Question

\(\sum _{k = 1}^{2026}sin^{-1}(cos\frac{kπ}{4}) =\)

• A. \(-\frac{π}{4}\)
• B. \(0\)
• C. \(\frac{π}{4}\)
• D. \(\frac{π}{2}\)

Hint:

The terms repeat with period 8 and each block of 8 sums to zero.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
\(\sin^{-1}(\cos\theta)\) returns a value in \([-\pi/2,\pi/2]\). We evaluate it for \(\theta=k\pi/4\), which repeats every 8 values of \(k\).

Step 2: One Full Period:
\(k=1\): \(\cos\frac\pi4=\frac{\sqrt2}{2}\), so the term is \(\frac\pi4\).
\(k=2\): \(\cos\frac\pi2=0\), term \(0\).
\(k=3\): \(\cos\frac{3\pi}4=-\frac{\sqrt2}2\), term \(-\frac\pi4\).
\(k=4\): \(\cos\pi=-1\), term \(-\frac\pi2\).
\(k=5\): \(\cos\frac{5\pi}{4}=-\frac{\sqrt2}2\), term \(-\frac\pi4\).
\(k=6\): \(\cos\frac{3\pi}2=0\), term \(0\).
\(k=7\): \(\cos\frac{7\pi}4=\frac{\sqrt2}2\), term \(\frac\pi4\).
\(k=8\): \(\cos2\pi=1\), term \(\frac\pi2\).

Step 3: Sum of a Period:
\[ \frac\pi4+0-\frac\pi4-\frac\pi2-\frac\pi4+0+\frac\pi4+\frac\pi2=0 \]

Step 4: Use the Period:
\(2026=8\times253+2\). The first 2024 terms add up to 0. The last two terms are those of \(k\equiv1,2\): \(\frac\pi4+0=\frac\pi4\).
Options (A) and (B) would need those leftover terms to cancel, and (D) would need an extra \(\pi/4\).

Final Answer:
The sum is \(\dfrac\pi4\), option (C). \[ \boxed{\text{(C) } \frac{\pi}{4}} \]