How can I calculate $\sin\left(10^{10^{100}} - 10\right)^\circ$?

Since $360=40\cdot 9$, $\gcd(40,9)=1$, $$40\mid 10^{10^{100}},$$ and $$10^{10^{100}}\equiv 1^{10^{100}}\equiv 1 \pmod 9,$$ by the Chinese Remainder Theorem, the residue of $10^{10^{100}}$ modulo $360$ will be the unique residue congruent to $0$ modulo $40$ and $1$ modulo $9$. This is $280$, so $$ \sin (10^{10^{100}}-10)^\circ = \sin 270^\circ = -1. $$

$$\begin{align} \underbrace{1000\ldots000}_{n \; \text{"}0\text{"s}} - 10 = 999\ldots990 = 90 \cdot \underbrace{11\ldots111}_{n-1\;\text{"}1\text{"s}} &= 90 \cdot (11+100 \cdot (\ldots)) \\&= 90 \cdot (3+4\cdot (\ldots)) \\&= 270+360 \cdot (\ldots) \end{align}$$

Consider the sequence $\mu_{n + 1} \equiv 10 \mu_{n} \pmod{360}$ with $\mu_0 = 1$, thus $\mu_n \equiv 10^n \pmod{360}$.

Therefore, $\mu_{n + 3} \equiv 1000 \mu_n \equiv 280 \mu_n \pmod{360}$.

But then, notice that:

$$\mu_{n + 4} \equiv 10 \mu_{n + 3} \equiv 10 \cdot 280 \mu_n \equiv 2800 \mu_n \equiv 280 \mu_n \equiv \mu_{n + 3} \pmod{360}$$

Indeed, this sequence reaches a steady state starting at $n = 3$. Thus, we conclude that:

$$\mu_n \equiv 10^n \equiv 280 \pmod{360} ~ ~ ~ ~ ~ ~ ~ \text{for} ~ n > 2$$

From the above result, it follows that for $n > 2$, the statement below holds:

$$\sin{(10^n - 10)} = \sin{\left ( (10^n - 10) ~ \text{mod} ~ 360 \right )} = \sin{\left ( 280 - 10 \right )} = \sin{(270)} = -1$$