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February 2nd, 2015, 12:45 PM   #1
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How do you prove this?

Prove: (tan(1))^2 + (tan(3))^2+....(tan(89))^2 = 45x(90-1) = 4005
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February 2nd, 2015, 01:00 PM   #2
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You are presumably going to form pairs$$\tan^2 x + \tan^2{\left( \tfrac\pi2 - x \right) }$$and see what that comes out as.
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February 2nd, 2015, 01:36 PM   #3
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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.
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February 2nd, 2015, 01:58 PM   #4
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What makes you think the original equation is correct?
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February 2nd, 2015, 02:03 PM   #5
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Quote:
Originally Posted by skipjack View Post
What makes you think the original equation is correct?
https://answers.yahoo.com/question/i...6111910AAJDR7W
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February 3rd, 2015, 04:05 PM   #6
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So you need an initial assumption, don't you? It's already been made implicitly.

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Originally Posted by USAMO Reaper View Post
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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