若tan(π/4-a)=-5/12 a属于(π/4,3π/4) cos(3π/2+2b)=7/25 b属于(0,π/4),求sin(a+b)的值

问题描述:

若tan(π/4-a)=-5/12 a属于(π/4,3π/4) cos(3π/2+2b)=7/25 b属于(0,π/4),求sin(a+b)的值

∵a属于(π/4,3π/4),∴(π/4-a)属于(-π/2,0)又∵tan(π/4-a)= -5/12 ,∴sin(π/4-a)= -5/13,cos(π/4-a)=12/13 ⑴∵ cos(3π/2+2b)=7/25,而cos(3π/2+2b)=cos[2π-(3π/2+2b)]=cos(π/2-2b)∴cos(π/2-2b)=7/25 ...