Question

已知向量m=(1,cosα-1/2),n=(sinα,1)且m⊥n,则sin2α等于

Answer 1

(1)m与n垂直,m .n = 0 于是1*sina +（cosa-0.5)*1 = 0,sina +cosa = 0.5

1+sin2a = 0.25 ,sin2a = -0.75

(2)cos(A-TT/4)=sqrt(2)/2*(cosA+sinA),sinA=4/5,cosA=-3/5,于是结果为sqrt(2)/10

S=0.5*b*c*sinA,于是b*c= 10

b^2 +c^2 -2*b*c*cosA=a^2,于是b^2+c^2 = 29

b+c=sqrt(29+20)=7