已知向量m=(sinx,-1);n=(cosx,3/2);f(x)=(m+n)•m;锐角△ABC中a,b,c分别为角A,B,C的对边;若5a=4(√2)c,b=7√2,f(B/2)=(3/10)√2,求边a,c.
m+n=(sinx+cosx,1/2);
故f(x)=(m+n)•m=(sinx+cosx)sinx-1/2=sin²x+sinxcosx-1/2=(1-cos2x)/2+(1/2)sin2x-1/2
=(1/2)(sin2x-cos2x)=(√2/2)sin(2x-π/4);
f(B/2)=(√2/2)sin(B-π/4)=(3/10)√2,故得sin(B-π/4)=3/5;B=π/4+arcsin(3/5);
a=(4/5)(√2)c,b=7√2,cosB=cos[π/4+arcsin(3/5)]=(√2/2)[sinarcsin(3/5)+cosarcsin(3/5)]
=(√2/2)[3/5+√(1-9/25)]=(√2/2)(3/5+4/5)=(7/10)√2;
由余弦定理,b²=a²+c²-2accosB,即有:
98=(32/25)c²+c²-(8/5)(√2)[(7/10)√2]c²=(57/25)c²-(56/25)c²=(1/25)c²;
故得c²=98×25;∴c=35√2;a=(4/5)(√2)(35√2)=56.