三角形abc的三个内角为a.b.c求当A为何值时cosA+cos(B+C)/2取最大值,并求出最大值

三角形abc的三个内角为a.b.c求当A为何值时cosA+cos(B+C)/2取最大值,并求出最大值
三角形abc的三个内角为a.b.c求当A为何值时cosA+cos(B+C)/2取最大值,并求出最大值

三角形abc的三个内角为a.b.c求当A为何值时cosA+cos(B+C)/2取最大值,并求出最大值
已知A+B+C=π
所以,B+C=π-A
令y=cosA+cos[(B+C)/2]=cosA+cos[(π-A)/2]=cosA+sin(A/2)
=1-2sin^2 (A/2)+sin(A/2)
令sin(A/2)=t,则t∈(0,1)
所以,y=-2t^2+t+1=-2[t^2-(t/2)+(1/16)]+(1/8)+1
=-2*[t-(1/4)]^2+(9/8)
所以,当t=1/4∈(0,1)时,y有最大值9/8
此时,sin(A/2)=t=1/4
===> A/2=arcsin(1/4)
===> A=2arcsin(1/4)
或者,A=π-2arcsin(1/4).

S= cosA +cos[(B+C)/2]
= cosA + sinA
= √2sin(A+π/4)
max S = √2
at A = π/4