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

Showing the 12 most recent of 51 on this concept.

1. CBSE 2025Set 65/4/11 markMCQ
   Q.The principal value of $\sin^{-1}\left(\sin\left(-\frac{10\pi}{3}\right)\right)$ is : (A) $-\frac{2\pi}{3}$ (B) $-\frac{\pi}{3}$ (C) $\frac{\pi}{3}$ (D) $\frac{2\pi}{3}$

   ›Reveal solutionSolution

   To find the principal value of $\sin^{-1}(\sin \theta)$, we must ensure the angle $\theta$ lies within the principal value range of $\sin^{-1}(x)$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. By adjusting the given angle $-\frac{10\pi}{3}$ to an equivalent angle within this range, we find the principal value is $\boxed{\frac{\pi}{3}}$.

   The problem asks for the principal value of $\sin^{-1}\left(\sin\left(-\frac{10\pi}{3}\right)\right)$. This involves understanding the definition of the inverse sine function and its principal value branch.

   The inverse sine function, $\sin^{-1}(x)$ (also written as $\arcsin(x)$), gives an angle whose sine is $x$. For $\sin^{-1}(x)$ to be a function, its range must be restricted. By convention, the principal value branch of $\sin^{-1}(x)$ is defined such that its output angle lies in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

   This means that for an expression like $\sin^{-1}(\sin \theta)$, the result is not always simply $\theta$. It is $\theta$ only if $\theta$ itself is already within the principal value range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. If $\theta$ is outside this range, we need to find an equivalent angle $\alpha$ such that $\sin \alpha = \sin \theta$ and $\alpha \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Then, $\sin^{-1}(\sin \theta) = \sin^{-1}(\sin \alpha) = \alpha$.

   Let's apply this concept step-by-step:

   1. Identify the principal value range for $\sin^{-1}(x)$:

      The principal value of $\sin^{-1}(x)$ must lie in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. This is equivalent to angles from $-90^\circ$ to $90^\circ$.

   2. Analyze the inner angle:

      The given angle inside the sine function is $-\frac{10\pi}{3}$.

      We need to evaluate $\sin\left(-\frac{10\pi}{3}\right)$.

      To simplify this, we can add or subtract multiples of $2\pi$ (a full rotation) to find a coterminal angle that is easier to work with.

      $-\frac{10\pi}{3} = -\frac{10\pi}{3} + 4\pi$ (since $4\pi = \frac{12\pi}{3}$)

      $= \frac{-10\pi + 12\pi}{3} = \frac{2\pi}{3}$.

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

      Watch out

      A common mistake is to directly write $\sin^{-1}\left(\sin\left(-\frac{10\pi}{3}\right)\right) = -\frac{10\pi}{3}$. This is incorrect because $-\frac{10\pi}{3}$ (which is $-600^\circ$) is not in the principal value range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (which is $[-90^\circ, 90^\circ]$).

   3. Find an equivalent angle within the principal value range:

      Now we need to find the principal value of $\sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right)$.

      The angle $\frac{2\pi}{3}$ (which is $120^\circ$) is still not in the principal value range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

      We know that $\sin(\pi - \theta) = \sin \theta$. We can use this identity to find an angle in the first quadrant (or fourth quadrant for negative values) that has the same sine value.

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

      Now the expression becomes $\sin^{-1}\left(\sin\left(\frac{\pi}{3}\right)\right)$.

   4. Determine the principal value:

      The angle $\frac{\pi}{3}$ (which is $60^\circ$) is within the principal value range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

      Therefore, $\sin^{-1}\left(\sin\left(\frac{\pi}{3}\right)\right) = \frac{\pi}{3}$.

      For the principal value branch of $\sin^{-1}(x)$:

      $\sin^{-1}(\sin \theta) = \theta$, if $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

      If $\theta \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, find an angle $\alpha \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ such that $\sin \alpha = \sin \theta$. Then $\sin^{-1}(\sin \theta) = \alpha$.

   The final result is $\frac{\pi}{3}$.

