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

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.