Ask Me Help Desk - Write 24 sin3x cos3x+10 cos^2 3x-5 in terms of one sin function

Well, that's how I do it, but there are quite 'ugly numbers'.

First off, use of the double angle of sine to get rid of the first cos function, and double angle of cosine to get rid of the cos squared. Then compound angle for the cos function.

$24\ sin(3x)\ cos(3x)\ +\ 10\ cos^2(3x-5)\\ \\ = 12\ sin(6x)\ +\ 5\ cos(6x-10)\ +\ 5\\ \\ = 12\ sin(6x)\ +\ 5\ (cos(6x)cos(10)\ +\ sin(6x)sin(10))\ +\ 5\\ \\ = 12\ sin(6x)\ +\ 5\ cos(6x)cos(10)\ +\ 5\ sin(6x)sin(10)\ +\ 5\\ \\ = (12\ +\ 5\ sin(10)) sin(6x) + (5\ cos(10)) cos(6x)\ +\ 5$

From there, I use the rule:
$a\ sin\theta + b cos\theta = R sin(\theta + \alpha)$

where $R^2 = a^2 + b^2$
and $tan \alpha = \frac{b}{a}$

$= (12\ +\ 5\ sin(10)) sin(6x)\ +\ (5\ cos(10))cos(6x)\ +\ 5\\ \\ = \sqrt{(12\ +\ 5\ sin(10))^2\ +\ (5cos(10))^2)}\ sin(6x\ +\ tan^{-1} $\frac{(5\ cos(10))}{(12\ +\ 5\ sin(10))}$)\ +\ 5\\ \\ = 16.95\ sin(6x\ +\ 2.93^o)\ +\ 5$