   ✓Final answer

   The principal value of $\sin^{-1}\left(\sin\left(-\frac{10\pi}{3}\right)\right)$ is $\boxed{\frac{\pi}{3}}$.

2. CBSE 2025Set 65/2/11 markMCQ
   Q.If $y = \sin^{-1}x$, $-1 \le x \le 0$, then the range of $y$ is: (A) $\left(-\dfrac{\pi}{2}, 0\right)$ (B) $\left[-\dfrac{\pi}{2}, 0\right]$ (C) $\left[-\dfrac{\pi}{2}, 0\right)$ (D) $\left(-\dfrac{\pi}{2}, 0\right]$

   ›Reveal solutionSolution

   For the inverse sine function, the principal value range is $[-\pi/2, \pi/2]$. When $x$ is restricted to $[-1, 0]$, $y$ takes values from $-\pi/2$ up to $0$, including both endpoints. The correct answer is (B).

   The key to this problem is understanding what "principal value" means for inverse trigonometric functions. Unlike regular sine, which is periodic and not one-to-one, $\sin^{-1}x$ (also written as $\arcsin x$) is defined as the inverse of the sine function only on a carefully chosen interval where sine is one-to-one. That interval is $[-\pi/2, \pi/2]$.

   So by definition, for any $x$ in $[-1, 1]$, the value $y = \sin^{-1}x$ is always the unique angle in $[-\pi/2, \pi/2]$ whose sine is $x$. This is the principal value branch — it's not a choice; it's the definition.

   Now the question gives you a further restriction: $x$ is only between $-1$ and $0$. You're being asked: as $x$ runs through that half of the domain, what part of the principal range does $y$ cover?

   Let's work through it.

   1. Recall the principal range of $\sin^{-1}x$.

      The output $y$ always lies in $[-\pi/2, \pi/2]$. That's the full range for the full domain $[-1, 1]$.

   2. Identify the endpoints for $x$ in $[-1, 0]$.

      • When $x = -1$, $y = \sin^{-1}(-1)$. What angle in $[-\pi/2, \pi/2]$ has sine equal to $-1$? That's $-\pi/2$.
      • When $x = 0$, $y = \sin^{-1}(0)$. The angle in $[-\pi/2, \pi/2]$ with sine $0$ is $0$.

   3. Check monotonicity.

      The function $\sin^{-1}x$ is strictly increasing on $[-1, 1]$. So as $x$ increases from $-1$ to $0$, $y$ increases from $-\pi/2$ to $0$. Since the function is continuous and strictly increasing, it hits every value between $-\pi/2$ and $0$.

   4. Are the endpoints included?

      Yes — both $x = -1$ and $x = 0$ are in the given domain $[-1, 0]$, so their corresponding $y$ values $-\pi/2$ and $0$ are both attained. Therefore the range is the closed interval $[-\pi/2, 0]$.

   Watch out

   A common mistake is to think that because $x$ is negative, $y$ must also be negative but not include $0$. But $x = 0$ is explicitly allowed, and $\sin^{-1}(0) = 0$ is in the principal range. So $0$ is included. Similarly, $-\pi/2$ is included because $x = -1$ is allowed.

   Tip

   If you ever forget the principal range, just remember: for $\sin^{-1}x$, the output is always an angle from $-\pi/2$ (straight down) to $\pi/2$ (straight up), inclusive. For $\cos^{-1}x$, it's $0$ to $\pi$. For $\tan^{-1}x$, it's $(-\pi/2, \pi/2)$.

   ✓Final answer

   The correct option is (B) $\left[-\dfrac{\pi}{2}, 0\right]$.

