Solving a Complicated Trig Problem I am preparing for AIME and I came across this problem which I need help solving:
$$\begin{eqnarray} 10^{10^{10}} \sin\left( \frac{109}{10^{10^{10}}} \right) - 9^{9^{9}} \sin\left( \frac{101}{9^{9^{9}}} \right) - 8^{8^{8}} \sin\left( \frac{17}{8^{8^{8}}} \right) + 7^{7^{7}} \sin\left( \frac{76}{7^{7^{7}}} \right) + 6^{6^{6}} \sin\left( \frac{113}{6^{6^{6}}} \right) \end{eqnarray} $$
There must be some simple way to do this for sure, but I don't know how to.

Since $\sin x\approx x$ for small $x$, this should be pretty darn close to $$109−101−17+76+113=180.$$
How close? By far the largest error comes from the $6^{6^6}\sin\frac{113}{6^{6^6}}$ term, because that argument to the sine is hugely larger than the other arguments (power towers grow fast). And we can estimate the relative difference between $\sin\frac{113}{6^{6^6}}$ and $\frac{113}{6^{6^6}}$ by the mean value theorem: the relative error is between $0$ and $1-\cos\frac{113}{6^{6^6}}$. And we can again estimate the cosine as $1-\cos x\approx \frac12x^2$ for small $x$, so the absolute difference from 180 should be close to
$$ 113 \cdot \frac12 \cdot \frac{113^2}{6^{2\cdot 6^6}} \approx \frac{113^3}{2\cdot 6^{2\cdot 6^6}} $$
where the $6^{2\cdot 6^6}$ in the denominator dominates absolutely everything.
Since $\log_{10}(6^{2\cdot 6^6}) = \log_{10}(6)\cdot 2\cdot 6^6 > 6^6 = 46656$, the answer is
$$ 179.\underbrace{999999\ldots999999}_{\text{at least 46,650 nines}}\ldots $$
