SOLUTION: Prove that: tan(pi/7)tan(2pi/7)tan(3pi/7)=sqrt(7)

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Question 1118654: Prove that: tan(pi/7)tan(2pi/7)tan(3pi/7)=sqrt(7)
Answer by ikleyn(52887) About Me  (Show Source):
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Consider complex number  

z = cos%28pi%2F7%29+%2B+i%2Asin%28pi%2F7%29.              (1)


It is the root of the degree 7 of 1 in complex domain,  so 


z%5E7 = 1 = %28cos%28pi%2F7%29+%2B+i%2Asin%28pi%2F7%29%29%5E7.      (2)


Apply the binomial decomposition to get


1 = cos%5E7%28pi%2F7%29+7i%2Acos%5E6%28pi%2F7%29%2Asin%28pi%2F7%29+21%2Ai%5E2%2Acos%5E5%28pi%2F7%29%2Asin%5E2%28pi%2F7%29+70%2Ai%5E3%2Acos%5E4%28pi%2F7%29%2Asin%5E3%28pi%2F7%29+70%2Ai%5E4%2Acos%5E3%28pi%2F7%29%2Asin%5E4%28pi%2F7%29+21%2Ai%5E5%2Acos%5E2%28pi%2F7%29%2Asin%5E5%28pi%2F7%29+ 7%2Ai%5E6%2Acos%28pi%2F7%29%2Asin%5E6%28pi%2F7%29+i%5E7%2Asin%5E7%28pi%2F7%29.


In the last equation extract the imaginary part, which is equal to zero:


0 = 7i%2Acos%5E6%28pi%2F7%29%2Asin%28pi%2F7%29 + 70%2Ai%5E3%2Acos%5E4%28pi%2F7%29%2Asin%5E3%28pi%2F7%29 + 21%2Ai%5E5%2Acos%5E2%28pi%2F7%29%2Asin%5E5%28pi%2F7%29 + i%5E7%2Asin%5E7%28pi%2F7%29.


Replace the degrees of "i" by i with the corresponding signs and then divide both sides by i%2Acos%5E7%28pi%2F7%29.  You will get


0 = 7%2Atan%28pi%2F7%29 - 70%2Atan%5E3%28pi%2F7%29 + 21%2Atan%28pi%2F7%29 - tan%5E7%28pi%2F7%29.


Divide by  -tan%28pi%2F7%29 both sides.  You will get


tan%5E6%28pi%2F7%29 - 21%2Atan%5E4%28pi%2F7%29 + 70%2Atan%5E2%28pi%2F7%29 - 7 = 0.    (3)


Thus you see that  tan%28pi%2F7%29 is the root of the polynomial equation 


x%5E6 - 21%2Ax%5E4 + 70%2Ax%5E2 - 7 = 0.                  (4)


If you do the same starting from  z = cos%282pi%2F7%29+%2B+i%2Asin%282pi%2F7%29  instead of  z = cos%28pi%2F7%29+%2B+i%2Asin%28pi%2F7%29, you will get by the same way that 

                   tan%282pi%2F7%29 is the root of the same polynomial equation.


Similarly,  If you do the same starting from  z = cos%283pi%2F7%29+%2B+i%2Asin%283pi%2F7%29  instead of  z = cos%28pi%2F7%29+%2B+i%2Asin%28pi%2F7%29, you will get by the same way that 

                   tan%283pi%2F7%29 is the root of the same polynomial equation.


 . . .    and so on   . . . 


Thus, the six numbers  tan%28pi%2F7%29,  tan%282pi%2F7%29,  tan%283pi%2F7%29, . . . ,  tan%286pi%2F7%29  all are the roots of the equation (4).


Then, according to Vieta's theorem,  the product of the roots  is equal to the constant term:


      tan%28pi%2F7%29.tan%282pi%2F7%29.tan%283pi%2F7%29. . . . .  tan%286pi%2F7%29 = -7.


Now take into account that  tan%28pi%2F7%29 = -tan%286pi%2F7%29,  tan%282pi%2F7%29 = -tan%285pi%2F7%29  and  tan%283pi%2F7%29 = -tan%284pi%2F7%29.


Based on it, you get


tan%5E2%28pi%2F7%29.tan%5E2%282pi%2F7%29.tan%5E2%283pi%2F7%29 = 7,


which implies


tan%28pi%2F7%29.tan%282pi%2F7%29.tan%283pi%2F7%29 = sqrt%287%29.


It is exactly what has to be proved.


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