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lmati · Apr 22, 2010, 04:50 PM

write 24 sin3x cos3x+10 cos^2 3x-5 in terms of one sin function

galactus · Apr 24, 2010, 04:08 AM

Is that

$24sin(3x)cos(3x)+10cos^{2}(3x-5)$ ?

lmati · Apr 24, 2010, 08:13 PM

Yes

Unknown008 · Apr 25, 2010, 05:03 AM

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$