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 February 2nd, 2015, 01:45 PM #1 Member   Joined: Jan 2015 From: Orlando, Florida Posts: 92 Thanks: 10 How do you prove this? Prove: (tan(1))^2 + (tan(3))^2+....(tan(89))^2 = 45x(90-1) = 4005
 February 2nd, 2015, 02:00 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,118 Thanks: 2369 Math Focus: Mainly analysis and algebra You are presumably going to form pairs$$\tan^2 x + \tan^2{\left( \tfrac\pi2 - x \right) }$$and see what that comes out as. Thanks from topsquark
 February 2nd, 2015, 02:36 PM #3 Member   Joined: Jan 2015 From: Orlando, Florida Posts: 92 Thanks: 10 I've tried pairing them like that: so each angle x except 45 (where tan^2(45)=1) forms a pair with 90-x. tan^2(x)+tan^2(90-x)=tan^2(x)+cot^2(x)=sin^2(x)/cos^2(x)+cos^2(x)/sin^2(x)= 1/(cos^2(x)sin^2(x)). Hmm.... not sure what to do from here Idk if this is the right way to proceed: 1/(sin(2x)/2)^2=4/sin^2(x) Now the sum comes down to 1+4(1/sin^2(1)+1/sin^2(3)+....1/sin^2(43)) No idea what to do from here idk if stuff telescopes or something.
 February 2nd, 2015, 02:58 PM #4 Global Moderator   Joined: Dec 2006 Posts: 18,433 Thanks: 1462 What makes you think the original equation is correct?
February 2nd, 2015, 03:03 PM   #5
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Quote:
 Originally Posted by skipjack What makes you think the original equation is correct?

February 3rd, 2015, 05:05 PM   #6
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So you need an initial assumption, don't you? It's already been made implicitly.

Quote:
 Originally Posted by USAMO Reaper I've tried pairing them . . . Now the sum comes down to 1+4(1/sin^2(1)+1/sin^2(3)+....1/sin^2(43)) No idea what to do from here . . .
You've already slipped up, but you've got the right idea.
Use the fact that 1/sin²(x) = csc²(x) = 1 + cot²(x).
Now you can try pairing again, but you'll need to be resourceful, as the pairs don't seem to be present at first.

This is a well-known problem, but I can't find a website where the full details of the above process are given.

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