f(a)=sin(a)^2+sin(a+b)^2+sin(a+c)^20度

Question

f(a)=sin(a)^2+sin(a+b)^2+sin(a+c)^2 0度

Answer 1

f(a)=sin(a)^2+sin(a+b)^2+sin(a+c)^2
f(0)=(sinb)^2+(sinc)^2
f(π/2)=1+(cosb)^2+(cosc)^2
如果f(a)取定值,则f(0)=f(π/2)
而f(0)+f(π/2)=3
所以,f(a)=3/2
=1/2(1-cos2a+1-cos(2a+2b)+1-cos(2a+2c))
=3/2-1/2(cos2a+cos(2a+2b)+cos(2a+2c))
=3/2-1/2(cos2a+2cos(2a+b+c)cos(b-c))
=3/2-1/2(cos2a+2cos2acos(b+c)cos(b-c)-2sin2asin(b+c)cos(b-c))
=3/2-1/2(cos2a(1+2cos(b+c)cos(b-c))-2sin2asin(b+c)cos(b-c))
上式取定值3/2,则：
cos2a(1+2cos(b+c)cos(b-c))-2sin2asin(b+c)cos(b-c))=0
如果要上式恒成立,则
令a=0时,有,1+2cos(b+c)cos(b-c)=0
而如果1+2cos(b+c)cos(b-c)=0,则
2sin2asin(b+c)cos(b-c)=0 恒成立
则sin(b+c)=0或者 cos(b-c)=0
如果cos(b-c)=0,则 1+2cos(b+c)cos(b-c)=1>0矛盾
所以 sin(b+c)=0
而c>b>=0,所以 b+c=π
代入 1+2cos(b+c)cos(b-c)=0
有 cos(b-c)=1/2
c>b
所以 c-b=π/6
这样 就有 c=7π/12,b=5π/12