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[Explore Stack Internal](https://stackoverflow.co/internal/?utm_medium=referral&utm_source=math-community&utm_campaign=side-bar&utm_content=explore-teams-compact-popover) # [Method to find sin(2π/7) sin ⁡ ( 2 π / 7 )](https://math.stackexchange.com/questions/311781/method-to-find-sin-2-pi-7) [Ask Question](https://math.stackexchange.com/questions/ask) Asked 13 years ago Modified [7 years, 8 months ago](https://math.stackexchange.com/questions/311781/method-to-find-sin-2-pi-7?lastactivity "2018-07-02 22:26:33Z") Viewed 4k times This question shows research effort; it is useful and clear 7 Save this question. Show activity on this post. I just thought a way to find sin2π7 sin ⁡ 2 π 7. Considering the equation x7\=1 x 7 \= 1 ⇒(x−1)(x6\+x5\+x4\+x3\+x2\+x\+1)\=0 ⇒ ( x − 1 ) ( x 6 \+ x 5 \+ x 4 \+ x 3 \+ x 2 \+ x \+ 1 ) \= 0 ⇒(x−1)\[(x\+1x)3\+(x\+1x)2−2(x\+1x)−1\]\=0 ⇒ ( x − 1 ) \[ ( x \+ 1 x ) 3 \+ ( x \+ 1 x ) 2 − 2 ( x \+ 1 x ) − 1 \] \= 0 We can then get the 7 solutions of x, but the steps will be very complicated, especially when solving cubic equation, and expressing x as a+bi. The imaginary part of the second root of x will be sin2π7 sin ⁡ 2 π 7. Besides this troublesome way, are any other approach? Thank you. - [trigonometry](https://math.stackexchange.com/questions/tagged/trigonometry "show questions tagged 'trigonometry'") [Share](https://math.stackexchange.com/q/311781 "Short permalink to this question") Share a link to this question Copy link [CC BY-SA 3.0](https://creativecommons.org/licenses/by-sa/3.0/ "The current license for this post: CC BY-SA 3.0") Cite Follow Follow this question to receive notifications [edited Feb 23, 2013 at 6:26](https://math.stackexchange.com/posts/311781/revisions "show all edits to this post") [](https://math.stackexchange.com/users/10400/will-jagy) [Will Jagy](https://math.stackexchange.com/users/10400/will-jagy) 148k88 gold badges158158 silver badges287287 bronze badges asked Feb 23, 2013 at 6:10 [](https://math.stackexchange.com/users/25841/jscb) [JSCB](https://math.stackexchange.com/users/25841/jscb) 13\.8k1515 gold badges6969 silver badges128128 bronze badges 8 - 4 It is an approach that works very nicely with 5 5 . With 7 7 , not so good, though it is a good way to prove that the regular 7 7 \-gon is not Euclidean constructible. I think that if you use the Cardano formula, you end up needing a cube root of a complex number, and that cube root cannot be found without knowing the required sine or a close relative. André Nicolas – [André