If 4 sin-1x-cos-1x - - then x is equal to

We know that 4 sin^ 1x + cos^ 1x = π ⇒ 3 sin^ 1x + sin^ 1x+ cos^ 1x = π ⇒ x = π π/2 = π/2 ⇒ sin^ 1x = π/6⇒ x = sin π/6 = 1/2. ...

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Properties Derived from Trigonometric Identities

If 4 sin-1x+c...

Question

$I f 4 s i n^{- 1} x + c o s^{- 1} x = π,$ then x is equal to

0

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\frac{1}{2}

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- \frac{\sqrt{3}}{2}

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\frac{1}{\sqrt{2}}

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\frac{1}{2}

\frac{\sqrt{3}}{2}

- \frac{1}{2}

- \frac{\sqrt{3}}{2}

s i n^{- 1} x - c o s^{- 1} x = \frac{π}{6}

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

{cos}^{- 1} x + {cos}^{- 1} y

A = \frac{1}{π} ⎡ ⎢ ⎢ ⎣ \begin{matrix} s i n^{- 1} (π x) & t a n^{- 1} (\frac{x}{π}) s i n^{- 1} (\frac{x}{π}) & c o t^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎦, B = ⎡ ⎢ ⎢ ⎣ \begin{matrix} - c o s^{- 1} (π x) & t a n^{- 1} (\frac{x}{π}) s i n^{- 1} (\frac{x}{π}) & t a n^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎦

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{s i n}^{- 1} (3 x / 5) + {s i n}^{- 1} (4 x / 5) = {s i n}^{- 1} x

{c o s}^{- 1} x + {s i n}^{- 1} (\frac{1}{2} x) = \frac{π}{6}

a \leq {t a n}^{- 1} (\frac{1 - x}{1 + x}) \leq b

0 \leq x \leq 1

(a, b) =

(0, π)

(0, π / 4)

(- π / 4, π / 4)

(π / 4, π / 2)

a \leq {({s i n}^{- 1} x)}^{3} + {({c o s}^{- 1} x)}^{3} \leq b

(\frac{π^{3}}{32}, \frac{7 π^{3}}{8})

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