For x-0-2 sin - 1cos x-

The correct option is B π2 x We have, sin #10;^ 1( cos x ) #10; #10; =sin #10;^ 1( sin( π2 x ) ) #10; #10; =π #10;2 x #10; ...

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Properties Derived from Trigonometric Identities

For x∈0,π/2...

Question

For $x \in (0, π / 2)$
${sin}^{- 1} (cos x) = ?$

π - x

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\frac{π}{2} - x

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π - \frac{x}{2}

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π - 2 x

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Solution

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A

0 < x < \frac{π}{2}

{sin}^{- 1} (c o s x) + {cos}^{- 1} (s i n x) = π - 2 x

{cos}^{- 1} x = \frac{π}{2} - {sin}^{- 1} x \forall x \in [0, 1]

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

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} x + a^{2} \sqrt{2} s i n x, & 0 \leq x < \frac{π}{4} x c o t x + b, & \frac{π}{4} \leq x < \frac{π}{2} b s i n 2 x - a c o s 2 x, & \frac{π}{2} \leq x \leq π \end{matrix}

[0, π]

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

g (x) = f (tan x) + f (cot x)

x \in (\frac{π}{2}, π)

f^{''} (x) < 0 \forall x \in (\frac{π}{2}, π),

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