Nicolas](https://math.stackexchange.com/users/6312/andr%C3%A9-nicolas "516,987 reputation") 2013-02-23 06:21:16 +00:00 [Commented Feb 23, 2013 at 6:21](https://math.stackexchange.com/questions/311781/method-to-find-sin-2-pi-7#comment675180_311781) - needs to be -1 at the end in the square brackets. Edited. Will Jagy – [Will Jagy](https://math.stackexchange.com/users/10400/will-jagy "148,377 reputation") 2013-02-23 06:26:55 +00:00 [Commented Feb 23, 2013 at 6:26](https://math.stackexchange.com/questions/311781/method-to-find-sin-2-pi-7#comment675182_311781) - The other bad news is that your cubic u3\+u2−2u−1 u 3 \+ u 2 − 2 u − 1 has three irrational real roots, which means there is no pretty way to separate the real and imaginary parts in Cardano's formula, [en.wikipedia.org/wiki/Casus\_irreducibilis](http://en.wikipedia.org/wiki/Casus_irreducibilis) Will Jagy – [Will Jagy](https://math.stackexchange.com/users/10400/will-jagy "148,377 reputation") 2013-02-23 06:32:52 +00:00 [Commented Feb 23, 2013 at 6:32](https://math.stackexchange.com/questions/311781/method-to-find-sin-2-pi-7#comment675185_311781) - 1 @ Will Jagy, You're right... JSCB – [JSCB](https://math.stackexchange.com/users/25841/jscb "13,768 reputation") 2013-02-23 06:39:19 +00:00 [Commented Feb 23, 2013 at 6:39](https://math.stackexchange.com/questions/311781/method-to-find-sin-2-pi-7#comment675189_311781) - 1 \+1: Might I recommend that you study algebraic number theory some time in the future. This type of calculations show up inside cyclotomic fields. Jyrki Lahtonen – [Jyrki Lahtonen](https://math.stackexchange.com/users/11619/jyrki-lahtonen "143,380 reputation") 2013-02-23 06:55:47 +00:00 [Commented Feb 23, 2013 at 6:55](https://math.stackexchange.com/questions/311781/method-to-find-sin-2-pi-7#comment675196_311781) \| [Show **3** more comments](https://math.stackexchange.com/questions/311781/method-to-find-sin-2-pi-7 "Expand to show all comments on this post") ## 3 Answers 3 Sorted by: [Reset to default](https://math.stackexchange.com/questions/311781/method-to-find-sin-2-pi-7?answertab=scoredesc#tab-top) This answer is useful 6 Save this answer. Show activity on this post. Just for laughs, we can at least in principle compute cos(π/7) cos ⁡ ( π / 7 ) by observing that sin3π7\=sin4π7 sin ⁡ 3 π 7 \= sin ⁡ 4 π 7 Using a combination of double-angle forumlae, we end up with a cubic equation for cos(π/7) cos ⁡ ( π / 7 ): 8cos3π7−4cos2π7−4cosπ7\+1\=0 8 cos 3 ⁡ π 7 − 4 cos 2 ⁡ π 7 − 4 cos ⁡ π 7 \+ 1 \= 0 This equation has one real solution which is cos(π/7) cos ⁡ ( π / 7 ). The bad news is that the expression is unwieldy