Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

3 Answers 3

up vote 15 down vote accepted

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

share|improve this answer
4  
Of course, this only holds for $n > 2$, but I am confident the problem at hand satisfies this condition. –  Thomas Jan 24 '13 at 5:58
4  
@Thomas: I didn't want to do all the work. :) –  Blue Jan 24 '13 at 9:27
5  
+1 for leaving the proof that $10^{100}>2$ to the OP. The comments made me laugh, thanks Thomas and Blue! –  Jonas Meyer Jan 28 '13 at 19:51
add comment

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 $$ \sin (10^{10^{100}}-10)^\circ = \sin 270^\circ = -1. $$

share|improve this answer
    
How do we know $40\mid 10^{10^{100}}$? –  Peter Olson Jan 24 '13 at 2:24
    
@PeterOlson: $10=2\cdot 5$ and $40=2^3\cdot 5$. –  Jonas Meyer Jan 24 '13 at 2:24
5  
@PeterOlson Because $40 \mid 10^3$... –  Erick Wong Jan 24 '13 at 2:25
add comment

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:

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.