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
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![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](/uploads/image/z/7015521-57-1.jpg?t=sin%283%CF%80-a%29%3D%E6%A0%B9%E5%8F%B72%E4%B9%98%E4%BB%A5cos%283%CF%80%2F2%2BB%29%2C%26%23160%3B%26%23160%3B%26%23160%3B+%E6%A0%B9%E5%8F%B73%E4%B9%98%E4%BB%A5cos%28-a%29%3D+-+%E6%A0%B9%E5%8F%B72%E4%B9%98%E4%BB%A5cos%28%CF%80%2BB%29%26%23160%3B%26%23160%3B+%2C0%EF%BC%9Ca%EF%BC%9C%CF%80%2C0%EF%BC%9CB%EF%BC%9C%CF%80%2C%E6%B1%82a%E5%92%8CB)
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