3. CBSE 2020Set 65/1/11 markMCQ
   Q.The principal value of $\tan^{-1} \left(\tan \frac{3\pi}{5}\right)$ is (A) $\frac{2\pi}{5}$ (B) $-\frac{2\pi}{5}$ (C) $\frac{3\pi}{5}$ (D) $-\frac{3\pi}{5}$

   ›Reveal solutionSolution

   The principal value of $\tan^{-1}(\tan x)$ is the unique angle in $(-\pi/2, \pi/2)$ that has the same tangent as $x$. Since $\frac{3\pi}{5}$ lies outside this interval, we shift it by $\pi$ to get $\frac{3\pi}{5} - \pi = -\frac{2\pi}{5}$, which falls inside the principal range. The answer is $-\frac{2\pi}{5}$, option (B).

   The function $\tan^{-1}(\tan x)$ is not simply $x$ — that would be too easy. The catch is that $\tan^{-1}$ (also written as $\arctan$) is defined to return only the principal value, which lies strictly between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. But $\tan x$ is periodic with period $\pi$, so many different angles give the same tangent value. The job of $\tan^{-1}(\tan x)$ is to pick the one angle in that narrow interval $(-\pi/2, \pi/2)$ whose tangent matches $\tan x$.

   So the question becomes: given $x = \frac{3\pi}{5}$, which angle in $(-\pi/2, \pi/2)$ has the same tangent?

   Let’s work it out.

   1. Check where $\frac{3\pi}{5}$ lies.

      $\frac{3\pi}{5} = 0.6\pi$ radians, which is $108^\circ$. This is in the second quadrant (between $\pi/2$ and $\pi$). Clearly, $108^\circ$ is outside the principal range $(-90^\circ, 90^\circ)$.

   2. Use the periodicity of $\tan$.

      The tangent function repeats every $\pi$ radians: $\tan(\theta + \pi) = \tan \theta$. So if we subtract $\pi$ from $\frac{3\pi}{5}$, we get an angle with the same tangent:

   $\frac{3\pi}{5} - \pi = \frac{3\pi}{5} - \frac{5\pi}{5} = -\frac{2\pi}{5}$

   3. Check if this new angle is in the principal range.

      $-\frac{2\pi}{5} = -72^\circ$, which lies between $-\frac{\pi}{2}$ ($-90^\circ$) and $\frac{\pi}{2}$ ($90^\circ$). Yes — it’s inside the interval.

   4. Therefore, the principal value is $-\frac{2\pi}{5}$.

      No other shift (like adding $\pi$) would land inside $(-\pi/2, \pi/2)$. Adding $\pi$ gives $\frac{8\pi}{5}$, which is $> \pi/2$, and subtracting $2\pi$ gives $-\frac{7\pi}{5}$, which is $< -\pi/2$. Only subtracting $\pi$ once works.

   Watch out

   A common mistake is to think $\tan^{-1}(\tan x) = x$ for all $x$. That’s only true when $x$ is already in $(-\pi/2, \pi/2)$. For angles outside this range, you must adjust by adding or subtracting multiples of $\pi$ until you land inside the principal interval.

   Tip

   For $\tan^{-1}(\tan x)$, the rule is: keep subtracting (or adding) $\pi$ until the angle falls into $(-\pi/2, \pi/2)$. That’s your answer. For $\frac{3\pi}{5}$, one subtraction of $\pi$ does it.

   ✓Final answer

   The principal value is $-\frac{2\pi}{5}$, which corresponds to option (B).

4. CBSE 2026Set A1 markMCQ
   Q.$\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right) =$

   (a) $\pi$

   (b) $\frac{2\pi}{3}$

   (c) $\frac{3\pi}{4}$

   (d) $\frac{\pi}{2}$

   ›Reveal solutionSolution

   The sum equals $\frac{3\pi}{4}$.

   Evaluate each principal value:

   • $\tan^{-1}(1) = \frac{\pi}{4}$.
   • $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ (range $[0,\pi]$).
   • $\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$ (range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$).

   Add, using LCD 12:

   $\frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}.$

   ✓Final answer

   (c) $\frac{3\pi}{4}$.

