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Pi calculation.......

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Can you please explain why the following identity is true? Where does it come from: pi/24 = cot^(-1)(2+sqrt(2)+sqrt(3)+sqrt(6)). Thanks for any help.

Dec 23, 2015

#1
+111321
+15

cot-1( 2 + sqrt(2) + sqrt(3) + sqrt(6) )  = pi/24

cot [ pi/24]  = 2 + sqrt(3) + sqrt(2) + sqrt(6)

cot [pi/24]  =

cot [(1/2)pi/12]] =

1/ tan [(1/2)pi/12)] =

1/ [sqrt ( 1 - cos(pi/12))/ (1 + cos(pi/12)]  =

sqrt( 1 + cos(pi/12)) / sqrt( 1 - cos(pi/12) ) =  [multiply top/bottom  by sqrt( 1 + cos(pi/12)) ]

[1 + cos (pi/12)] / sqrt( 1 - cos^2(pi/12)) =

[1 + cos(pi/12)] / sqrt( sin^2(pi/12)) =

[1 + cos(pi/12) ] / sin(pi/12)  =

1/sin(pi/12) + cos(pi/12) / sin(pi/12)  =

csc[pi/12] + cot[pi/12] =

1/ sin [ pi/12] + 1 / [tan [pi/12]  =

1 / [ sin[ pi/4 - pi/6]] + 1 / [ tan[ pi/4 - pi/6] ]

1/ [ sin(pi/4)cos(pi/6) - sin(pi/6)cos(pi/4)]  +  1/ ( [tan(pi/4] - tan[pi/6]] / [1 + tan(pi/4)tan(pi/6)] )  =

1 / [ sqrt(2)sqrt(3)/4] - (sqrt(2)/4) ] + [ 1 + 1/sqrt(3)]/ [1 - 1/sqrt(3)]  =

4 / [ sqrt(6) - sqrt(2)]  +  [ (sqrt(3) + 1)/ sqrt(3))]   / ( [ sqrt(3) - 1]/ sqrt(3) )

4 [ sqrt(6) + sqrt(2)]/ 4  +  [ sqrt(3) + 1] / [sqrt(3) - 1]  =  [ multiply the second term by [sqrt(3) + 1] on top/bottom ]

sqrt(6) + sqrt(2)  + [sqrt(3) + 1]^2 / 2  =

sqrt(6) + sqrt(2) + [3 + 2sqrt(3) + 1] / 2  =

sqrt(6) + sqrt(2) + [ 4 + 2 sqrt(3)] / 2  =

sqrt(6) + sqrt(2) + 2 + sqrt(3)  =

2 + sqrt(3) + sqrt(2) + sqrt(6)    ..........and the left side  =  the right side

Dec 23, 2015
edited by CPhill  Dec 23, 2015
edited by CPhill  Dec 23, 2015
edited by CPhill  Dec 23, 2015

#1
+111321
+15

cot-1( 2 + sqrt(2) + sqrt(3) + sqrt(6) )  = pi/24

cot [ pi/24]  = 2 + sqrt(3) + sqrt(2) + sqrt(6)

cot [pi/24]  =

cot [(1/2)pi/12]] =

1/ tan [(1/2)pi/12)] =

1/ [sqrt ( 1 - cos(pi/12))/ (1 + cos(pi/12)]  =

sqrt( 1 + cos(pi/12)) / sqrt( 1 - cos(pi/12) ) =  [multiply top/bottom  by sqrt( 1 + cos(pi/12)) ]

[1 + cos (pi/12)] / sqrt( 1 - cos^2(pi/12)) =

[1 + cos(pi/12)] / sqrt( sin^2(pi/12)) =

[1 + cos(pi/12) ] / sin(pi/12)  =

1/sin(pi/12) + cos(pi/12) / sin(pi/12)  =

csc[pi/12] + cot[pi/12] =

1/ sin [ pi/12] + 1 / [tan [pi/12]  =

1 / [ sin[ pi/4 - pi/6]] + 1 / [ tan[ pi/4 - pi/6] ]

1/ [ sin(pi/4)cos(pi/6) - sin(pi/6)cos(pi/4)]  +  1/ ( [tan(pi/4] - tan[pi/6]] / [1 + tan(pi/4)tan(pi/6)] )  =

1 / [ sqrt(2)sqrt(3)/4] - (sqrt(2)/4) ] + [ 1 + 1/sqrt(3)]/ [1 - 1/sqrt(3)]  =

4 / [ sqrt(6) - sqrt(2)]  +  [ (sqrt(3) + 1)/ sqrt(3))]   / ( [ sqrt(3) - 1]/ sqrt(3) )

4 [ sqrt(6) + sqrt(2)]/ 4  +  [ sqrt(3) + 1] / [sqrt(3) - 1]  =  [ multiply the second term by [sqrt(3) + 1] on top/bottom ]

sqrt(6) + sqrt(2)  + [sqrt(3) + 1]^2 / 2  =

sqrt(6) + sqrt(2) + [3 + 2sqrt(3) + 1] / 2  =

sqrt(6) + sqrt(2) + [ 4 + 2 sqrt(3)] / 2  =

sqrt(6) + sqrt(2) + 2 + sqrt(3)  =

2 + sqrt(3) + sqrt(2) + sqrt(6)    ..........and the left side  =  the right side

CPhill Dec 23, 2015
edited by CPhill  Dec 23, 2015
edited by CPhill  Dec 23, 2015
edited by CPhill  Dec 23, 2015
#2
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Thank you CPhill. Brilliant!.

Dec 23, 2015