Ask Uncle Colin: A Cosec Proof

Dear Uncle Colin

I’m stuck on a trigonometry proof: I need to show that $\mathrm{cosec}(x)-\sin (x)\ge 0$ for $0<x<\pi$. How would you go about it?

- Coming Out Short of Expected Conclusion

Hi, COSEC, and thank you for your message! As is so often the case, there are several ways to approach this.

The most obvious one

The first approach I would try would be to turn the left hand side into a single fraction: $\frac{1}{\sin (x)}-\sin (x)\equiv \frac{1-{\sin }^2(x)}{\sin (x)}$.

The top of that is ${\cos }^2(x)$, so you have $\frac{{\cos }^2(x)}{\sin (x)}$.

In the specified region, $\cos (x)$ is non-negative (it is zero at $x=\frac{\pi }{2}$), while $\sin (x)>0$ (because the endpoints are excluded). Therefore, you have a non-negative number divided by a positive number, which is non-negative, as required.

Working from the range of $\sin (x)$

A really neat alternative is to note that, in the given domain, $0<\sin (x)\le 1$.

Dividing that through by $\sin (x)$, which is ok everywhere, because it’s positive in that domain, we get $0<1\le \mathrm{cosec}(x)$, which tells us that $\sin (x)\le 1\le \mathrm{cosec}(x)$, so $\sin (x)\le \mathrm{cosec}(x)$, which means $\mathrm{cosec}(x)-\sin (x)>0$. $◼$.

Hope that helps!

- Uncle Colin