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MHTCET 19 Apr 2026 Shift 1 Paper

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Question : 118 of 150

Marks: +1, -0

\sum_{k=1}^{2026} \sin ^{-1}\left(\cos \frac{k \pi}{4}\right)=

$-\frac{\pi}{4}$
0
$\frac{\pi}{4}$
$\frac{\pi}{2}$

Solution:

Concept:

Use the principal value range of

\sin^{-1}x

, which is

\left[-\frac{\pi}{2},\frac{\pi}{2}\right]

, and the periodic nature of

\cos\frac{k\pi}{4}

.

Explanation:

The values of

\cos\frac{k\pi}{4}

repeat every 8 terms.

For

k=1,2,3,4,5,6,7,8

, the values of

\sin^{-1}\left(\cos\frac{k\pi}{4}\right)

are:

\frac{\pi}{4},\,0,\,-\frac{\pi}{4},\,-\frac{\pi}{2},\,-\frac{\pi}{4},\,0,\,\frac{\pi}{4},\,\frac{\pi}{2}

.

The sum of these 8 terms is

0

.

Now,

2026 = 8 \times 253 + 2

.

So the first

2024

terms form

253

complete cycles, giving sum

0

.

The remaining two terms correspond to

k \equiv 1,2 \pmod{8}

:

\sin^{-1}\left(\cos\frac{\pi}{4}\right) + \sin^{-1}\left(\cos\frac{2\pi}{4}\right) = \frac{\pi}{4} + 0 = \frac{\pi}{4}

.

Answer:

The required sum is

\frac{\pi}{4}

, which is Option C.

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