must simplify [3cos(x) - 12sin^2(x)cos(x)] into 3cos(3x), help please

Question

anonymous · Answer

lets see other reply s...

anonymous · Answer

hmm i can get so many variations but no idea how to get 3cos(3x)

anonymous · Answer

oh yes it is !

anonymous · Answer

can yu please show me how!! i know the rule cos (x+y) or something is in it i think but im just so lost

callisto · Answer

Wow!
$$3cosx - 12sin^2 x cosx$$$$= 3 ( cosx -4sin^2 xcosx)$$$$=3[cosx - 2(2sinxcosx)sinx]$$$$=3(cosx-2sin2xsinx)$$$$=3[cosx-2(-\frac{1}{2}cos(2x+x)-cos(2x-x))]$$simplify ... Amazing :D

anonymous · Answer

i got logged out, sorry

anonymous · Answer

ok first off we need an expression for $\cos(3x)$ in terms of $\cos(x)$ 
then we can write the above in terms of $\cos(x)$ and see if they match up 
that will be the strategy i think

anonymous · Answer

$$\cos(2x)=2\cos^2(x)-1$$ and so 
$$\cos(3x)=\cos(2x+x)=$$ actually @Callisto  way a lot shorter, nvm

anonymous · Answer

but if you can start with the fact that 
$$\cos(3x)=4\cos^3(x)-3\cos(x)$$  then it is some simple algebra
if you have to derive that formula, it takes a bunch of steps

anonymous · Answer

nice work  @Callisto

callisto · Answer

It's not shorter... just different approach...
$$3cos3x$$$$=3cos(2x+x)$$$$=3(cos2xcosx-sin2xsinx)$$$$=3(1-2sin^2x)cosx - 3(2sinxcosx)(sinx)$$$$=3cosx-6sin^2xcosx-6sin^2xcosx$$$$=...$$Amazing :)
I should have thought from the result :|

anonymous · Answer

thanks guys!