Evaluate the sum: sum_{k=1}^{2026} sin^{-1}(c | vedclass

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(C) Let $f(k) = \sin^{-1}(\cos \frac{k\pi}{4})$. Using the identity $\sin^{-1}(\cos x) = \frac{\pi}{2} - x$ for $x \in [0, \pi]$, we note that the fun

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$\frac{\pi}{4}$

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(C) Let $f(k) = \sin^{-1}(\cos \frac{k\pi}{4})$. Using the identity $\sin^{-1}(\cos x) = \frac{\pi}{2} - x$ for $x \in [0, \pi]$, we note that the function $f(k)$ is periodic with period $8$ because $\cos(\frac{(k+8)\pi}{4}) = \cos(\frac{k\pi}{4} + 2\pi) = \cos(\frac{k\pi}{4})$.Sum of one period $S_8 = \sum_{k=1}^{8} \sin^{-1}(\cos \frac{k\pi}{4})$.Calculating values for $k=1$ to $8$: $\sin^{-1}(\cos \frac{\pi}{4}) = \frac{\pi}{4}$, $\sin^{-1}(\cos \frac{2\pi}{4}) = 0$, $\sin^{-1}(\cos \frac{3\pi}{4}) = -\frac{\pi}{4}$, $\sin^{-1}(\cos \pi) = -\frac{\pi}{2}$, $\sin^{-1}(\cos \frac{5\pi}{4}) = -\frac{\pi}{4}$, $\sin^{-1}(\cos \frac{6\pi}{4}) = 0$, $\sin^{-1}(\cos \frac{7\pi}{4}) = \frac{\pi}{4}$, $\sin^{-1}(\cos 2\pi) = \frac{\pi}{2}$.Sum $S_8 = \frac{\pi}{4} + 0 - \frac{\pi}{4} - \frac{\pi}{2} - \frac{\pi}{4} + 0 + \frac{\pi}{4} + \frac{\pi}{2} = 0$.Since $2026 = 8 	imes 253 + 2$, the total sum is $253 	imes S_8 + f(1) + f(2) = 0 + \frac{\pi}{4} + 0 = \frac{\pi}{4}$.

Evaluate the sum: $\sum_{k=1}^{2026} \sin^{-1}(\cos \frac{k\pi}{4})$.

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