#即pai1,以知#/4<a<3#/4,0<B<#/4,cos(#/4-a)=3/5,sin(3#/4+B)=5/13 求sin(a+B)(答案是56/65,需要过程)2, 化简ctg(7#+a)cos(5#+a)/tg(7#/2-a)sin(3#/2+a)
热心网友
1.π/4 sin(π/4-a)=-4/50 cos(3π/4+B)=-12/13于是：sin(a+B)=-cos[(3π/4+B)-(π/4-a)]=-cos(3π/4+B)cos(π/4-a)-sin(3π/4+B)sin(π/4-a)=-(-12/13)(3/5)-(5/13)(-4/5)=56/652.ctg(7π+a)cos(5π+a)/tg(7π/2-a)sin(3π/2+a)=cota(-cosa)/(-cota)(-cosa)=-1