Question

已知sinx=根号5/5,siny=根号10/10,且x,y为钝角,求x+y

Answer 1

x,y为钝角π/2＜x＜π,π/2＜y＜π则π＜x+y＜2πsinx=√5/5,siny=√10/10所以cosx=-√[1-(√5/5)^2]=-2√5/5cosy=-√[1-(√10/10)^2]=-3√10/10所以cos(x+y)=cosxcosy-sinxsiny=(-2√5/5)*(-3√10/10)-(√5/5)*(√1...