cos(a+β)=3/5,cos(a-β)=12/13,0

Question

cos(a+β)=3/5,cos(a-β)=12/13,0

数学

作业帮用户2016-12-16

Answer 1

∵0∴0sin(a-β)=√[1-cos^2(a-β)]=5/13.
sin(a+β)=√[1-cos^2(a+β)]=4/5.
cos2a=cos[(a+β)+(a-β)]=cos(a+β)*cos(a-β)-sin(a+β)*sin(a-β)=16/65.