Simplify : $\sin^{-1}[\sin10]$

> [!TLDR] > Writes $10$ as $3\pi+\phi$ with $\phi$ small, uses $\sin(3\pi+\phi)=-\sin\phi$, and identifies the angle in $[-\pi/2,\pi/2]$ with the same sine. > > 1. The principal value of $\sin^{-1}$ lies in $\left[-\dfrac\pi2,\dfrac\pi2\right]$, but $10$ radians is far outside this range (since $10>\pi$), so we must find an angle in this range with the same sine as $10$. > 2. Write $10=3\pi+\phi$ where $\phi=10-3\pi\approx10-9.4248=0.5752$ rad. > 3. $\sin(10)=\sin(3\pi+\phi)=\sin3\pi\cos\phi+\cos3\pi\sin\phi=0\cdot\cos\phi+(-1)\sin\phi=-\sin\phi$ (using $\sin3\pi=0,\ \cos3\pi=-1$). > 4. So $\sin(10)=-\sin\phi=\sin(-\phi)$, since $\sin$ is an odd function. …

Simplify : sin⁻¹[sin10]

Question 70 of 71

\sin^{-1}[\sin10]

Puducherry TnboardTamil Nadu HSC (DGE) Board 2026Subjective· 2mImportance★★★★★

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Concept understanding — Principal Value Evaluation of Inverse Trigonometric Functions

The principal value of an inverse trigonometric function at a point $x$ is the value of the inverse function at $x$ that lies in its principal-value-branch range (the tables in the two concepts above). When solving $\text{(inverse function)}(x)=y$ , there may be infinitely many angles $\theta$ with the right ratio, but exactly one of them lies in the principal range — that one is the principal value.

Note

Tie-breaking rule. If two candidate values are numerically equal but opposite in sign (e.g. solving $\cos y=-\tfrac12$ might tempt $y=\pm\tfrac{2\pi}3$ ), the principal value is taken to be the positive one, subject to it actually lying in the function's principal range.

Reference table (principal domain → range of the inverse):

Function	Principal domain	Range	Inverse	Domain	Range of principal value
sine	$\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]$	$[-1,1]$	$\sin^{-1}$	$[-1,1]$	$\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]$
cosine	$[0,\pi]$	$[-1,1]$	$\cos^{-1}$	$[-1,1]$	$[0,\pi]$
tangent	$\left(-\tfrac{\pi}2,\tfrac{\pi}2\right)$	$\mathbb{R}$	$\tan^{-1}$	$\mathbb{R}$	$\left(-\tfrac{\pi}2,\tfrac{\pi}2\right)$
cosecant	$\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]\setminus\{0\}$	$\mathbb{R}\setminus(-1,1)$	$\text{cosec}^{-1}$	$\mathbb{R}\setminus(-1,1)$	$\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]\setminus\{0\}$
secant	$[0,\pi]\setminus\left\{\tfrac{\pi}2\right\}$	$\mathbb{R}\setminus(-1,1)$	$\sec^{-1}$	$\mathbb{R}\setminus(-1,1)$	$[0,\pi]\setminus\left\{\tfrac{\pi}2\right\}$
cotangent	$(0,\pi)$	$\mathbb{R}$	$\cot^{-1}$	$\mathbb{R}$	$(0,\pi)$

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Writes $10$ as $3\pi+\phi$ with $\phi$ small, uses $\sin(3\pi+\phi)=-\sin\phi$ , and identifies the angle in $[-\pi/2,\pi/2]$ with the same sine.

The principal value of $\sin^{-1}$ lies in $\left[-\dfrac\pi2,\dfrac\pi2\right]$ , but $10$ radians is far outside this range (since $10>\pi$ ), so we must find an angle in this range with the same sine as $10$ .
Write $10=3\pi+\phi$ where $\phi=10-3\pi\approx10-9.4248=0.5752$ rad.
$\sin(10)=\sin(3\pi+\phi)=\sin3\pi\cos\phi+\cos3\pi\sin\phi=0\cdot\cos\phi+(-1)\sin\phi=-\sin\phi$ (using $\sin3\pi=0,\ \cos3\pi=-1$ ).
So $\sin(10)=-\sin\phi=\sin(-\phi)$ , since $\sin$ is an odd function. …

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Q.Simplify : sin⁡−1[sin⁡10]\sin^{-1}[\sin10]sin−1[sin10]

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Q.Simplify : $\sin^{-1}[\sin10]$

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