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First off, the proof is crap and doesn't make sense, so ignore it.

haha....i feel like an idiot, answering my own question.... Anyways, new question: By expressing sin x and cos x in terms of t = tan(x/2), show that the equation asinx + bcosx = c has real roots only if a^2 + b^2 >= c^2.

lol.....don't forget to post a new question.

xD....yeah, I guess that you could look at it that way. My proof was like this: Let a^log_a(b) = y. log_a(a^log_a(b)) = log_a(y) log_a(b) * log_a(a) = log_a(y) log_a(b) = log_a(y) b = y. Therefore, a^log_a(b) = b. Yours is a better alternative to solving it though.

I{1->0} 2^(log_e(x)) Notation: log_a(...) means log base a of .... . Note: Writing it on paper makes things a lot less confusing. Consider log_e(x). log_e(x) = log_2(x) / log_2(e) --(change base) log_e(x) = (1/log_2(e)) * log_2(x) log_e(x) = log_2(x^(1/log_2(e))) Therefore, I{1->0}...

If a/b >= c/a, a^2/bc >= 1 (true because a>1 and c>1) log(a^2/bc) >= log 1 (base e) log(a^2/bc) >= 0 Since a^2/bc = (a/b)*(a/c), log((a/b)*(a/c)) >= 0 log(a/b) + log(a/c) >= 0 log(a/b) >= -log(a/c) log(a/b) >= log(c/a) a/b >= c/a. New Question: Given that a+b+c = 1 and a+b+c >= 3(abc)^1/3...

P(X=4) = 12C4((1-p)^8)(p^4) P(X=3) = 12C3((1-p)^9)(p^3) Since P(X=4) = 3 * P(X=3), 495*((1-p)^8)(p^4) = 3*220*((1-p)^9)(p^3) 495p = 660(1-p) 495p+660p = 660 p = 660/1155 p = 4/7 New Question: Prove that (a^2-b^2)(c^2-d^2) <= (ac-bd)^2

Proof: If a^2+b^2+c^2 > ab+ac+bc, then (ab)^2+(bc)^2+(ca)^2 > ab^2c + bc^2a + ca^2b (ab)^2+(bc)^2+(ca)^2 > abc(a+b+c) (a^2b^2 +b^2c^2+c^2a^2) / (a+b+c) > abc