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How can I calculate the sine of a googolplex minus 10 degrees?

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

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    $\begingroup$ Of course, this only holds for $n > 2$, but I am confident the problem at hand satisfies this condition. $\endgroup$ – Thomas Jan 24 '13 at 5:58
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    $\begingroup$ @Thomas: I didn't want to do all the work. :) $\endgroup$ – Blue Jan 24 '13 at 9:27
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    $\begingroup$ +1 for leaving the proof that $10^{100}>2$ to the OP. The comments made me laugh, thanks Thomas and Blue! $\endgroup$ – Jonas Meyer Jan 28 '13 at 19:51
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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 $$ \sin (10^{10^{100}}-10)^\circ = \sin 270^\circ = -1. $$

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  • $\begingroup$ How do we know $40\mid 10^{10^{100}}$? $\endgroup$ – Peter Olson Jan 24 '13 at 2:24
  • $\begingroup$ @PeterOlson: $10=2\cdot 5$ and $40=2^3\cdot 5$. $\endgroup$ – Jonas Meyer Jan 24 '13 at 2:24
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    $\begingroup$ @PeterOlson Because $40 \mid 10^3$... $\endgroup$ – Erick Wong Jan 24 '13 at 2:25
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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:

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

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