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GUJCET 2026 Math Paper

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Question : 13 of 40

Marks: +1, -0

\sin^{-1} (\sin \frac{3\pi}{5}) = \dots \dots \dots

$\frac{2\pi}{5}$
$\frac{\pi}{5}$
$\frac{3\pi}{5}$
$\frac{4\pi}{5}$

Solution:

Concept:

The inverse sine function gives an angle from its principal range

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

.

Explanation:

Here

\frac{3\pi}{5}

is greater than

\frac{\pi}{2}

, so it is outside the principal range.

Use the identity

\sin(\pi-\theta)=\sin\theta

.

\sin\left(\frac{3\pi}{5}\right)=\sin\left(\pi-\frac{3\pi}{5}\right)=\sin\left(\frac{2\pi}{5}\right)

.

Since

\frac{2\pi}{5}

lies in

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

,

\sin^{-1}\left(\sin\frac{2\pi}{5}\right)=\frac{2\pi}{5}

.

Answer:

\sin^{-1}\left(\sin\frac{3\pi}{5}\right)=\frac{2\pi}{5}

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