Recent content by harry654

Solution: 0, 1, -2010

Thank you. So all integers for which is fraction equal to integer is 0 and 1?

Good day! I have problem: Find all integers for which is fraction (n3+2010)/(n2+2010) equals to integer. I can find 0 and 1 and I tried prove that any integers don't exist, but I didnt contrive it. Could someone help me with it?

Hi tiny-tim! If α=135 and β=30 then sin²α + sin²β < 1 and apply α+β>90 so what isn't correct?

because I thought that I can say cos(α-β) is positive because sin(α-β) is not 0 but I found out that I can not :D OK thank you again:)

Assume that apply sin(α-β) is not 0 then question is : Is cos(α-β) positive?

Anyway I have question. Does it apply when sin(α-β) isn't 0 so cos(α-β) is positive or no?

Yes THANK YOU tiny-tim! I understand this and I thank you for your patience and assistance.

Yes I know what do you mean, but when I use picture so it isn't correct mathematical proof so I don't know ...

How can I prove sin²α + sin²β > 1 when α+β > 90°?

how? into sin²(α-β)b²/sin²β ≤ (a²+b²)sin²(α-β) and so b²/sin²β ≤ (a²+b²) thats isn't true because when α=β we divide 0

Could someone help me? I am desperate:(

apply sin²α + sin²β = 1 when α+β = 90° and from that α+β > 90 so sin²α + sin²β >1 so when I prove sin²α + sin²β >1, I proved that (a²+b²)cos(α-β) ≤ 2abcos(α-β)?

certainly β or α >45° so sin²α or sin²β > 0,5 so sin²α+sin²β >1 but how explain it mathematically

oh, I have feeling that I never finish this prove :( sin²α + sin²β = 1 when α+β=90 so inequality doesn't apply I think