Why does $\tan(\theta)\sin^2(\theta) = \theta^{3}$? I'm reviewing a solution of an assignment I've got in my physics II class, and at one point the manual says this:
because $\sin^2(\theta)\tan(\theta) << 1:$
$\sin(\theta)\approx\tan(\theta)\approx\theta$
$\theta^3=5.73\text{x}10^{-3}$
$\theta=0.179\text{rad}$
Plugging that value back into the formula gives:
$\sin^2(0.179)\tan(0.179)=0.00573574$
but I still don't understand how they came up with that. I get that if $\sin(\theta)$ and $\tan(\theta)$ are equal to $\theta$, $\sin^2(\theta)\tan(\theta)$ is $\theta^2\theta$, which is $\theta^3$, but $\sin(\theta)$ and $\tan(\theta)$ are not just magically equal to their angles. My next best guess is that it's saying because $\sin^2(\theta)\tan(\theta)$ will always be less that one, then sin and tan are equal to their angles, but that does not make any sense to me either. Thanks in advance for any help you can give me to help me understand this.
 A: You already received the explanation : close to $x=0$, both $\sin(x)$ and $\tan(x)$ are $\simeq x$.
Let me go a bit further, using more terms in Taylor expansion $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^6\right)$$ $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$ $$\sin^2(x)=x^2-\frac{x^4}{3}+\frac{2 x^6}{45}+O\left(x^7\right)$$ $$\tan(x)\sin^2(x)=x^3+\frac{x^7}{15}+O\left(x^8\right)\approx x^3\left(1+\frac{x^4}{15}\right)$$ You see how small is the error. Just for fun, plot the function and its approximations for $0 \leq x \leq 1$; you will be surprized to notice how close  are the two curves.
A: They’re not saying that they’re equal to the angle, only that they’re approximately equal.  
The Taylor series for $\sin\theta$ around $0$ is $\theta+\frac{\theta^3}3+\dots$, and that for $\tan\theta$ is $\theta+\frac{\theta^3}6+\dots$, so for very small $\theta$ we can drop the higher-order terms as an approximation. If $\sin^2\theta\tan\theta$ is small, then so is $\theta$, so you can use these approximations.
A: This is because
of the only
trig limit that matters:
$\lim_{x \to 0} \frac{\sin x}{x}
= 1
$.
This follows from the expansion
$\sin(x)
=x+\frac{x^3}{6}+...
$.
From this you can deduce that
$\lim_{x \to 0} \frac{1-\cos(x)}{x^2}
=\frac12
$
and
$\lim_{x \to 0} \frac{\tan x}{x}
= 1
$.
