I'm tutoring for a college math class and we're doing putnam problems next week and this one stumped me:
Find the minimum value of $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$ for real numbers $x$.
[Math] Find the minimum value of $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$ for real numbers $x$
trigonometry
Best Answer
Let $\sin{x}+\cos{x}=\sqrt{2}\sin{(x+\frac{\pi}{4})}=a$, then $a$ can take any value between $-\sqrt{2}$ and $\sqrt{2}$. We have $\sin{x}\cos{x}=\frac{a^2-1}{2}$. Thus
\begin{align} \left||\sin{x}+\cos{x}+\tan{x}+\cot{x}+\sec{x}+\csc{x} \right|& =\left|\sin{x}+\cos{x}+\frac{1}{\sin{x}\cos{x}}+\frac{\sin{x}+\cos{x}}{\sin{x}\cos{x}} \right| \\ & =\left|a+\frac{2+2a}{a^2-1} \right| \\ & =\left|a+\frac{2}{a-1} \right| \end{align}
If $a<1$, let $b=1-a>0$, then $a+\frac{2}{a-1}=1-(b+\frac{2}{b}) \leq 1-2\sqrt{2}<0$, so $\left|a+\frac{2}{a-1} \right| \geq 2\sqrt{2}-1$.
($b+\frac{2}{b} \geq 2\sqrt{2}$ by AM-GM inequality or square $\geq 0$)
Equality holds when $b=\sqrt{2}, a=1-\sqrt{2}, x=\sin^{-1}{(\frac{1-\sqrt{2}}{\sqrt{2}})}-\frac{\pi}{4}$.
If $1 \leq a \leq \sqrt{2}$, then $a+\frac{2}{a-1}>\frac{2}{a-1}>2$, so $\left|a+\frac{2}{a-1} \right|>2>2\sqrt{2}-1$.
Thus the minimum value is $2\sqrt{2}-1$, achieved when $x=\sin^{-1}{(\frac{1-\sqrt{2}}{\sqrt{2}})}-\frac{\pi}{4}$.