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Geometry Level 4

tan ⁡ 6 2 π 9 − 33 tan ⁡ 4 2 π 9 + 27 tan ⁡ 2 2 π 9 + 27 = ? \large \tan^6 \dfrac{2\pi}9 - 33\tan^4 \dfrac{2\pi}9 + 27\tan^2 \dfrac{2\pi}9 + 27 = \, ?


The answer is 30.

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Watsapol Kerdlarppol by Watsapol Kerdlarppol , 22, Thailand

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2 solutions

Pi Han Goh
Oct 15, 2016

Let A = 2 π 9 A = \dfrac{2\pi}9 , then tan ⁡ ( 9 A ) = 0 \tan (9A) = 0 . Using the generalized compound tangent formula :

0 = tan ⁡ ( 9 A ) = ( 9 1 ) ( tan ⁡ A ) − ( 9 3 ) ( tan ⁡ A ) 3 + ( 9 5 ) ( tan ⁡ A ) 5 − ( 9 7 ) ( tan ⁡ A ) 7 + ( 9 9 ) ( tan ⁡ A ) 9 1 − ( 9 2 ) ( tan ⁡ A ) 2 + ( 9 4 ) ( tan ⁡ A ) 4 − ( 9 6 ) ( tan ⁡ A ) 6 + ( 9 8 ) ( tan ⁡ A ) 8 . 0 = \tan (9A) = \dfrac{ {9 \choose 1} (\tan A) - { 9 \choose 3} (\tan A)^3 + { 9 \choose 5} (\tan A)^5 - {9 \choose 7} (\tan A)^7 + {9 \choose 9} (\tan A)^9 }{1 - {9 \choose 2} (\tan A)^2 + { 9 \choose 4} (\tan A)^4 - {9 \choose 6} (\tan A)^6 + {9\choose 8} (\tan A)^8 } \; .

So the numerator of the fraction above must be equal to 0. Dividing that expression by tan ⁡ A ≠ 0 \tan A \ne 0 , gives

9 − 84 ( tan ⁡ A ) 2 + 126 ( tan ⁡ A ) 4 − 36 ( tan ⁡ A ) 6 + ( tan ⁡ A ) 8 = 0 ⇔ ( tan ⁡ 2 A − 3 ) ( ( tan ⁡ A ) 6 − 33 ( tan ⁡ A ) 4 + 27 ( tan ⁡ A ) 2 − 3 ) = 0 . 9-84 (\tan A)^2+126 (\tan A)^4-36 (\tan A)^6+(\tan A)^8 = 0 \Leftrightarrow (\tan^2 A - 3) ((\tan A)^6-33 (\tan A)^4+27 (\tan A)^2-3) = 0 \; .

Since tan ⁡ 2 A − 3 ≠ 0 \tan^2 A - 3 \ne 0 , we cancel divide the equation by tan ⁡ 2 A − 3 \tan^2 A - 3 . The result follows.

4 Helpful 0 Interesting 0 Brilliant 0 Confused

+1 Your solution is a great read:

  1. Nice observation of the tangent formula. However, this would only work if the coefficients were "suitably" chosen.

Thanks for contributing and helping other members aspire to be like you!

Calvin Lin Staff - 4 years, 8 months ago

3 Helpful 0 Interesting 0 Brilliant 0 Confused

+1 Your solution is a great read:

  1. Instead of "Just prove the identities", what you mean is "Just use the identities".
  2. This "just happens" to work out. What other approaches could we use?

Thanks for contributing and helping other members aspire to be like you!

Calvin Lin Staff - 4 years, 8 months ago

Yup, this is essentially correct. But it would be much easier to read through your steps if you elaborated a little more on a few points. For example,

[1] It's better to explictly mention that tan ⁡ ( 2 π / 9 ) \tan (2\pi /9) is not a well known number, but tan ⁡ ( 3 × 2 π / 9 ) = − 3 \tan(3\times2\pi/9) = -\sqrt3 is a simple constant. And because you only know the value of tan ⁡ ( 3 A ) \tan(3A) , but not tan ⁡ A \tan A , this motivates you to use the triple angle identity , tan ⁡ ( 3 X ) = 3 tan ⁡ ( X ) − tan ⁡ 3 ( X ) 1 − 3 tan ⁡ 2 ( X ) \tan(3X) = \dfrac{3\tan(X) - \tan^3(X)}{1 - 3\tan^2(X)} .

[2] There isn't a need to convert 2 π / 9 2\pi /9 radian into 40 degrees, as it not needed in the first place. It is however, useful to state that you converted the angles to degrees from the start because you felt more comfortable to work with degrees, as opposed to radians.

Pi Han Goh - 4 years, 8 months ago

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