+0

# Pi calculation.......

0
819
2

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.

Guest Dec 23, 2015

#1
+81027
+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
Sort:

#1
+81027
+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
0

Thank you CPhill. Brilliant!.

Guest Dec 23, 2015

### 11 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details