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Let x,y,z belong [-1,1] be such that sin^-1 x+sin^-1 y+sin^-1 z=3pi/2 Find the value of , x^2018+y^2019+z^2020

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Solution: For any $xin[-1,1]$ , the maximum value of $sin^-1x$ is $pi /2$

and it attains this value at $x=1$

$herefore$ $sin^-1xleq pi /2$ , $sin^-1yleq pi /2$ , $sin^-1zleq pi /2$ $forall$ $x,y,zin[-1,1]$

$Rightarrow$ $sin^-1x+sin^-1y+sin^-1zleq 3pi /2$ $forall$ $x,y,zin[-1,1]$

$Rightarrow$ $sin^-1x+sin^-1y+sin^-1z=3pi /2$

$Rightarrow$ $sin^-1x=pi /2$ , $sin^-1y=pi /2$ , $sin^-1z=pi /2$

$Rightarrow$ $x=y=z=1$

$Rightarrow$ $x^2018+y^2019+z^2020=1^2018+1^2019+1^2020=3$

Posted by

Deependra Verma

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