at best: cosπ7\=16⎛⎝⎜1\+72/312(−1\+3i3–√)−−−−−−−−−−−−√3\+72(−1\+3i3–√)−−−−−−−−−−−−√3⎞⎠⎟ cos ⁡ π 7 \= 1 6 ( 1 \+ 7 2 / 3 1 2 ( − 1 \+ 3 i 3 ) 3 \+ 7 2 ( − 1 \+ 3 i 3 ) 3 ) The imaginary part of this expression is of course zero. The real part, however, ends up being expressed in terms of a sine and cosine of another angle, and I think the point of an exercise like this is to not do that. Anyway, I hope this adds to the discussion above. [Share](https://math.stackexchange.com/a/311792 "Short permalink to this answer") Share a link to this answer Copy link [CC BY-SA 3.0](https://creativecommons.org/licenses/by-sa/3.0/ "The current license for this post: CC BY-SA 3.0") Cite Follow Follow this answer to receive notifications answered Feb 23, 2013 at 6:54 [](https://math.stackexchange.com/users/53268/ron-gordon) [Ron Gordon](https://math.stackexchange.com/users/53268/ron-gordon) 142k1616 gold badges200200 silver badges325325 bronze badges 1 - I considered figuring out how to derive this equation with double/triple angle sine/cosine formulas. +1 for saving me the trouble. Jyrki Lahtonen – [Jyrki Lahtonen](https://math.stackexchange.com/users/11619/jyrki-lahtonen "143,380 reputation") 2013-02-23 07:02:52 +00:00 [Commented Feb 23, 2013 at 7:02](https://math.stackexchange.com/questions/311781/method-to-find-sin-2-pi-7#comment675203_311792) [Add a comment](https://math.stackexchange.com/questions/311781/method-to-find-sin-2-pi-7 "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \| This answer is useful 2 Save this answer. Show activity on this post. Target=\> sin2π7 sin ⁡ 2 π 7 set 1–√7\=ζk7 1 7 \= ζ 7 k ζk7\=e2kiπ7 ζ 7 k \= e 2 k i π 7 as seventh roots of unity so, the statements applies ζ17\+ζ27\+ζ37\+ζ47\+ζ57\+ζ67\+1\=0 ζ 7 1 \+ ζ 7 2 \+ ζ 7 3 \+ ζ 7 4 \+ ζ 7 5 \+ ζ 7 6 \+ 1 \= 0 ζk7\=cos2kπ7±isin2kπ7 ζ 7 k \= cos ⁡ 2 k π 7 ± i sin ⁡ 2 k π 7 ζ−k7\=cos2kπ7∓isin2kπ7 ζ 7 − k \= cos ⁡ 2 k π 7 ∓ i sin ⁡ 2 k π 7 ζk7\+ζ−k7\=2cos2kπ7 ζ 7 k \+ ζ 7 − k \= 2 cos ⁡ 2 k π 7 ζk7−ζ−k7\=±2isin2kπ7 ζ 7 k − ζ 7 − k \= ± 2 i sin ⁡ 2 k π 7 ζ2k7−2\+ζ−2k7\=−4sin22kπ7 ζ 7 2 k − 2 \+ ζ 7 − 2 k \= − 4 sin 2 ⁡ 2 k π 7 2cos4kπ7−2\=−4sin22kπ7 2 cos ⁡ 4 k π 7 − 2 \= − 4 sin 2 ⁡ 2 k π 7 2cos4kπ7−2−−−−−−−−−−−√\=±2isin2kπ7 2 cos ⁡ 4 k π 7 − 2 \= ± 2 i sin ⁡ 2 k π 7 the ζ17\+ζ27\+ζ37\+ζ47\+ζ57\+ζ67\+1\=0 ζ 7 1 \+ ζ 7 2 \+ ζ 7 3 \+ ζ 7 4 \+ ζ 7 5 \+ ζ 7 6 \+ 1 \= 0 gives us irreducible cubic polynomial t3\+t2−2t−1\=0 t 3 \+ t 2 − 2 t − 1 \= 0 where tk\=2cos2kπ7 t k \= 2 cos ⁡ 2 k π 7 applying lagrange resolvent for cubic −1\=t1\+t2\+t3 − 1 \= t 1 \+ t 2 \+ t 3 r1\=t1\+ζ13t2\+ζ23t3 r 1 \= t 1 \+ ζ 3 1 t 2 \+ ζ 3 2 t 3 r2\=t1\+ζ23t2\+ζ13t3 r 2 \= t 1 \+ ζ 3 2 t 2 \+ ζ 3 1 t 3 where ζ23 ζ 3 2 is the cuberoot of unity, Applying Discrete Fourier Transform inverse 3t1\=−1\+r1\+r2 3 t 1 \= − 1 \+ r 1 \+ r 2 3t2\=−1\+ζ23r1\+ζ13r2 3 t 2 \= − 1 \+ ζ 3 2 r 1 \+ ζ 3 1 r 2 3t3\=−1\+ζ13r1\+ζ23r2 3 t 3 \= − 1 \+ ζ 3 1 r 1 \+ ζ 3 2 r 2 on the action of lagrange method, lagrange resolvent gives r31\=7(3ζ13−1) r 1 3 \= 7 ( 3 ζ 3 1 − 1 ) r32\=7(3ζ23−1) r 2 3 \= 7 ( 3 ζ 3 2 − 1 ) now... r1\=7(3ζ13−1)−−−−−−−−−√3 r 1 \= 7 ( 3 ζ 3 1 − 1 ) 3 r2\=7(3ζ23−1)−−−−−−−−−√3 r 2 \= 7 ( 3 ζ 3 2 − 1 ) 3 ζ13\=−1\+−3−−−√2 ζ 3 1 \= − 1 \+ − 3 2 ζ23\=−1−−3−−−√2 ζ 3 2 \= − 1 − − 3 2 r1\=72(1\+3−3−−−√)−−−−−−−−−−−√3 r 1 \= 7 2 ( 1 \+ 3 − 3 ) 3 r1\=72(1−3−3−−−√)−−−−−−−−−−−√3 r 1 \= 7 2 ( 1 − 3 − 3 ) 3 3t1\=−1\+r1\+r2 3 t 1 \= − 1 \+ r 1 \+ r 2 3t2\=−1\+ζ23r1\+ζ13r2 3 t 2 \= − 1 \+ ζ 3 2 r 1 \+ ζ 3 1 r 2 3t3\=−1\+ζ13r1\+ζ23r2 3 t 3 \= − 1 \+ ζ 3 1 r 1 \+ ζ 3 2 r 2 3t1\=−1\+72(1\+3−3−−−√)−−−−−−−−−−−√3\+72(1−3−3−−−√)−−−−−−−−−−−√3 3 t 1 \= − 1 \+ 7 2 ( 1 \+ 3 − 3 ) 3 \+ 7 2 ( 1 − 3 − 3 ) 3 3t2\=−1\+−1−−3−−−√272(1\+3−3−−−√)−−−−−−−−−−−√3\+−1\+−3−−−√272(1−3−3−−−√)−−−−−−−−−−−√3 3 t 2 \= − 1 \+ − 1 − − 3 2 7 2 ( 1 \+ 3 − 3 ) 3 \+ − 1 \+ − 3 2 7 2 ( 1 − 3 − 3 ) 3 3t3\=−1\+−1\+−3−−−√272(1\+3−3−−−√)−−−−−−−−−−−√3\+−1−−3−−−√272(1−3−3−−−√)−−−−−−−−−−−√3 3 t 3 \= − 1 \+ − 1 \+ − 3 2 7 2 ( 1 \+ 3 − 3 ) 3 \+ − 1 − − 3 2 7 2 ( 1 − 3 − 3 ) 3 2cos2π7\=−1\+72(1\+3−3−−−√)−−−−−−−−−−−√3\+72(1−3−3−−−√)−−−−−−−−−−−√33 2 cos ⁡ 2 π 7 \= − 1 \+ 7 2 ( 1 \+ 3 − 3 ) 3 \+ 7 2 ( 1 − 3 − 3 ) 3 3 2cos4π7\=−1\+−1−−3√272(1\+3−3−−−√)−−−−−−−−−−−√3\+−1\+−3√272(1−3−3−−−√)−−−−−−−−−−−√33 2 cos ⁡ 4 π 7 \= − 1 \+ − 1 − − 3 2 7 2 ( 1 \+ 3 − 3 ) 3 \+ − 1 \+ − 3 2 7 2 ( 1 − 3 − 3 ) 3 3 2cos6π7\=−1\+−1\+−3√272(1\+3−3−−−√)−−−−−−−−−−−√3\+−1−−3√272(1−3−3−−−√)−−−−−−−−−−−√33 2 cos ⁡ 6 π 7 \= − 1 \+ − 1 \+ − 3 2 7 2 ( 1 \+ 3 − 3 ) 3 \+ − 1 − − 3 2 7 2 ( 1 − 3 − 3 ) 3 3 2cos4kπ7−2−−−−−−−−−−−√\=±2isin2kπ7 2 cos ⁡ 4 k π 7 − 2 \= ± 2 i sin ⁡ 2 k π 7 t2k−2−−−−−−√\=2isin2kπ7 t 2 k − 2 \= 2 i sin ⁡ 2 k π 7 t2−2−−−−−√\=2isin2π7 t 2 − 2 \= 2 i sin ⁡ 2 π 7 t4−2−−−−−√\=2isin4π7 t 4 − 2 \= 2 i sin ⁡ 4 π 7 t6−2−−−−−√\=2isin6π7 t 6 − 2 \= 2 i sin ⁡ 6 π 7 t4\=t3 t 4 \= t 3 t6\=t1 t 6 \= t 1 t2−2−−−−−√\=2isin2π7 t 2 − 2 \= 2 i sin ⁡ 2 π 7 −1\+−1−−3√272(1\+3−3−−−√)−−−−−−−−−−−√3\+−1\+−3√272(1−3−3−−−√)−−−−−−−−−−−√33−2−−−−−−−−−−−−−−−−−−−−−−−−−−−⎷\=2isin2π7 − 1 \+ − 1 − − 3 2 7 2 ( 1 \+ 3 − 3 ) 3 \+ − 1 \+ − 3 2 7 2 ( 1 − 3 − 3 ) 3 3 − 2 \= 2 i sin ⁡ 2 π 7 −7\+−1−−3√272(1\+3−3−−−√)−−−−−−−−−−−√3\+−1\+−3√272(1−3−3−−−√)−−−−−−−−−−−√33−−−−−−−−−−−−−−−−−−−−−−−−⎷\=2isin2π7 − 7 \+ − 1 − − 3 2 7 2 ( 1 \+ 3 − 3 ) 3 \+ − 1 \+ − 3 2 7 2 ( 1 − 3 − 3 ) 3 3 \= 2 i sin ⁡ 2 π 7 3(−7\+−1−−3√272(1\+3−3−−−√)−−−−−−−−−−−√3\+−1\+−3√272(1−3−3−−−√)−−−−−−−−−−−√3)9−−−−−−−−−−−−−−−−−−−−−−−−−−⎷\=2isin2π7 3 ( − 7 \+ − 1 − − 3 2 7 2 ( 1 \+ 3 − 3 ) 3 \+ − 1 \+ − 3 2 7 2 ( 1 − 3 − 3 ) 3 ) 9 \= 2 i sin ⁡ 2 π 7 3(−7\+−1−−3√272(1\+3−3−−−√)−−−−−−−−−−−√3\+−1\+−3√272(1−3−3−−−√)−−−−−−−−−−−√3)−−−−−−−−−−−−−−−−−−−−−−−−−⎷3\=2isin2π7 3 ( − 7 \+ − 1 − − 3 2 7 2 ( 1 \+ 3 − 3 ) 3 \+ − 1 \+ − 3 2 7 2 ( 1 − 3 − 3 ) 3 ) 3 \= 2 i sin ⁡ 2 π 7 so, sin2π7\=−3(−7\+−1−−3√272(1\+3−3−−−√)−−−−−−−−−−−√3\+−1\+−3√272(1−3−3−−−√)−−−−−−−−−−−√3)−−−−−−−−−−−−−−−−−−−−−−−−−−−⎷6 sin ⁡ 2 π 7 \= − 3 ( − 7 \+ − 1 − − 3 2 7 2 ( 1 \+ 3 − 3 ) 3 \+ − 1 \+ − 3 2 7 2 ( 1 − 3 − 3 ) 3 ) 6 [Share](https://math.stackexchange.com/a/2838999 "Short permalink to this answer") Share a link to this answer Copy link [CC BY-SA 4.0](https://creativecommons.org/licenses/by-sa/4.0/ "The current license for this post: CC BY-SA 4.0") Cite Follow Follow this answer to receive notifications answered Jul 2, 2018 at 22:26 [](https://math.stackexchange.com/users/523526/flam-rakanz) [Flam Rakanz](https://math.stackexchange.com/users/523526/flam-rakanz) 10911 silver badge55 bronze badges [Add a comment](https://math.stackexchange.com/questions/311781/method-to-find-sin-2-pi-7 "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \| This answer is useful 1 Save this answer. Show activity on this post. Using [this](https://math.stackexchange.com/questions/309271/two-trigonometric-identities-using-complex-variables/309301#309301) or Point\#24 24 of [this](http://mathworld.wolfram.com/Multiple-AngleFormulas.html) , sin7x\=7s−56s3\+112s5−64s7 sin ⁡ 7 x \= 7 s − 56 s 3 \+ 112 s 5 − 64 s 7 where s\=sinx s \= sin ⁡ x If sin7x\=0,7x\=nπ sin ⁡ 7 x \= 0 , 7 x \= n π where n n is any integer. So, x\=nπ7 x \= n π 7 where n\=0,1,2,3,4,5,6 n \= 0 , 1 , 2 , 3 , 4 , 5 , 6 So, the roots of 7s−56s3\+112s5−64s7\=0−−−\>(1) 7 s − 56 s 3 \+ 112 s 5 − 64 s 7 \= 0 − − − \> ( 1 ) are sinnπ7 sin ⁡ n π 7 where n\=0,1,2,⋯5,6 n \= 0 , 1 , 2 , ⋯ 5 , 6 So, the roots of 64s6−112s4\+56s2−7\=0−−−\>(2) 64 s 6 − 112 s 4 \+ 56 s 2 − 7 \= 0 − − − \> ( 2 ) are sinnπ7 sin ⁡ n π 7 where n\=1,2,⋯5,6 n \= 1 , 2 , ⋯ 5 , 6 So, the roots of 64t3−112t2\+56t−7\=0−−−\>(3) 64 t 3 − 112 t 2 \+ 56 t − 7 \= 0 − − − \> ( 3 ) are sin2nπ7 sin 2 ⁡ n π 7 where n\=1,2,4 n \= 1 , 2 , 4 or 3,5,6 3 , 5 , 6 as sin(7−r)π7\=sin(π−rπ7)\=sinrπ7 sin ⁡ ( 7 − r ) π 7 \= sin ⁡ ( π − r π 7 ) \= sin ⁡ r π 7 If we choose n\=1,2,4 n \= 1 , 2 , 4 observe that sin24π7−sin22π7\=2sinπ7cos3π7\>0 