Dear Uncle Colin

I'm stuck on a trigonometry proof: I need to show that $\cosec(x) - \sin(x) \ge 0$ 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 $x=\piby 2$), 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, $0 \lt \sin(x) \le 1$.

Dividing that through by $\sin(x)$, which is ok everywhere, because it's positive in that domain, we get $0 \lt 1 \le \cosec(x)$, which tells us that $\sin(x) \le 1 \le \cosec(x)$, so $\sin(x) \le \cosec(x)$, which means $\cosec(x) - \sin(x) > 0$. $\blacksquare$.

Hope that helps!

- Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.