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Mathematics 13 Online
OpenStudy (anonymous):

the expression sin^2 thita =( x^2+y^2)/ 2xy is possible if a) x=y b) x>y c) x=-y d) x

OpenStudy (souvik):

\[-1\le \sin \theta \le 1\] \[\sin ^{2} \theta \le 1\]

OpenStudy (anonymous):

that means \[\sin^2 \le 1 \] \[x^2 + y^2 /2xy \le 1 \]

OpenStudy (anonymous):

\[x^2+ y^2 \le 2xy \] then ?

OpenStudy (souvik):

\[x ^{2}-2xy+y ^{2}\le0\]

OpenStudy (souvik):

\[(x-y)^{2}\le0\] a square quantity can not be less than zero... so (x-y)^2=0

OpenStudy (anonymous):

that the expression holds true if x=y right !

OpenStudy (souvik):

yes..

OpenStudy (anonymous):

okei thank you very much ! @souvik

OpenStudy (souvik):

you are welcome..:)

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