※ 引述《lilygarfield (一定是我不夠努力)》之銘言：
: y = √1-cos(theta) + √1+sin(theta)
: 或是
: y = √1-sin(theta) + √1+cos(theta)
: 應該怎麼解呢？
: 極值...
To find max y, we may assume -cosθ=cos u >=0, sinθ=sin u >=0
cos u = c, sin u =s
y=√(1+c) + √(1+s)
y^2=1+c+1+s+2√(1+c+s+cs)
c+s<=√2
cs<=1/2
the above maxima BOTH occur at c=s=1/√2
max y^2 = 2+√2 + 2√(3/2 + √2)
min is similar
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無梗題 bad question
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