Sin- - 3 -sin -1 - 1 2 - -

The correct option is A 1 Range of sin ^ 1 (x) is [ π nbsp; / 2 ,π nbsp; / 2 ] ∴ nbsp; sin ^ 1 ( 1 / 2 ) = π nbsp; / 6 ...

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Standard XII

Mathematics

Cos(A+B)Cos(A-B)

sin[ π 3 -sin...

Question

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

\frac{1}{2}

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\frac{1}{3}

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\frac{1}{4}

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1

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Solution

The correct option is A $1$
Range of ${s i n}^{- 1} (x)$ is $[- \frac{π}{2}, \frac{π}{2}]$
$∴ {s i n}^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$
Now, the equation reduces to
$s i n (\frac{π}{3} - (- \frac{π}{6})) = s i n (\frac{3 π}{6}) = s i n (\frac{π}{2}) = 1$
Option D

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Similar questions

Q. Find the value of

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

Q. Evaluate

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

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}]

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. If

{sin}^{- 1} \frac{3}{x} + {sin}^{- 1} \frac{4}{x} = \frac{π}{2}

, then

x

is equal to

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sin[π3−sin−1(−12)]=

The correct option is A 1Range of sin−1(x) is [−π2,π2]∴sin−1(−12)=−π6Now, the equation reduces tosin(π3−(−π6))=sin(3π6)=sin(π2)=1Option D

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

The correct option is A $1$
Range of ${s i n}^{- 1} (x)$ is $[- \frac{π}{2}, \frac{π}{2}]$
$∴ {s i n}^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$
Now, the equation reduces to
$s i n (\frac{π}{3} - (- \frac{π}{6})) = s i n (\frac{3 π}{6}) = s i n (\frac{π}{2}) = 1$
Option D