Q.Find the principal value of the following: $\sin^{-1}\left(-\frac{1}{2}\right)$

Concept: reduce the angle modulo 2pi, then use sin^-1(sin theta) = theta for theta in [-pi/2, pi/2].

39 pi / 5 = 7.8 pi. Subtract 6 pi (three full turns, which does not change the cosine):

39pi/5 - 30pi/5 = 9pi/5.

cos(9pi/5) = cos(2pi - pi/5) = cos(pi/5).

Now write cos(pi/5) as a sine:

cos(pi/5) = sin(pi/2 - pi/5) = sin(5pi/10 - 2pi/10) = sin(3pi/10).

Since 3pi/10 = 54 degrees lies in [-pi/2, pi/2], the principal value is

sin^-1( sin(3pi/10) ) = 3pi/10.

ANSWER: B

The core idea here is that the expression inside the cosine is a sum of an inverse sine and an inverse cosine of the same number. There is a fundamental relationship between $\sin^{-1}x$ and $\cos^{-1}x$: for any $x$ in $[-1, 1]$, they add up to $\frac{\pi}{2}$. That’s the key that collapses the problem instantly.

Once you see that, the rest is just evaluating $\cos\left(\frac{\pi}{2} + \sin^{-1}\frac{3}{4}\right)$ — but wait, we need to be careful: the given expression is $\cos\left[2\sin^{-1}\frac{3}{4} + \cos^{-1}\frac{3}{4}\right]$, not $\cos\left[\sin^{-1}\frac{3}{4} + \cos^{-1}\frac{3}{4}\right]$. So we have an extra $\sin^{-1}\frac{3}{4}$ inside. Let’s handle it step by step.

1. Apply the inverse identity For $x = \frac{3}{4}$, which lies in $[-1, 1]$, we know:

This is a standard result: the sum of an angle and its complementary angle (in the sense of sine and cosine) is a right angle.

2. Rewrite the argument The argument of the cosine is:

Substitute the sum from step 1:

3. Evaluate the cosine Now we need:

Use the cosine addition formula: $\cos\left(\frac{\pi}{2} + \theta\right) = -\sin\theta$.

So:

That gives $-\frac{3}{4}$, which is option (A). But wait — is that the final answer? Let’s check domain restrictions carefully.

4. Re-examine the domain of $\cos^{-1}\frac{3}{4}$

   $\cos^{-1}\frac{3}{4}$ is defined and its principal value lies in $[0, \pi]$. Specifically, $\cos^{-1}\frac{3}{4}$ is an angle in $(0, \frac{\pi}{2})$ because $\frac{3}{4} > 0$. Similarly, $\sin^{-1}\frac{3}{4}$ is in $(0, \frac{\pi}{2})$. So $2\sin^{-1}\frac{3}{4} + \cos^{-1}\frac{3}{4}$ is a sum of positive angles. Let’s check if it exceeds $\pi$?

   $\sin^{-1}\frac{3}{4} \approx 0.848$ rad, so $2 \times 0.848 = 1.696$ rad, and $\cos^{-1}\frac{3}{4} \approx 0.723$ rad. Sum $\approx 2.419$ rad, which is less than $\pi$ (3.1416). So the argument is within $[0, \pi]$, and cosine is defined. So the calculation stands.

   But the options include "does not exist". Why? Possibly because the problem expects you to notice that the expression simplifies to $-\frac{3}{4}$, which is indeed option (A). However, let’s verify if there’s any hidden issue: the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ is valid for $x \in [-1,1]$, but note that $\cos^{-1}x$ is defined to give values in $[0, \pi]$, and $\sin^{-1}x$ in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. For $x = \frac{3}{4}$, both are positive, so the sum is $\frac{\pi}{2}$. No problem.

   So the answer should be $-\frac{3}{4}$. But the problem lists (D) "does not exist". Could it be that the expression inside the cosine is not defined? No, it is defined. Perhaps the trick is that $\cos^{-1}\frac{3}{4}$ is not the same as $\arccos\frac{3}{4}$? It is.

