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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

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

Similar Questions

If $\sin^{-1}(\tan \frac{\pi}{4}) - \sin^{-1}(\sqrt{\frac{3}{x}}) = \frac{\pi}{6}$, then $x$ is a root of the equation:

The equation $\cos ^{-1}(1-x)-2 \cos ^{-1} x=\frac{\pi}{2}$ has

The sum of the maximum and the minimum values of $2(\cos ^{-1} x)^2-\pi \cos ^{-1} x+\frac{\pi^2}{4}$ is

For $\alpha, \beta, \gamma \neq 0$. If $\sin ^{-1} \alpha+\sin ^{-1} \beta+\sin ^{-1} \gamma=\pi$ and $(\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3 \alpha \beta$,then $\gamma$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module