Conjecture: Does $\tan(\pi/2 - 10^{-k})$ approach powers of 10?

Question

I was playing around with my calculator and I stumbled about the most curious phenomenon. I noticed that $\tan(\pi/2-0.1)$, $\tan(\pi/2-0.01)$, $\tan(\pi/2-0.001)$ etc. seem to approach powers of $10$. Here are some calculations: $$\tan(\pi/2-0.1) \approx 9.96664$$ $$\tan(\pi/2-0.01) \approx 99.9966$$ $$\tan(\pi/2-0.001) \approx 999.999$$ $$\tan(\pi/2-0.0001) \approx 9999.99$$

And after that, my calculator rounds the numbers at powers of $10$. Is the conjecture $\lim_{n \to \infty}{\tan(\pi/2-10^{-n})-10^n} = 0$ true ? I've tried to prove it, but to no avail. I think that maybe expanding $\tan$ as Taylor series might be useful, but I haven't gotten anywhere. Does anyone have any ideas ?

Answer 1

Let's consider what happens to $\tan(\pi/2 - x)$ as $x \to 0$ in general. You can rewrite the tangent as

$$\frac{\sin(\pi/2 -x)}{\cos(\pi/2 - x)}.$$

Now we have the Taylor series approximations

$$\sin(\pi/2 - x) = 1 - \frac 1 2 x^2 + O(x^4)$$

and

$$\cos(\pi/2 - x) = x + O(x^3).$$

Thus, the quotient is well approximated by

$$\frac 1 x - \frac 1 2 x + O(x).$$

Therefore, $\tan(\pi/2 - x) - \frac 1 x \to 0$ as $x \to 0$; nothing special about powers of $10$. We can give explicit error estimates too, using the approximation above. If all you care about is the limit, then knowing $\sin(\pi/2 - x) = 1 + \text{error}$ is enough.

Answer 2

\begin{align} \tan(\pi/2-x)&=\cot x=\frac{\cos x}{\sin x}=\frac{1-x^2/2+\cdots} {x-x^3/6+\cdots}\\ &=\frac1x(1-x^2/2+\cdots)(1+x^3/6+\cdots) =\frac1x-\frac x3+\cdots. \end{align}

Answer 3

Hint: Look at the limits.

• First, observe that $\tan(\pi/2-t)=\frac{\cos(t)}{\sin(t)}$.

• For $t$ close to zero, $\cos(t)\approx1$ and $\sin(t)\approx t$.

• Therefore, you're observing that the value of $\tan(\pi/2-t)\approx\frac{1}{t}$.

Answer 4

Sure. Applying L'Hopital's rule twice, one gets:\begin{align}\lim_{x\to0}\tan\left(\frac\pi2-x\right)-\frac1x&=\lim_{x\to0}\cot(x)-\frac1x\\&=\lim_{x\to0}\frac{x\cos(x)-\sin(x)}{x\sin(x)}\\&=\lim_{x\to0}\frac{-x\sin(x)+\cos(x)-\cos(x)}{x\cos(x)+\sin(x)}\\&=-\lim_{x\to0}\frac{x\sin(x)}{x\cos(x)+\sin(x)}\\&=-\lim_{x\to0}\frac{x\cos(x)+\sin(x)}{-x\sin(x)+2\cos(x)}\\&=0.\end{align}