What is the principal value of sin -1 sin 2- 3 -

The correct option is C π3 sin^ 1 (sin (2π3 ) ) as 2π3ϵ [π2, 3π2 ] ∴sin^ 1 (sin x) = π x in xϵ [π2, 3π2 ] sin^ 1 (sin (2π3 ) ...

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Byju's Answer

Standard XII

Mathematics

Domain and Range of Basic Inverse Trigonometric Functions

What is the p...

Question

What is the principal value of ${sin}^{- 1} (sin \frac{2 π}{3})$ ?

\frac{π}{4}

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\frac{π}{2}

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\frac{π}{3}

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\frac{2 π}{3}

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Solution

The correct option is C $\frac{π}{3}$
${sin}^{- 1} (sin (\frac{2 π}{3}))$ as $\frac{2 π}{3} ϵ [\frac{π}{2}, \frac{3 π}{2}]$
$∴ {sin}^{- 1} (sin x) = π - x$ in $x ϵ [\frac{π}{2}, \frac{3 π}{2}]$
${sin}^{- 1} (sin (\frac{2 π}{3})) = π - \frac{2 π}{3} = \frac{π}{3}$ .

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Similar questions

Q. Assertion :

{sin}^{- 1} (sin (\frac{2 π}{3})) = \frac{π}{3}

Reason:

{sin}^{- 1} (sin θ) = θ;

where

θ \in [- \frac{π}{2}, \frac{π}{2}]

If $z = \frac{- 2}{1 + i \sqrt{3}}$ , then the value of arg(z) is

Inverse circular functions,Principal values of ${sin}^{- 1} x, {c o s}^{- 1} x, {tan}^{- 1} x$ .
${tan}^{- 1} x + {tan}^{- 1} y = {tan}^{- 1} \frac{x + y}{1 - x y}$ , $x y < 1$
$π + {tan}^{- 1} \frac{x + y}{1 - x y}$ , $x y > 1$ .
Evaluate the following :
(a) $sin [\frac{π}{3} - {sin}^{- 1} (- \frac{1}{2})]$
(b) $sin [\frac{π}{2} - {sin}^{- 1} (- \frac{\sqrt{3}}{2})]$

Q. (i) Evaluate

{sin}^{- 1} (sin \frac{2 π}{3})

(ii) Show that:

{tan}^{- 1} (\frac{cos x}{1 - sin x}) = \frac{π}{4} - \frac{x}{2}

Q. Match the equation in

z

C o l u m n - I

with the corresponding value of

a r g (z)

c o l u m n - I I

$C o l u m n - I$ (equation in z)	$C o l u m n - I I$ (principal value of arg(z))
$z^{2} - z + 1 = 0$	$- 2 π / 3$
$z^{2} + z + 1 = 0$	$- π / 3$
${2 z}^{2} + 1 + i \sqrt{3} = 0$	$π / 3$
${2 z}^{2} + 1 - i \sqrt{3} = 0$	$2 π / 3$

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What is the principal value of sin−1(sin2π3)?

The correct option is C π3sin−1(sin(2π3)) as 2π3ϵ[π2,3π2]∴sin−1(sinx)=π−x in xϵ[π2,3π2]sin−1(sin(2π3))=π−2π3=π3.

What is the principal value of ${sin}^{- 1} (sin \frac{2 π}{3})$ ?

The correct option is C $\frac{π}{3}$
${sin}^{- 1} (sin (\frac{2 π}{3}))$ as $\frac{2 π}{3} ϵ [\frac{π}{2}, \frac{3 π}{2}]$
$∴ {sin}^{- 1} (sin x) = π - x$ in $x ϵ [\frac{π}{2}, \frac{3 π}{2}]$
${sin}^{- 1} (sin (\frac{2 π}{3})) = π - \frac{2 π}{3} = \frac{π}{3}$ .