# Why does applying a small angle approximation to equivalent forms of $\frac{\cos^2 \theta}{\sin \theta\tan \theta}$ yield different results?

In a question about small angle approximations I had answered, I simplified the trigonometric expression using identities and then applied the small and approximation.

I hadn't made any arithmetic mistakes, but the answer I got was different to the mark scheme, where they applied the small angle approximation straight away with no simplification.

My question is: Why do these different approaches yield different answers?

Edit: The question is:

$$\frac{\cos^2 \theta}{\sin \theta\tan \theta}$$

$$\theta^{-2}-{\frac 12}$$

The mark scheme says:

$${\frac 14}{\theta^2}+\theta^{-2}-1$$

• It might help if you could post both approaches, including the simplifications you made. Also, why are you sure that you have no errors? – Dirk Apr 24 at 8:48
• @Dirk. My teacher and I went through it many times and we found nothing arithmetically wrong. I will add the question, although it may not look pretty – Adam Cummings Apr 24 at 8:50
• Your LaTeX looked reasonably pretty (congratulations!), but I think I made it a little prettier. :) Be that as it may ... You should also include your simplification, so that everyone can be as certain as you and the teacher that you made no mistake there. – Blue Apr 24 at 9:20
• We cannot judge your work if you don't explain it. – Yves Daoust Apr 24 at 9:22

The correct approximation using Taylor expansion is $$\theta^{-2}-\dfrac76+O(\theta^2)$$. However, since your approximation is not taking higher order terms into account, it undoubtedly depends on the expression you start with, even if they are equivalent.

For instance, you could write

$$\frac{\cos^2 \theta}{\sin\theta\tan\theta}\simeq\frac{(1-\frac12\theta^2)^2}{\theta\cdot\theta}=\theta^{-2}-1+\frac{\theta^2}4$$

Or

$$\frac{\cos^2 \theta}{\sin\theta\tan\theta}=\frac{\cos^3\theta}{\sin^2\theta}\simeq\frac{(1-\frac12\theta^2)^3}{\theta^2}=\theta^{-2}-\frac{3}{2}+\frac{3\theta^2}4-\frac{\theta^4}{8}$$

Note that only the leading term $$\theta^{-2}$$ is correct.

Here is a plot to compare the variants. The difference with $$\frac{\cos^2\theta}{\sin \theta\tan \theta}$$ is plotted.

From the top:

• orange: $$\theta^{-2}-\frac12$$
• green: $$\theta^{-2}-1+\frac{\theta^2}4$$
• blue: $$\theta^{-2}-\dfrac76$$
• purple: $$\dfrac{(1-\frac12\theta^2)^3}{1-(1-\frac12\theta^2)^2}$$, obtained from $$\dfrac{\cos^3\theta}{1-\cos^2\theta}$$, yet another equivalent expression
• red: $$\theta^{-2}-\frac{3}{2}+\frac{3\theta^2}4-\frac{\theta^4}{8}$$

Note that all these approximations are local: they are valid near $$0$$, and the farther from $$0$$, the worst the approximation. With Taylor expansion, you can add more terms to get a better approximation on a larger (but still finite) interval. Notice the behavior of the purple solution on a larger interval is better: it's because it's not a polynomial approximation but a rational function.

Note also that the difference, which is plotted, is (locally) much lower than the leading term $$\theta^{-2}$$.

If the goal is to get a better approximation on a bounded interval, there are other methods, such as uniform approximation by a polynomial (on a bounded interval) using Chebyshev's equioscillation theorem, Padé approximants, splines...

• I don't want to add a new answer that would be almost identical to this one, but I would try to emphasize the following point: $$\frac{\cos\theta}{\tan^2\theta}=\frac{\cos^2\theta}{\sin\theta\tan\theta}=\frac{\cos^3\theta}{\sin^2\theta}.$$ While the denominator is always approximated by $\theta^2$ when ignoring higher order terms, the numerator is obviously different in each case. – Ennar Apr 24 at 9:30
• @Ennar And the first is certainly the simplifiation the OP did. – Jean-Claude Arbaut Apr 24 at 9:32
• Yes. Hopefully looking at these expressions side by side will help them. – Ennar Apr 24 at 9:33
• Thanks you, this is very helpful. Apologies for the barrage of questions, but which of these result is "more correct"? (If that makes any sense). If there are multiple ways of obtaining the approximation, why has the mark scheme only specified one as correct? – Adam Cummings Apr 24 at 10:28
• @AdamCummings None of these approximations is very good (you can expect one correct term, and even that may fail), but if you want the "best", it's the one with the constant term closest to $-7/6$ (you could also try $\frac{\cos^3 x}{1-\cos^2 x}$, slightly better). If you want several terms, the good way is the Taylor formula. Now, regarding the mark scheme: as dumb as it seems, you had only to apply the approximation without simplification. – Jean-Claude Arbaut Apr 24 at 10:36