:1-2sinacosa=1/25 sinacosa=12/25 ∴sin^2 a+cos^2 a=(sina+cosa)^2-2sinacosa=(sina+cosa)^2-24/25 ...