If sin−1x+sin−1y+sin−1z=23π, then (x500+y500+z500)−(x501+y501+z501) is

We know that the range of sin−1x is [−2π,2π]Given:sin−1x+sin−1y+sin−1z=23π⇒sin−1x+sin−1y+sin−1z=2π+2π+2π⇒sin−1x=2πsin−1y=2πsin−1z=2π⇒x=sin2π,y=sin2π,z=sin2π⇒x=1,y=1,z=1(x500+y500+z500)−(x501+y501+z501)=(1500+1500+1500)−(1501+1501+1501)=(1+1+1)−(1+1+1)=3−3=0

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If ^{-1}x + ^{-1}y + ^{-1}z = {3}{2}, then (x^{500} + y^{50

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If $sin^{- 1} x + sin^{- 1} y + sin^{- 1} z = \frac{3 π}{2}$ , then $(x^{500} + y^{500} + z^{500}) - (x^{501} + y^{501} + z^{501})$ is

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Lucas Discussed

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

, then

(x^{500} + y^{500} + z^{500}) - (x^{501} + y^{501} + z^{501})

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Text solutionVerified

We know that the range of

sin^{- 1} x

[- \frac{π}{2}, \frac{π}{2}]

Given:

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

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

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

\Rightarrow x = sin \frac{π}{2}, y = sin \frac{π}{2}, z = sin \frac{π}{2}

\Rightarrow x = 1, y = 1, z = 1

(x^{500} + y^{500} + z^{500}) - (x^{501} + y^{501} + z^{501})

= (1^{500} + 1^{500} + 1^{500}) - (1^{501} + 1^{501} + 1^{501})

= (1 + 1 + 1) - (1 + 1 + 1) = 3 - 3 = 0

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Question Text	If $sin^{- 1} x + sin^{- 1} y + sin^{- 1} z = \frac{3 π}{2}$ , then $(x^{500} + y^{500} + z^{500}) - (x^{501} + y^{501} + z^{501})$ is
Answer Type	Text solution:1
Upvotes	150

If sin−1x+sin−1y+sin−1z=23π​, then (x500+y500+z500)−(x501+y501+z501) is

Text solutionVerified

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If $sin^{- 1} x + sin^{- 1} y + sin^{- 1} z = \frac{3 π}{2}$ , then $(x^{500} + y^{500} + z^{500}) - (x^{501} + y^{501} + z^{501})$ is