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

trigonometry

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:

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

The answer I got was:

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

The mark scheme says:

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

Best Answer

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: $\theta^{-2}-\frac{3}{2}+\frac{3\theta^2}4-\frac{\theta^4}{8}$

enter image description here

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...

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