Find the value of $\sin^{-1}\left[\sin\left(\dfrac{5\pi}{4}\right)\right]$.

> [!TLDR] > Rewrites $5\pi/4$ as an angle whose sine matches an angle inside the principal branch $[-\pi/2,\pi/2]$ of $\sin^{-1}$. > > 1. The principal value branch of $\sin^{-1}x$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, and $\sin^{-1}(\sin\theta)=\theta$ only when $\theta$ already lies in this interval. > 2. Here $\theta=\dfrac{5\pi}{4}$, which does NOT lie in $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, so we cannot cancel directly. > 3. Write $\dfrac{5\pi}{4}=\pi+\dfrac{\pi}{4}$, so $\sin\dfrac{5\pi}{4}=\sin\left(\pi+\dfrac{\pi}{4}\right)=-\sin\dfrac{\pi}{4}$ (using $\sin(\pi+\theta)=-\sin\theta$). > 4. Also $-\sin\dfrac{\pi}{4}=\sin\left(-\dfrac{\pi}{4}\right)$, so $\sin\dfrac{5\pi}{4}=\sin\left(-\dfrac{\pi}{4}\right)$. …

Find the value of sin⁻¹[sin((5π)/(4))].

Question 55 of 71

\sin^{-1}\left[\sin\left(\dfrac{5\pi}{4}\right)\right]

Puducherry TnboardTamil Nadu HSC (DGE) Board 2020Subjective· 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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Rewrites $5\pi/4$ as an angle whose sine matches an angle inside the principal branch $[-\pi/2,\pi/2]$ of $\sin^{-1}$ .

The principal value branch of $\sin^{-1}x$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ , and $\sin^{-1}(\sin\theta)=\theta$ only when $\theta$ already lies in this interval.
Here $\theta=\dfrac{5\pi}{4}$ , which does NOT lie in $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ , so we cannot cancel directly.
Write $\dfrac{5\pi}{4}=\pi+\dfrac{\pi}{4}$ , so $\sin\dfrac{5\pi}{4}=\sin\left(\pi+\dfrac{\pi}{4}\right)=-\sin\dfrac{\pi}{4}$ (using $\sin(\pi+\theta)=-\sin\theta$ ).
Also $-\sin\dfrac{\pi}{4}=\sin\left(-\dfrac{\pi}{4}\right)$ , so $\sin\dfrac{5\pi}{4}=\sin\left(-\dfrac{\pi}{4}\right)$ . …

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Q.Find the value of sin⁡−1[sin⁡(5π4)]\sin^{-1}\left[\sin\left(\dfrac{5\pi}{4}\right)\right]sin−1[sin(45π​)].

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Q.Find the value of $\sin^{-1}\left[\sin\left(\dfrac{5\pi}{4}\right)\right]$ .

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