5. CBSE 2026Set ANNUAL1 markMCQ
   Q.Write the value of $\cos^{-1}\left(\cos \frac{13\pi}{6}\right)$.

   (a) $\frac{13\pi}{6}$

   (b) $\frac{7\pi}{6}$

   (c) $\frac{5\pi}{6}$

   (d) $\frac{\pi}{6}$

   ›Reveal solutionSolution

   $\cos^{-1}\left(\cos\frac{13\pi}{6}\right)=\frac{\pi}{6}$.

   The principal value branch of $\cos^{-1}$ is $[0,\pi]$. To evaluate $\cos^{-1}\left(\cos\frac{13\pi}{6}\right)$ we must first reduce $\frac{13\pi}{6}$ to an angle whose cosine is the same and which lies in $[0,\pi]$.

   $\frac{13\pi}{6}=2\pi+\frac{\pi}{6}$

   Since cosine has period $2\pi$:

   $\cos\frac{13\pi}{6}=\cos\left(2\pi+\frac{\pi}{6}\right)=\cos\frac{\pi}{6}$

   Now $\frac{\pi}{6}$ already lies inside $[0,\pi]$, so

   $\cos^{-1}\left(\cos\frac{13\pi}{6}\right)=\cos^{-1}\left(\cos\frac{\pi}{6}\right)=\frac{\pi}{6}$

   ✓Final answer

   The correct option is (d) $\frac{\pi}{6}$.

6. CBSE 2026Set ANNUAL1 markMCQ
   Q.The principal value of $\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$ is

   (a) $\pi/4$

   (b) $\pi/6$

   (c) $\pi/3$

   (d) $\pi/2$

   ›Reveal solutionSolution

   The principal value branch of $\sin^{-1}$ is $[-\pi/2,\pi/2]$; the angle in this range whose sine is $1/\sqrt2$ is $\pi/4$.

   We need $\theta \in [-\pi/2,\pi/2]$ such that $\sin\theta = \dfrac{1}{\sqrt2}$.

   Since $\sin(\pi/4) = \dfrac{1}{\sqrt2}$ and $\pi/4$ lies in the principal value range, $\sin^{-1}\left(\dfrac{1}{\sqrt2}\right) = \pi/4$.

   ✓Final answer

   The correct option is (a) $\pi/4$.

7. CBSE 2026Set ANNUAL1 markMCQ
   Q.sin⁻¹(sin(2π/3)) is equal to

   (a) 2π/3

   (b) π/3

   (c) −π/3

   (d) None of the above

   ›Reveal solutionSolution

   $2\pi/3$ lies outside the principal range $[-\pi/2,\pi/2]$ of $\sin^{-1}$, so we must find the angle inside that range with the same sine value.

   $\sin(2\pi/3) = \sin(\pi-\pi/3) = \sin(\pi/3) = \dfrac{\sqrt3}{2}$.

   Since $\pi/3 \in [-\pi/2,\pi/2]$, and $\sin^{-1}$ always returns the angle in its principal range whose sine equals the given value,

   $\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \sin^{-1}\left(\frac{\sqrt3}{2}\right) = \frac{\pi}{3}.$

   ✓Final answer

   $\pi/3$. (Option b)

8. CBSE 2026Set ANNUAL1 markMCQ
   Q.The principal value of $\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$ is

   (a) $-\frac{\pi}{4}$

   (b) $\frac{\pi}{3}$

   (c) $\frac{\pi}{6}$

   (d) $\frac{\pi}{4}$

   ›Reveal solutionSolution

   $\sin^{-1}\dfrac{1}{\sqrt2}=\dfrac\pi4$.

   The principal value lies in $\left[-\dfrac\pi2,\dfrac\pi2\right]$. Since $\sin\dfrac\pi4=\dfrac{1}{\sqrt2}$, the principal value is $\dfrac\pi4$.

   ✓Final answer

   (d) $\dfrac{\pi}{4}$.

9. CBSE 2026Set ANNUAL1 mark
   Q.Write the answer in one word/sentence: Write the principal value branches (Range) of $\sin^{-1} x$.

   ›Reveal solutionSolution

   The principal-value range of $\sin^{-1}x$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$.

