Q.The value of $\sin^{-1}\left(\cos\frac{33\pi}{5}\right)$ is
(A) $\frac{3\pi}{5}$
(B) $\frac{-7\pi}{5}$
(C) $\frac{\pi}{10}$
(D) $\frac{-\pi}{10}$

Concept: Inverse sine principal value — $\sin^{-1}(x)$ returns an angle in $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Step 1: Simplify the inner angle.

$\frac{33\pi}{5} = 6\pi + \frac{3\pi}{5}$. Since $\cos$ has period $2\pi$,

$\cos\frac{33\pi}{5} = \cos\frac{3\pi}{5}$.

Step 2: Express cosine as sine of a complementary angle.

$\cos\frac{3\pi}{5} = \sin\left(\frac{\pi}{2} - \frac{3\pi}{5}\right) = \sin\left(-\frac{\pi}{10}\right)$.

Step 3: Check principal value range. …

We need to evaluate $\sin^{-1}\left(\cos\frac{33\pi}{5}\right)$. The inverse sine function, $\sin^{-1}(x)$, returns an angle $\theta$ such that $\sin\theta = x$ and $\theta \in [-\pi/2, \pi/2]$. This restricted range is the principal value branch. So our job is not just to find any angle whose sine equals $\cos(33\pi/5)$, but the unique one inside that interval.

The first instinct is to simplify the inner cosine. Since $33\pi/5$ is a large angle, we reduce it using the periodicity of cosine: $\cos(\theta + 2\pi n) = \cos\theta$. Let's find an equivalent angle between $0$ and $2\pi$.

1. Reduce the angle modulo $2\pi$

   $33\pi/5 = 6.6\pi$. Subtract multiples of $2\pi = 10\pi/5$:

   $33\pi/5 - 2\pi \cdot 3 = 33\pi/5 - 30\pi/5 = 3\pi/5$.

   So $\cos\frac{33\pi}{5} = \cos\frac{3\pi}{5}$.

2. Convert cosine to sine using a complementary angle identity

   We have $\cos\frac{3\pi}{5} = \sin\left(\frac{\pi}{2} - \frac{3\pi}{5}\right) = \sin\left(\frac{5\pi}{10} - \frac{6\pi}{10}\right) = \sin\left(-\frac{\pi}{10}\right)$.

   So $\cos\frac{33\pi}{5} = \sin\left(-\frac{\pi}{10}\right)$.

3. Check if this angle lies in the principal range of $\sin^{-1}$

   The principal range is $[-\pi/2, \pi/2]$. Here $-\pi/10$ is about $-18^\circ$, which is well inside that interval.

   Therefore, $\sin^{-1}\left(\sin\left(-\frac{\pi}{10}\right)\right) = -\frac{\pi}{10}$. …

Method: Evaluating $\sin^{-1}(\cos\alpha)$ for a large angle $\alpha$

The goal is to end with an angle inside $\sin^{-1}$'s principal range $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. Reduce the big angle, convert cosine to sine, then read off the principal value.

Steps

Step 1: Reduce the inner angle modulo $2\pi$.

Subtract whole multiples of $2\pi$ from $\alpha$ using periodicity, $\cos(\alpha+2\pi n)=\cos\alpha$, to bring it into $[0,2\pi)$. This does not change the cosine value.

Step 2: Convert cosine into a sine.

Since the outer function is $\sin^{-1}$, rewrite the cosine as a sine using

Now the expression reads $\sin^{-1}(\sin\beta)$ for some angle $\beta$.

Step 3: Force $\beta$ into the principal range. …

Common Mistakes

Mistake 1: Feeding a cosine value straight into $\sin^{-1}$ without converting.

Why it's wrong: $\sin^{-1}$ expects a sine value; leaving it as $\cos\tfrac{3\pi}{5}$ leads nowhere. Correct approach: convert with $\cos\phi=\sin\left(\tfrac{\pi}{2}-\phi\right)$ first, then invert.

Mistake 2: Not reducing $\tfrac{33\pi}{5}$ modulo $2\pi$ before working.

Why it's wrong: leaving a large angle invites arithmetic slips. Correct approach: $\tfrac{33\pi}{5}-6\pi=\tfrac{3\pi}{5}$, using $\cos$'s $2\pi$ period. …