The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

summing arctangents

I have the following example tan⁡−1(8)+tan⁡−1(2)+tan⁡−1(23)\displaystyle \tan^{-1}\left ( 8 \right ) +\tan^{-1}\left ( 2 \right )+\tan^{-1}\left ( \frac{2}{3} \right ).

By using arctangent addition, I've gotten to arctan⁡(0)=nπ\arctan(0) = n\pi, but the answer is π\pi, so how do I eliminate all other solutions?
Reply 1
Avatar for B_9710
B_9710
18
Consider complex numbers say z1,z2,z3 z_1, z_2, z_3 such that z1z2z3=w z_1 z_2 z_3 = w , where arg(z1)=arctan⁡8,arg(z2)=arctan⁡2,arg(z3)=arctan⁡23 \text{arg}(z_1)= \arctan 8, \text{arg}(z_2)= \arctan 2, \text{arg}(z_3) = \arctan \frac{2}{3} and consider the argument of w w .
(edited 7 years ago)
Reply 2
Avatar for Ayman!
Ayman!
OP
15
Original post by B_9710
Consider complex numbers say z1,z2,z3 z_1, z_2, z_3 such that z1z2z3=w z_1 z_2 z_3 = w , where arg(z1)=arctan⁡8,arg(z2)=arctan⁡2,arg(z3)=arctan23 arg(z_1)= \arctan 8, arg(z_2)= \arctan 2, arg(z_3) = arctan \frac{2}{3} and consider the argument of w w .


I understand the complex number argument approach, but I'm wonderinh what I'm doing wrong with addition. :redface:
Reply 3
Avatar for Zacken
Zacken
22
Original post by Ayman!
I have the following example tan⁡−1(8)+tan⁡−1(2)+tan⁡−1(23)\displaystyle \tan^{-1}\left ( 8 \right ) +\tan^{-1}\left ( 2 \right )+\tan^{-1}\left ( \frac{2}{3} \right ).

By using arctangent addition, I've gotten to arctan⁡(0)=nπ\arctan(0) = n\pi, but the answer is π\pi, so how do I eliminate all other solutions?


Well, note that for example, arctan⁡8,arctan⁡2,arctan⁡23\arctan 8, \arctan 2, \arctan \frac{2}{3} are all positive terms, so you know that the sum is bounded below by 0<nπ0 < n\pi, furthermore, note that each term is bounded above by π2\frac{\pi}{2} (obviously), so the sum is bounded above by 3π2\frac{3\pi}{2} and the only integers that satisfies 0<n<320 < n < \frac{3}{2} is n=1n=1.
(edited 7 years ago)
Original post by Zacken
Well, note that for example, arctan⁡8,arctan⁡2,arctan⁡23\arctan 8, \arctan 2, \arctan \frac{2}{3} are all positive terms, so you know that the sum is bounded below by 0<nπ0 < n\pi, furthermore, note that each term is bounded above by π2\frac{\pi}{2} (obviously), so the sum is bounded above by 3π2\frac{3\pi}{2} and the only integers that satisfies 0<n<3π20 < n < \frac{3\pi}{2} is n=1n=1.


That second term of your inequality should be n*pi, not just n, surely?
Reply 5
Avatar for Zacken
Zacken
22
Original post by HapaxOromenon3
That second term of your inequality should be n*pi, not just n, surely?


Whoop, sorry. Typo there, I meant 0<n<320 < n < \frac{3}{2}. Thanks.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you