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Minimizing Trigo value (Posted on 2014-03-02) Difficulty: 3 of 5
Find the minimum value of | sin x + cos x + tan x + cot x + sec x + csc x | for any real number x.

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 4.0000 (2 votes)

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Solution Wherefore art thou 2.65344585..? Comment 7 of 7 |
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 broll’s earlier post.]



  Posted by Harry on 2014-03-09 18:40:39
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