Express $(1+\cos(x-1))^3$ as a trigonometric polynomial in x. I keep doing this problem and somehow I keep messing up the constants, and it just jumbles up in my head.
$$(1+\cos(x-1))^3$$ $$= (1+\cos(x-1))(1+\cos(x-1))(1+\cos(x-1))$$ $$= (1+\cos(x-1))(1+2\cos(x-1)+\cos^2(x-1))$$ $$= 1+3\cos(x-1)+3\cos^2(x-1)+\cos^3(x-1)$$
Now I hope I've got this correct so far, but after this I making some kind of mistake.
$$=1+3(\cos1\cos x+\sin1\sin x) + 3/2+3/2\cos2(x-1) +3/4(\cos1\cos x+\sin1\sin x) +1/4\cos3(x-1)$$
But then how do I do $\cos2(x-1)$ and $\cos3(x-1)$ ?
Hint: Expand the "inside" of the parenthesis first into $1+\cos(1)\cos(x)+\sin(1)\sin(x)$. Now think of this as $A+B+C$ and expand $(A+B+C)^3$ via usual algebra, for example there will be a term $6ABC$ (and several others). After that if you put the suggested things back for $A,B,C$ it will be a polynomial in $\cos(x),\sin(x).$