I'm not seeing any trigonometry....

One thing to note is that

Pr[winning at least 3 prizes] = 1 - Pr[winning 2 or less prizes]

which is less work than you seem to have done.

given $p=\dfrac 1 4$ we get a (10, 1/4) binomial distribution with

$Pr[\mbox{k prizes}] = \begin{pmatrix}10 \\k\end{pmatrix} p^k (1-p)^{10-k}$

and you can quickly find the probability of winning 2 or less prizes as

$(3/4)^{10} + 10 (1/4) (3/4)^9+ 45 (1/4)^2 (3/4)^8\approx 0.5256$ (it's exactly $\frac{137781}{262144}$)

and thus

$Pr[\mbox{winning 3 or more prizes}] \approx 1 - 0.5256 = 0.4744$