The value of sin -2 - sin-1 -32 - is

The correct option is A 12 We have sin [π2 sin^ 1 ( √(3)2 ) ] = sin [π2 + sin^ 1 (√(3)2 ) ] = sin [π2 + π3 ] = sin [5π6 ] = cos (π3 ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Basic Trigonometric Identities

The value of ...

Question

The value of $sin [\frac{π}{2} - {sin}^{- 1} (- \frac{\sqrt{3}}{2})]$ is

\frac{1}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- \frac{1}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{1}{2}$
We have $sin [\frac{π}{2} - {sin}^{- 1} (- \frac{\sqrt{3}}{2})]$
$= sin [\frac{π}{2} + {sin}^{- 1} (\frac{\sqrt{3}}{2})]$
$= sin [\frac{π}{2} + \frac{π}{3}] = sin [\frac{5 π}{6}]$
$= cos (\frac{π}{3}) = \frac{1}{2}$ .

Suggest Corrections

Similar questions

Q. If

{sin}^{- 1} \frac{1}{2} + {sin}^{- 1} \frac{2}{3} = {sin}^{- 1} x

, then the value of

x

is-

Q. Find the value of

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

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) $sin [\frac{π}{2} - {sin}^{- 1} (- \frac{\sqrt{3}}{2})]$

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

Q. Find the value of

cos [\frac{π}{2} - {sin}^{- 1} (\frac{1}{3})]

Join BYJU'S Learning Program

The value of sin[π2−sin−1(−√32)] is

The correct option is A 12We have sin[π2−sin−1(−√32)]=sin[π2+sin−1(√32)]=sin[π2+π3]=sin[5π6]=cos(π3)=12.

The value of $sin [\frac{π}{2} - {sin}^{- 1} (- \frac{\sqrt{3}}{2})]$ is

The correct option is A $\frac{1}{2}$
We have $sin [\frac{π}{2} - {sin}^{- 1} (- \frac{\sqrt{3}}{2})]$
$= sin [\frac{π}{2} + {sin}^{- 1} (\frac{\sqrt{3}}{2})]$
$= sin [\frac{π}{2} + \frac{π}{3}] = sin [\frac{5 π}{6}]$
$= cos (\frac{π}{3}) = \frac{1}{2}$ .