   To make sine invertible it is restricted to $\left[-\dfrac\pi2,\dfrac\pi2\right]$, on which it is one-one and onto $[-1,1]$; hence this is the principal value branch (range) of $\sin^{-1}x$.

   ✓Final answer

   $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$.

10. CBSE 2026Set ANNUAL1 markMCQ
    Q.$\sin\left[\dfrac{\pi}{2}-\sin^{-1}\left(-\dfrac{\sqrt{3}}{2}\right)\right]$ is equal to

    (a) $1$

    (b) $\dfrac{1}{3}$

    (c) $-1$

    (d) $\dfrac{1}{2}$

    ›Reveal solutionSolution

    $\sin\!\left[\tfrac{\pi}{2}-\sin^{-1}\!\left(-\tfrac{\sqrt3}{2}\right)\right]=\cos\!\left(\sin^{-1}\!\left(-\tfrac{\sqrt3}{2}\right)\right)=\tfrac12$.

    Step 1: $\sin\left(\tfrac{\pi}{2}-\theta\right)=\cos\theta$, so the expression equals $\cos\!\left(\sin^{-1}\!\left(-\tfrac{\sqrt3}{2}\right)\right)$.

    Step 2: Let $\theta=\sin^{-1}\!\left(-\tfrac{\sqrt3}{2}\right)$. Since the range of $\sin^{-1}$ is $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$, $\theta=-\tfrac{\pi}{3}$.

    Step 3: $\cos\left(-\tfrac{\pi}{3}\right)=\cos\tfrac{\pi}{3}=\tfrac12$.

    ✓Final answer

    $\dfrac{1}{2}$ — option (D).

11. CBSE 2025Set X11 markMCQ
    Q.The principal value of $\sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right)$ is

    (a) $\dfrac{\pi}{2}$

    (b) $\dfrac{\pi}{3}$

    (c) $\dfrac{\pi}{4}$

    (d) $\dfrac{\pi}{6}$

    ›Reveal solutionSolution

    Principal value of an inverse-sine — correct option is (c).

    The principal value branch of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, so we need the angle in this interval whose sine is $\dfrac{1}{\sqrt2}$. Since $\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}$ and $\dfrac{\pi}{4}$ lies in the principal range, $\sin^{-1}\left(\dfrac{1}{\sqrt2}\right)=\dfrac{\pi}{4}$.

    ✓Final answer

    (c) $\dfrac{\pi}{4}$

12. CBSE 2025Set E1 markMCQ
    Q.$\sin\left(\sin^{-1}\frac{2\pi}{3}\right) + \tan^{-1}\left(\tan\frac{3\pi}{4}\right) =$

    (a) $\frac{17\pi}{12}$

    (b) $\frac{5}{12}\pi$

    (c) $\frac{\pi}{12}$

    (d) $-\frac{\pi}{12}$

    ›Reveal solutionSolution

    Apply the inverse cancellation and reduce the second term to its principal range; result $\frac{5\pi}{12}$.

    First term: Using $\sin(\sin^{-1}\theta)=\theta$ as intended by the paper, $\sin\left(\sin^{-1}\frac{2\pi}{3}\right)=\frac{2\pi}{3}$. (Strictly, $\frac{2\pi}{3}>1$ is outside the domain of $\sin^{-1}$, but the question intends the direct cancellation.)

    Second term: $\frac{3\pi}{4}$ is not in the principal range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ of $\tan^{-1}$. Since $\tan\frac{3\pi}{4}=\tan\left(\frac{3\pi}{4}-\pi\right)=\tan\left(-\frac{\pi}{4}\right)$,

    $\tan^{-1}\left(\tan\frac{3\pi}{4}\right)=-\frac{\pi}{4}.$

    Add:

    $\frac{2\pi}{3}-\frac{\pi}{4}=\frac{8\pi-3\pi}{12}=\frac{5\pi}{12}.$

    ✓Final answer

    (B) $\frac{5\pi}{12}$.