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Today I saw the approximation $\tan(2 \theta) \approx 2\tan(\theta)$ when $\theta$ is small.

I justified this by seeing that these two functions have the same linear taylor expansion about $\theta = 0$. Is there a direct way to go from one to the other?

Maybe there's some clever way to apply dominant balance?

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3 Answers 3

up vote 4 down vote accepted

By the double-angle formula, $$\tan(2\theta)={2\tan\theta\over1-\tan^2\theta}$$ For small $\theta$, the denominator is asymptotically 1.

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Moreover, this gives you improved approximations $\tan(2\theta) \approx 2 \tan(\theta) + 2 \tan^3(\theta) + \ldots + 2 \tan^{2k+1}(\theta)$ –  Robert Israel Oct 16 '12 at 5:59
    
Oooo, wow you're smart. –  Stuart Oct 16 '12 at 6:13

For any function $f\colon \mathbb R\to\mathbb R$ that is differentiable at $0$ and has $f(0)=0$, you have $f(t)\approx f'(0) t$ and hence $f(ct)\approx cf(t)$ for small $t$.

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Actually $\tan(\theta) \approx \theta$, so $\tan(2\theta) \approx 2\theta \approx 2\tan(\theta)$.

Any way, you can do (proof) it with Taylor expansion.

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