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

  • 1
    $\begingroup$ Note that with $\theta=0$ you get a result of $n$, showing there is no upper bound on the sequence; with $\theta=\frac{\pi}2$ there is at least one portion of each term that is zero. Also, consider running the formula you mention in reverse on two consecutive terms... $\endgroup$
    – abiessu
    Jul 6, 2020 at 13:12
  • $\begingroup$ Out of curiosity, where is this problem from? $\endgroup$
    – YiFan Tey
    Jul 6, 2020 at 13:16
  • $\begingroup$ @abiessu Can you please elaborate on your last point? $\endgroup$
    – Amadeus
    Jul 6, 2020 at 13:20
  • 1
    $\begingroup$ @YiFan I am preparing for an Indian engineering entrance exam, and this is from a book by an Indian author. Although I presume that it is not an original question and was taken from somewhere else. $\endgroup$
    – Amadeus
    Jul 6, 2020 at 13:22
  • $\begingroup$ @l1mbo Thank you! This seems like a very interesting (and challenging) question for an engineering entrance exam. $\endgroup$
    – YiFan Tey
    Jul 6, 2020 at 13:28

3 Answers 3


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

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

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

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

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


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.




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

you get then geometric series.


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