# Find the minimum for a trigonometric function

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..

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If you put $y=x+\frac{\pi}6$ then you get $\cos y-(\tan y-\frac1{\tan y})$. – Martin Sleziak Apr 24 '11 at 7:19
Sorry, I should have written $\cos y-\tan y-\frac1{\tan y}$ or $\cos y-(\tan y+\frac1{\tan y})$. – Martin Sleziak Apr 24 '11 at 8:05
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.) – Martin Sleziak Apr 24 '11 at 21:38
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 y – Julie Apr 25 '11 at 2:47
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. wolframalpha.com/input/?i=cos%28x%29-tan%28x%29-cot%28x%29 – Martin Sleziak Apr 25 '11 at 5:56

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.