[กขฃคฅฆง]

1. $sin1^o sin3^o sin5^o ...sin89^o$

$=\dfrac{sin1^o sin2^o sin3^o ...sin89^o}{sin2^o sin4^o sin6^o ...sin88^o} $

$=\dfrac{sin1^o cos1^o sin2^o cos2^o ...sin44^o cos44^o sin45^o }{sin2^o sin4^o sin6^o ...sin88^o} $

$=\dfrac{sin2^o sin4^o sin6^o ...sin88^o sin45^o }{2^{44}sin2^o sin4^o sin6^o ...sin88^o} $

$=2^{-\frac{89}{2} }$

2.วิธีไม่ใช้ AM-GM

$(a^4+b^4-2a^2b^2) + (c^4+d^4-2c^2d^2) + (2a^2b^2+2c^2d^2-4abcd) = 0$

$(a^2-b^2)^2 + (c^2-d^2)^2 + 2(ab-cd)^2 = 0$

$a^2=b^2 , c^2=d^2 , ab=cd$

$a=b=c=d$