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amardeepreddy (0)

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distance between two points {a cos (+2/3), asin (+2/3) }  and { a cos (+/3) , a sin(+/3)} is ??????????????  plz explain it................
    
sankydreams (1005)

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first of all
theta + 2pie/3
= theta +120
=theta + 90 + 30
cos(theta + 30 +90) = -sin(theta +30)
sin(theta + 30 +90) = cos(theta + 30)

using distance formula
dist = sqrt of [{acos(theta + 30) - acos(theta + 60)}^2 + {asin(theta + 30) - asin(theta + 60)}^2]
take a^2 common from both brackets
also let theta +30 =A and theta + 60 =B

dist = a^2 * sqrt of [(cosA - cos B)^2 + (sinA - sin B)^2]
= a^2 * sqrt of [2-2( sinAsinB + cosAcosB ) ]
= a^2 * sqrt of[2 -2(cos (A - B))]
now cos(A-B) = cos(-30)=cos30=(sqrt of 3)/2
put value and get the answer
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elessar_iitkgp (2390)

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The polar coordinates of the two points is (a, +2/3) and (a,+/3)
d = r12 + r22 - 2r1r2cos(1-2) = a2+a2 - 2a2cos/3 = a



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elessar_iitkgp (2390)

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If you are not familiar with polar coordinates and this formula, pick up SL Loney. The proof of this result is given somewhere in the begining.



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