A geometry problem by Mahan Farzan

Geometry Level 5

$\large \tan^2 1^{\circ}+\tan^2 2^{\circ} + \cdots + \tan^2 89^{\circ} = \, ?$

Give your answer to 1 decimal place.

The answer is 5310.3.

1 solution

Mahan Farzan
Sep 2, 2016

$\tan(90\theta)=\frac{C^{90}_1t-C^{90}_3t^3+...-C^{90}_{87}t^{87}+C^{90}_{89}t^{89}} {1-C^{90}_2t^2+...+C^{90}_{88}t^{88}-t^{90}}\ \ (t=\tan\theta,x=t^2)$

$1-C^{90}_2x+...+C^{90}_{88}x^{44}-x^{45}=0$ has roots $\tan^21^\circ,\tan^23^\circ,...,\tan^289^\circ.$

$C^{90}_1-C^{90}_3x+...-C^{90}_{87}x^{43}+C^{90}_{89}x^{44}=0$ has roots $\tan^22^\circ,\tan^24^\circ,...,\tan^288^\circ.$

$\tan^21^\circ+\tan^22^\circ+\tan^23^\circ+...+\tan^289^\circ =C^{90}_{88}+\frac{C^{90}_{87}}{C^{90}_{89}}=\frac{15931}3.\#$

We can also use the identity

$\displaystyle\sum_{k=1}^{n-1} \tan^{2}\left(\dfrac{k\pi}{2n}\right) = \dfrac{(n - 1)(2n - 1)}{3}$

where in this case we have $n = 90$ . A proof of this identity is given here .

Brian Charlesworth - 4 years, 9 months ago

yes you're right but I prefer my solution

Mahan Farzan - 4 years, 9 months ago

A geometry problem by Mahan Farzan

The answer is 5310.3.

1 solution

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