一道三角函数题
已知在三角形ABC中,sin A=3/5,cos B=5/13.
求cos C
参考答案:cos c=cos[180-(a+b)]
=cos 180*cos(a+b)+sin 180*sin(a+b)
=-cos(a+b)
=sin a * sin b-cos a *cos b
=3/5 * 12/13 - 4/5 *5/13
=16/65
已知在三角形ABC中,sin A=3/5,cos B=5/13.
求cos C
参考答案:cos c=cos[180-(a+b)]
=cos 180*cos(a+b)+sin 180*sin(a+b)
=-cos(a+b)
=sin a * sin b-cos a *cos b
=3/5 * 12/13 - 4/5 *5/13
=16/65