Question

If $S_n = \sum_{r=1}^{n} r!$ then for $n > 6$ then $\sin^{-1} \left( \sin \left(S_n -7 \left[ \frac{S_n}{7} \right] \right) \right)$ is:

• A. 5 - 2\pi
• B. 2\pi - 5
• C. 6 - 2\pi
• D. \pi - 4

Answer 1

Answer: (A)

5 - 2\pi

Answer 2

The expression $S_n - 7 \left[ \frac{S_n}{7} \right]$ represents the remainder when $S_n$ is divided by 7. For $n > 6$, any term $r!$ where $r \ge 7$ is divisible by 7, meaning $r! \equiv 0 \pmod{7}$. Therefore, for $n > 6$, $S_n \equiv 1! + 2! + 3! + 4! + 5! + 6! \pmod{7}$.

Calculating the values modulo 7: $1! \equiv 1 \pmod{7}$ $2! \equiv 2 \pmod{7}$ $3! \equiv 6 \pmod{7}$ $4! = 24 \equiv 3 \pmod{7}$ $5! = 120 \equiv 1 \pmod{7}$ $6! = 720 \equiv 6 \pmod{7}$

Summing these remainders: $S_n \equiv 1 + 2 + 6 + 3 + 1 + 6 = 19 \equiv 5 \pmod{7}$. So, the expression simplifies to $\sin^{-1}(\sin(5))$.

The principal value range for $\sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. We need to find a value $y$ such that $\sin(y) = \sin(5)$ and $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. We know that $\sin(x) = \sin(x - 2k\pi)$ for any integer $k$. Let $k=1$, then we consider $5 - 2\pi$. Approximating $\pi \approx 3.14159$, we get $2\pi \approx 6.28318$. So, $5 - 2\pi \approx 5 - 6.28318 = -1.28318$. The range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is approximately $[-1.5708, 1.5708]$. Since $-1.5708 < -1.28318 < 1.5708$, the value $5 - 2\pi$ lies within the principal range. Therefore, $\sin^{-1}(\sin(5)) = 5 - 2\pi$.