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Basic Inverse Trigonometric Functions

If x ∈ 3π/2...

Question

If $x \in (\frac{3 π}{2}, 2 π)$ then the value of the expression $s i n^{- 1} [c o s (c o s^{- 1} (c o s x) + s i n^{- 1} (s i n x))]$ , is

- \frac{π}{2}

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\frac{π}{2}

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0

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π

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Solution

The correct option is C $\frac{π}{2}$
$∵ x \in (\frac{3 π}{2}, 2 π)$

Now, $c o s^{- 1} (c o s x) = 2 π - x$

and $s i n^{- 1} (s i n x) = x - 2 π$

$∴ c o s^{- 1} (c o s x) + s i n^{- 1} (s i n x) = 0$

Therefore,
$s i n^{- 1} [c o s {c o s^{- 1} (c o s x) + s i n^{- 1} (s i n x)}]$
$= s i n^{- 1} {c o s (0)} = s i n^{- 1} (1) = \frac{π}{2}$

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A

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