Find the following- sin- - 2 -sin -1 - 3 2 -

We have, sin[π2 sin^ 1( √(3)2)] =sin[π2 sin^ 1(sin(4π3))] =sin[π2 4π3] =sin[3π 8π6] =sin[ 5π6] = sin[5π6] = 12 Hence, this is ...

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Byju's Answer

Standard XII

Mathematics

Domain

Find the foll...

Question

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

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Solution

We have,
$sin [\frac{π}{2} - {sin}^{- 1} (\frac{- \sqrt{3}}{2})]$
$= sin [\frac{π}{2} - {sin}^{- 1} (sin (\frac{4 π}{3}))]$
$= sin [\frac{π}{2} - \frac{4 π}{3}]$
$= sin [\frac{3 π - 8 π}{6}]$
$= sin [\frac{- 5 π}{6}]$
$= - sin [\frac{5 π}{6}]$
$= - \frac{1}{2}$

Hence, this is the answer.

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Q. Find the simplified value of the following expressions

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

Q. Find the value of

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

Q. Solve the following :
sin

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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})] =

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Find the following:sin[π2−sin−1(−√32)]

We have,sin[π2−sin−1(−√32)]=sin[π2−sin−1(sin(4π3))]=sin[π2−4π3]=sin[3π−8π6]=sin[−5π6]=−sin[5π6]=−12Hence, this is the answer.

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

We have,
$sin [\frac{π}{2} - {sin}^{- 1} (\frac{- \sqrt{3}}{2})]$
$= sin [\frac{π}{2} - {sin}^{- 1} (sin (\frac{4 π}{3}))]$
$= sin [\frac{π}{2} - \frac{4 π}{3}]$
$= sin [\frac{3 π - 8 π}{6}]$
$= sin [\frac{- 5 π}{6}]$
$= - sin [\frac{5 π}{6}]$
$= - \frac{1}{2}$

Hence, this is the answer.