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[M2] 考考大家

考考大家

http://lsforum.net/board/viewthread.php?tid=152106&highlight=
How many ways can you think of to solve this max & min problem?

PS [the post is closed, but I found this interesting when I re-viewed it]
   

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[隱藏]
one of them is differentiation
http://www.wolframalpha.com/inpu ... 29%2F%28cosx%2B2%29
so 0>=x>=-4/3
any other?

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未有時間去諗, 目測用半角公式後, 可以看出 max = 0

但 min 反而未見有化簡到...

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回覆 3# cwk7093942 的帖子

哈哈
min係最難搵
其實你可以試下將trigo變成非trigo
因為sinx cosx都係[-1,1]

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0>= - t^2/[(0.5+t)^2+3/4] = -1/(0.5+t^-1)^2+3/4}>=-4/3    where  t=tan (pi/2-x/2) & t^-1 applicable when t not equal 0.
interesting question!

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Solving
x^2+y^2=1
and  y-1=m(x+2)
m=0 or m=-4/3
so (sinx-1)/(cosx+2) = slope of the line segment joining (-2,1) and point on x^2+y^2=1
so -4/3 <= (sinx-1)/(cosx+2) <=0
so max =0 and min =-4/3

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(sinx-1)/(cosx+2) =m   
(m^2+1)(cosx)^2+2(1+2m)mcosx+(1+2m)^2-1=0  treat as quadratic in cosx
D>=0 for real x
Simplifying -m^2-4m>=0  0>=m(3m+4)
Hence 0=>m>=-4/3
sub. m back to check and get x

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2 tangents from (-2,1) to unit circle centre (0,0) Obviously upper one is horizontal i.e. slope=0 so max value =0
the lower tangent slanting downward from L to R has minimum slope (-ve)
let -x be angle turning from the upper tangent clockwise (-ve) to the line joining (-2,1) to (0,0)
noting radius is perpendicular to tangent and noting the 2 equivalent triangles and tan x = 1/2
so minimum slope tan (-2x) = 2(-1/2)/[1-(-1/2)^2] = -4/3 so min value = -4/3
This requires very little computation, one can actually get the answer with mental calculation!

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2 congruent triangles ...too sleepy last night

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