   Let’s check the possibility: sometimes, in Indian exams, they consider that $\cos^{-1}x$ is defined only for $x \in [-1,1]$, which is fine. So I suspect the intended answer is indeed $-\frac{3}{4}$, and option (D) is a distractor. But the problem explicitly says "does not exist" — maybe they want you to realize that the argument simplifies to $\frac{\pi}{2} + \sin^{-1}\frac{3}{4}$, and then $\cos(\frac{\pi}{2} + \theta) = -\sin\theta$, which is $-\frac{3}{4}$. So (A) is correct.

   However, to be thorough: is there any chance that $\sin^{-1}\frac{3}{4} + \cos^{-1}\frac{3}{4} = \frac{\pi}{2}$ is not true for principal values? It is always true. So the calculation is solid.

Thus the answer is $-\frac{3}{4}$, which corresponds to option (A).

Step 1 — Innermost block.

Then

Step 2 — The periodic sine factor.

using the $2\pi$-periodicity of the sine function. So this factor is simply $1$ and leaves the rest of the argument unchanged.

Step 3 — Collapse the outer bracket.

With the sine factor equal to $1$, the bracket reduces to the coefficient $\dfrac{1}{\sqrt3}$:

Step 4 — Evaluate the principal value.

so $\tan^{-1}\dfrac{1}{\sqrt3}=\dfrac{\pi}{6}$. (Option (D), $\pi$, is not even in the principal range of $\tan^{-1}$, so it can be rejected outright; $\pi/3$ would need the argument $\sqrt3$, and $0$ would need argument $0$.)

ANSWER: B

The key here is to check the domain of the inverse trigonometric functions before doing anything else. Many students jump straight into using identities like $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$, but that identity only holds when $x$ is in the domain of both functions.

For $\sin^{-1}x$, the domain is $[-1, 1]$. For $\cos^{-1}x$, the domain is also $[-1, 1]$. The given input is $\frac{\pi}{3} \approx 1.047$, which is greater than 1. So neither $\sin^{-1}\frac{\pi}{3}$ nor $\cos^{-1}\frac{\pi}{3}$ is a real number. The expression is therefore undefined.

Let’s walk through it step by step.

1. Check the domain of $\sin^{-1}x$

   The inverse sine function $\sin^{-1}x$ (also written as $\arcsin x$) is defined only for $x \in [-1, 1]$.

   Here $x = \frac{\pi}{3} \approx 1.047 > 1$, so $\sin^{-1}\frac{\pi}{3}$ does not exist as a real number.

2. Check the domain of $\cos^{-1}x$

   Similarly, $\cos^{-1}x$ (or $\arccos x$) is defined only for $x \in [-1, 1]$.

   Again $x = \frac{\pi}{3} > 1$, so $\cos^{-1}\frac{\pi}{3}$ also does not exist.

3. Consequence for the sum

   Since both terms are undefined, their sum $\sin^{-1}\frac{\pi}{3} + \cos^{-1}\frac{\pi}{3}$ is undefined.

   Therefore $\cos\left(\sin^{-1}\frac{\pi}{3} + \cos^{-1}\frac{\pi}{3}\right)$ is also undefined.

Step 1 — Substitute. Let

Since $\theta \in [0,\pi]$, $\sin\theta \ge 0$, so

Step 2 — Simplify the right-hand side.

So the equation becomes

Step 3 — Apply the principal-value restriction. The identity $\sin^{-1}(\sin\alpha) = \alpha$ holds only when $\alpha$ lies in the principal range of $\sin^{-1}$:

Here $\alpha = 2\theta$, and $2\theta \ge 0$, so the requirement is

Step 4 — Translate back to $x$. $\cos$ is decreasing on $[0, \pi]$, so $0 \le \theta \le \frac{\pi}{4}$ gives

Step 5 — Verify with a test value. Take $x = 1/\sqrt2$ ($\theta = \pi/4$): LHS $= 2(\pi/4) = \pi/2$; RHS $= \sin^{-1}\!\left(2\cdot\frac{1}{\sqrt2}\cdot\frac{1}{\sqrt2}\right) = \sin^{-1}(1) = \pi/2$ ✓.

Now take $x = 0$ ($\theta = \pi/2$), which lies in options (A), (B), (C): LHS $= \pi$, RHS $= \sin^{-1}(0) = 0$ ✗ — so every interval containing $x=0$ is wrong. This confirms (D).

ANSWER: D