-1

How to evaluate $$\tan^4{10°}+\tan^4{50°}+\tan^4{70°} = ?$$

It should equals 59.

But I don't know how? I also think that we should use power-reduction formulas.

Alexander Gruber
  • 26,963
Liza
  • 31
  • 1
    Please show your effort..... Then we will happy to help...:-) – Aatmaj Feb 15 '21 at 15:50
  • Does this https://math.stackexchange.com/a/770557/876009 help you Liza? –  Feb 15 '21 at 15:57
  • See https://math.stackexchange.com/questions/175736/evaluate-tan220-circ-tan240-circ-tan280-circ – lab bhattacharjee Feb 15 '21 at 17:57
  • 1
    Once a question has received an answer which is either accepted or with a positive score, it is no longer within the poster's power to remove the question. Therefore, please don't change it to invalidate other people's work. –  Feb 18 '21 at 13:27
  • Please do not make question edits which invalidate existing (valid) answers. – PM 2Ring Feb 18 '21 at 15:43

1 Answers1

3

Per the triple-angle identity $$\tan 3x = \frac{3\tan x-\tan^3x}{1-3\tan^2x}$$ $\tan 10^\circ$, $-\tan50^\circ$ and $\tan 70^\circ$ are the three roots of $$ \sqrt3 (3t- t^3)=1-3t^2$$

Square to get $3t^6+33t^2 = 27t^4+1$, and square again to get the cubic equation $$9t^{12} -531t^8 +1035t^4-1=0$$ in $t^4$, with roots $\tan^410^\circ$, $\tan^450^\circ$ and $\tan^470^\circ$ . Thus $$\tan^4{10°}+\tan^4{50°}+\tan^4{70°} =\frac{531}9=59$$

Quanto
  • 97,352
  • Thank you very much. It is interesting answer, but I can't understand how did you got the last equation from (9t^12−531t^8+1035t^4−1=0) this – Liza Feb 15 '21 at 17:25
  • @Liza - I used the Vieta’s formula for the sum of roots $\tan^4\theta$ https://en.wikipedia.org/wiki/Vieta%27s_formulas – Quanto Feb 15 '21 at 17:32
  • 1
    Ingenuous! +1!!!!!! – nonuser Feb 15 '21 at 19:39