sin(3π-a)=根号2乘以cos(3π/2+B), 根号3乘以cos(-a)= - 根号2乘以cos(π+B) ,0<a<π,
问题描述:
sin(3π-a)=根号2乘以cos(3π/2+B), 根号3乘以cos(-a)= - 根号2乘以cos(π+B) ,0<a<π,
0<B<π,求a和B
答
sin(3π-a)=根号2乘以cos(3π/2+B),即sina=√2sinB,
根号3乘以cos(-a)= - 根号2乘以cos(π+B),即√3cosa=√2cosB.
得sin^2a+3cos^2a=2,即cos^2a=1/2.得a=π/4,B=π/6.