sin 2 ⁡ 4 π 7 − sin 2 ⁡ 2 π 7 \= 2 sin ⁡ π 7 cos ⁡ 3 π 7 \> 0 (Using sin2A−sin2B\=sin(A\+B)sin(A−B) sin 2 ⁡ A − sin 2 ⁡ B \= sin ⁡ ( A \+ B ) sin ⁡ ( A − B )) Similarly, sin22π7−sin2π7\>0 sin 2 ⁡ 2 π 7 − sin 2 ⁡ π 7 \> 0 So, sin24π7\>sin22π7\>sin2π7 sin 2 ⁡ 4 π 7 \> sin 2 ⁡ 2 π 7 \> sin 2 ⁡ π 7 Using [Cardano's method](http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method), we can solve the Cubic equation (3) ( 3 ) [Share](https://math.stackexchange.com/a/311819 "Short permalink to this answer") Share a link to this answer Copy link [CC BY-SA 3.0](https://creativecommons.org/licenses/by-sa/3.0/ "The current license for this post: CC BY-SA 3.0") Cite Follow Follow this answer to receive notifications [edited Apr 13, 2017 at 12:20](https://math.stackexchange.com/posts/311819/revisions "show all edits to this post") [](https://math.stackexchange.com/users/-1/community) [Community](https://math.stackexchange.com/users/-1/community)Bot 1 answered Feb 23, 2013 at 8:15 [](https://math.stackexchange.com/users/33337/lab-bhattacharjee) [lab bhattacharjee](https://math.stackexchange.com/users/33337/lab-bhattacharjee) 281k2121 gold badges213213 silver badges339339 bronze badges [Add a comment](https://math.stackexchange.com/questions/311781/method-to-find-sin-2-pi-7 "Use comments to ask for more information or suggest improvements. 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Using [this](https://math.stackexchange.com/questions/309271/two-trigonometric-identities-using-complex-variables/309301#309301) or Point\#24 of [this](http://mathworld.wolfram.com/Multiple-AngleFormulas.html) , sin ⁡ 7 x \= 7 s − 56 s 3 \+ 112 s 5 − 64 s 7 where s \= sin ⁡ x If sin ⁡ 7 x \= 0 , 7 x \= n π where n is any integer. So, x \= n π 7 where n \= 0 , 1 , 2 , 3 , 4 , 5 , 6 So, the roots of 7 s − 56 s 3 \+ 112 s 5 − 64 s 7 \= 0 − − − \> ( 1 ) are sin ⁡ n π 7 where n \= 0 , 1 , 2 , ⋯ 5 , 6 So, the roots of 64 s 6 − 112 s 4 \+ 56 s 2 − 7 \= 0 − − − \> ( 2 ) are sin ⁡ n π 7 where n \= 1 , 2 , ⋯ 5 , 6 So, the roots of 64 t 3 − 112 t 2 \+ 56 t − 7 \= 0 − − − \> ( 3 ) are sin 2 ⁡ n π 7 where n \= 1 , 2 , 4 or 3 , 5 , 6 as sin ⁡ ( 7 − r ) π 7 \= sin ⁡ ( π − r π 7 ) \= sin ⁡ r π 7 If we choose n \= 1 , 2 , 4 observe that sin 2 ⁡ 4 π 7 − sin 2 ⁡ 2 π 7 \= 2 sin ⁡ π 7 cos ⁡ 3 π 7 \> 0 (Using sin 2 ⁡ A − sin 2 ⁡ B \= sin ⁡ ( A \+ B ) sin ⁡ ( A − B )) Similarly, sin 2 ⁡ 2 π 7 − sin 2 ⁡ π 7 \> 0 So, sin 2 ⁡ 4 π 7 \> sin 2 ⁡ 2 π 7 \> sin 2 ⁡ π 7 Using [Cardano's method](http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method), we can solve the Cubic equation ( 3 )