Question

1)a、b、C为三角形ABC的内角,所对的边分别为a、b、c、且(a^2+c^2-b^2)/(a^2+b^2-c^2)=

Answer 1

1)用角的关系做简单

(a^2+c^2-b^2)/(a^2+b^2-c^2)=C/(2a-c)

【2ac(a^2+c^2-b^2)/2ac】/【2ab(a^2+b^2-c^2)/2ab】=sinC/(2sinA-sinC)

ccosB/bcosC=sinC/(2sinA-sinC)

sinCcosB/sinBcosC=sinC/(2sinA-sinC)

cosB(2sinA-sinC)=sinBcosC

2cosBsinA=cosBsinC+sinBcosC=sin(B+C)=sinA

cosB=1/2;

SinB+sin(C-A)=2Sin2A

sin(A+C)+sin(C-A)=2sin2A

2sinCcosA=4sinAcosA

sinC=2sinA

c=2a

(cosA不等于0,若等于,则sinC=2);

由cosB=1/2和c=2a不能确定一个三角形,所以无解!

2)f(0)=a+bsin0+cos0=1

a=0

f(π/4)=a+bsin(π/2)+cos(π/2)=1

a+b=1,b=1

f(x)=sin2x+cos2x=√2sin(2x+π/4)

当X属于[0,π/4]时,x=π/8时,f(x)最大为√2

矛盾!