Question

已知0<=a<=兀,0<=b<=兀/4,且a+b=2兀/3

求∶y={[1-cos(兀-2a)]/[(cota/2)-(tana/2)]}-[cos(兀/4-b)]^2的最大值，并求出相应的a、b的值.

Answer 1

y={[1-cos(兀-2a)]/[(cota/2)-(tana/2)]}-[cos(兀/4-b)]^2
=[1+cos(2a)]/【{[cos(a/2)]^2-[sin(a/2)]^2}/[cos(a/2)sin(a/2)]】-[√2/2(cosb+sinb)]^2
=2(cosa)^2/[cosa/(1/2sina)]-1/2(1+sin2b)
=sinacosa-1/2sin2b-1/2
=1/2(sin2a-sin2b)-1/2
=1/2[2cos(a+b)*sin(a-b)]-1/2
=cos2/3π*sin(a-b)-1/2
=-1/2sin(a-b)-1/2
ymax=-1/2*(-1)-1/2=0
sin(a-b)=-1
(a-b)=-π/2,
a+b=2π/3
解得，a=π/12,b=7π/12
(与已知条件不符)