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| Minimizing Trigo value (Posted on 2014-03-02) |
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Find the minimum value of | sin x + cos x + tan x + cot x + sec x + csc x | for any real number x.
Wherefore art thou 2.65344585..?
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Comment 7 of 7 | 
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Calculus approach..
Denoting the function by |f(x)|, and abbreviating sin x and cos x
by s and c: f(x) = s + c + s/c + c/s + 1/c + 1/s,
Any minimum of |f(x)| will occur when f(x)2 is minimum, i.e.
where 2f(x)f(x) is zero. Since f(x) itself is never zero, f(x) = 0.
f(x) = c s + 1/c2 1/s2 + s/c2 c/s2 = 0
=> (s2c3 s3c2 + s2 c2 + s3 c3)/(s2c2) = 0
=> (c s)(s + 1)(c + 1)(sc - s - c) = 0
Taking each factor in turn:
c s = 0 => s = c = + 1/sqrt(2), giving |f| = |2(1 + sqrt(2))|
so stationary values of ~ 6.2426(min) & 2.2426(max).
s + 1 = 0 => s = -1, c = 0, so f not defined.
c + 1 = 0 => c = -1, s = 0, so f not defined.
sc s c = 0 => sin x + cos x = sin x cos x (1)
Writing x = 5*pi/4 + X: sin x = -(sin X + cos X)/sqrt(2) (2)
cos x = -(cos X sin X)/sqrt(2) (3)
(1) now becomes -sqrt(2)*cos X = cos2 X sin2 X
which simplifies to cos2 X + sqrt(2)*cos X ½ = 0
with real solution cos X = 1 1/sqrt(2) (4)
Thus X = + 2n*pi + arccos(1 1/sqrt(2))
and x = 5*pi/4 + 2n*pi + arccos(1 1/sqrt(2))
giving x = 2.65344585.... , 5.20053578..., ...
where |f(x)| has its overall minimum value of 1.82842712...
[Using surds: (4) => sin X = -sqrt(2sqrt(2) - 1 )
then (2), (3) => sin x = (1 sqrt(2) + sqrt(2sqrt(2) 1)/2
and cos x = (1 sqrt(2) - sqrt(2sqrt(2) 1)/2
giving |f|min = 2sqrt(2) 1 agreeing with brolls earlier post.]
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Posted by Harry
on 2014-03-09 18:40:39 |
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