4
$\begingroup$

Find the local minimum of the following function: $$\tan\left(x+\frac{2\pi}{3}\right)-\tan\left(x+\frac{\pi}{6}\right)+\cos\left(x+\frac{\pi}{6}\right)$$

I am wondering how can I simply this function..

  • 4
    If you put $y=x+\frac{\pi}6$ then you get $\cos y-(\tan y-\frac1{\tan y})$.2011-04-24
  • 0
    Sorry, I should have written $\cos y-\tan y-\frac1{\tan y}$ or $\cos y-(\tan y+\frac1{\tan y})$.2011-04-24
  • 0
    This yields $f'(y)=\frac{\cos^2y-\sin^2y}{\cos^2y\sin^2y}-\sin y$; however I do not know how to solve $f'(y)=0$. (It leads to an equation of 5th degree.)2011-04-24
  • 0
    sorry, this is a high school question, and derivative may not be used for the purpose. Is there any simpler way to write -cot y-tan y+cos y2011-04-25
  • 1
    You can rewrite it just using sines and cosines as $\cos y - \frac1{\cos y\sin y}$ but I do not see anyway getting the minima from this. http://www.wolframalpha.com/input/?i=cos%28x%29-tan%28x%29-cot%28x%292011-04-25
  • 0
    it has local minimum at some point but no global minimum as we can see from Wolframalpha. I just wonder how does it work to find the graph, the min, max2011-04-26
  • 0
    You can certainly plot it and find the local minimum. It looks to be around 2.435+pi/6 (remember the offset between your function and the plot) and you can zoom in to improve the accuracy. Only you know if that is an acceptable approach.2011-09-02

1 Answers 1

1

I would go at this brute force. Following @Martin's plot, minima occur near $3\pi/4 + n\pi$ with $n$ being an integer. Take your function $f(y) = \cos y - \tan y - \cot y$ and expand it with a Taylor series at these values; $$f({3\pi\over 4}+z) = f({3\pi\over 4}) + f'({3\pi\over 4})z + f''({3\pi\over 4}){z^2\over 2} .. $$ Solve for $z$ such that the derivative of this function is zero, giving you a minimum at $$ z = -{f'(3\pi/4) \over f''(3\pi/4)}.$$ This gives you to first order the location of the minimum. The value of the function at the minimum can then be found by plugging back into the function.

This gives you an approximate solution. You can improve the order of the approximation if you like.