Geometric Derivation of Small Angle Approximation: $\tan\theta\approx\theta$. Is this valid?

Question

This is my first question here...

I considered a circle with radius $r$ and a small angle $\theta$ subtended at the center. I realized that for a very small $\theta$, the arc length $s$ can be approximated as a straight line, forming the opposite side of a right triangle with adjacent side $r$.

$s = r\theta$ (where $\theta$ is in radians).

For $\theta\ll$ the arc $s$ can be approximated as a straight line.

Consider a right triangle with opposite side $s$ (approximated arc length) and adjacent side $r$.

$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$.

Therefore, $\tan\theta\approx\frac{s}r$.

For $\theta$ in degrees, we multiply by $\frac\pi{180}: \tan\theta\approx\frac{\theta\pi}{180}$.

What do you think about the validity of this derivation? I'm new to trigonometry, so please feel free to point out any mistakes....

"Approximation" needs some clarification: what is your error target (over what range of input angles)? — Eric Towers, Nov 27 '24 at 21:18
Welcome to Math SE. FYI, this small angle approximation is quite well known, with it coming from that $\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1$, so $\sin\theta\approx\theta$ for small theta, and $\cos(0)=1$, so $\cos\theta\approx 1$ for small $\theta$. As the related A geometric proof for the "small angle approximation" for the sine, cosine and tangent states, a more specific small angle approximation for $\cos\theta$ is $1-\frac{\theta^2}{2}$. Note that I found this other post using an ... — John Omielan, Nov 27 '24 at 21:31
(cont.) Approach0 search, from which you can see among the results that there are quite other posts which mention the $\tan\theta\approx\theta$ approximation. — John Omielan, Nov 27 '24 at 21:33
@EricTowers: I calculated that it approximated tan < 6° upto 3 decimal places, tan < 10° upto 2 places and tan < 50° upto a single decimal point... — Paarth Katiyarr, Nov 28 '24 at 11:26
@MoisheKohan I got that my geometric argument isn't like a formal proof. I'm trying to develop a more intuitive understanding of the approximation before going into more rigorous methods like Taylor series, etc.. — Paarth Katiyarr, Nov 28 '24 at 11:28

score 1 · Answer 1 · answered Nov 28 '24 at 14:23

Consider a right triangle $ABC$ with a unit leg $AB$ and an acute angle $\angle BAC=\alpha$. Note that a circular sector of radius $AB=1$ and angle $\alpha$ lies inside $ABC$, while $ABC$ lies inside a circular sector of radius $AC=1/\cos\alpha$ and angle $\alpha$. Comparison of the areas gives then: $$ \alpha\le \tan\alpha \le {\alpha\over\cos^2\alpha}. $$

Geometric Derivation of Small Angle Approximation: $\tan\theta\approx\theta$. Is this valid?

1 Answers1