Ask Uncle Colin: A Trig Inequality

Dear Uncle Colin,

I have to find the values of $x$, between 0 and $\pi$ inclusive, where $2\cos (x)>\sec (x)$. My answer was $0\le x<\frac{\pi }{4}$, but the answer also includes $\frac{\pi }{2}<x<\frac{3}{4}\pi$. I don’t understand why!

Stuck Evaluating Confusing And Nasty Trig

Hi, SECANT, and thanks for your message!

What I suspect you’ve done is multiply both sides by $\cos (x)$ to get $2{\cos }^2(x)>1$, said $\cos (x)>\frac{1}{\sqrt{2}}$ when $0\le x<\frac{\pi }{4}$.

That’s true, so far as it goes, but it’s not the whole story. Multiplying both sides by $\cos (x)$ is fraught with danger, because it is not guaranteed to be positive.

What you should do instead

Instead, I would bring everything to the left-hand side:

$2\cos (x)-\sec (x)>0$

Then turn it into a single fraction:

$\frac{2{\cos }^2(x)-1}{\cos (x)}>0$

Now, look for sign changes: when ${\cos }^2(x)=\frac{1}{2}$, the fraction is 0 and we have a change of sign. This occurs when $x=\frac{1}{2}\pi$ or $x=\frac{3}{4}\pi$.

We also have a change of sign when $\cos (x)=0$, at $x=\frac{\pi }{2}$. This gives us three interesting points to consider: $\frac{\pi }{4}$, $\frac{\pi }{2}$ and $\frac{3}{4}\pi$.

Between $0$ (inclusive) and $\frac{\pi }{4}$ (exclusive), the inequality holds true. It is untrue between $\frac{\pi }{4}$ and $\frac{\pi }{2}$, undefined at $x=\frac{\pi }{2}$ and then true until $\frac{3}{4}\pi$ (which is excluded).

The final answer is $0\le x<\frac{\pi }{4}$ or $\frac{\pi }{2}<x<\frac{3}{4}\pi$.

Hope that helps!

- Uncle Colin