$\cos\theta\cos2\theta\cos3\theta + \cos2\theta\cos3\theta\cos4\theta + ...$

Question

Evaluate: $$\cos\theta\cos2\theta\cos3\theta + \cos2\theta\cos3\theta\cos4\theta + …$$ upto $n$ terms

I tried solving the general term $\cos n\theta\cos (n+1)\theta\cos (n+2)\theta$.First, I applied the formula $2\cos\alpha\cos\beta = \cos(\alpha+\beta)+\cos(\alpha-\beta)$ on the two extreme terms. After solving I applied this once again and after further solving arrived at $$\frac{1}{4}[\cos(3n+3)\theta + \cos(n+1)\theta+\cos(n+3)\theta+\cos(n-1)\theta]$$

which I simplified to

$$\frac{\cos n\theta}{2}[\cos\theta+\cos(2n+3)\theta]$$

After this I am stuck as to what else I could do so as to make the telescope or something else to easily calculate the sum using some fact from trigonometry. Or maybe this is a dead end. And help or hints would be appreciated, thanks

Answer 1

$$\cos(n-1)t\cdot\cos nt\cdot\cos(n+1)t$$

$$=\dfrac{\cos nt(\cos2t+\cos2n t)}2$$

$$=\dfrac{\cos2t\cos nt}2+\dfrac{\cos nt+\cos3nt}4$$

$$=\dfrac{2\cos2t+1}4\cdot\cos nt+\dfrac{\cos3nt }4$$

Use $\sum \cos$ when angles are in arithmetic progression

Answer 2

Let's write $c_n:=\cos(n\theta),s_n:=\sin(n\theta)$ to simplify notation. We observe:$$ \begin{split}c_nc_{n+1}c_{n+2}&=c_{n+1}(c_{n+1}c_1+s_{n+1}s_1)(c_{n+1}c_1-s_{n+1}s_1)\\&=c_{n+1}(c_{n+1}^2c_1^2-s_{n+1}^2s_1^2)\\&=c_{n+1}(c_{n+1}^2(c_1^2+s_1^2)-s_1^2)\\&=c_{n+1}^3-c_{n+1}s_1.\end{split} $$ Hence, we see that $$\begin{split}\sum_{k=1}^nc_kc_{k+1}c_{k+2}=\sum_{k=2}^nc_k^3-s_1\sum_{k=2}^nc_k.\end{split}$$ You can find a reference for the latter sum here. For the first one, we may use the complex exponential representation $$\cos(\theta)=\frac12\left(e^{i\theta}+e^{-i\theta}\right)$$ in order to deduce that $$\cos(k\theta)^3=\frac18\left(e^{3ik\theta}+3e^{ik\theta}+3e^{-ik\theta}+e^{-3ik\theta}\right).$$

Summing this is now just a sum of four different geometric series, which should be easy to do (of course, you want to convert it back into sines and cosines at the end). This method can also be used for the second sum, actually.

Answer 3

hint

$$\cos((n-1)t)\cos(nt)\cos((n+1)t)=$$

$$\frac 18(e^{i(n-1)t}+e^{-i(n-1)t})(e^{int}+e^{-int})(e^{i(n+1)t}+e^{-i(n+1)t})$$

you get then geometric series.