Proof and example of the formulas for $\sin\left(\sum_{i=1}^\infty\theta_i\right)$ and $\cos\left(\sum_{i=1}^\infty\theta_i\right)$

Was reading the page on "List of trigonometric identities" on Wikipedia and came across an identity for sines and cosines of sums of infinitely many angles whose sum converges absolutely towards a single value. The identities are as follows:

I can't for the life of me find a proof of these identities online nor are they cited from anywhere on the wikipedia page and where would this be useful could someone give a problem for this.

Any help would be appreciated

Like most trigonometric inequalities, you can show it using that $$e^{ix}=\cos (x) +i \sin (x)$$.

In particular, expand both sides of $$e^{i(\sum_j \theta_j)}=\prod_j e^{i\theta_j}$$

• Could you please elaborate Commented Jan 28 at 8:38

$$\sin\left(\sum_{r=1}^{\infty} \theta_r \right) = Im\left(\prod_{r=1}^{\infty}e^{i\theta_r}\right)$$

Now we really just need to simplify this product using Euler's identity. We can write it down as $$P=Im\left(\prod_{r=1}^{\infty} \cos(\theta_r)+i\sin(\theta_r)\right)$$

Now since $$\lim_{r \to \infty} \sin(\theta_r) =0$$, we must use finitely many $$\sin$$ factors in each term and since $$\lim_{r \to \infty} \cos(\theta_r) =1$$, we may use infinitely many $$\cos$$ factors in a term.

When we write this product as a sum of factors, all imaginary terms within the product are those which have different $$\sin$$ multiplied with each other an odd number of times, since $$i^{4k+2}=-1$$. Further since $$i^{4k+1}=i$$ and $$i^{4k+3}=-i$$, for non-negative $$k$$ the terms which have $$\sin$$ factors $$4k-1$$ times will be negative.

So when we multiply the factors and write out the imaginary terms, it will appear as a sum of products of $$\sin$$ of an angle multiplied with $$\cos$$ of all other angles, taken one $$\sin$$ at a time, then three $$\sin$$, then five $$\sin$$ at a time, and so on, for all possible combinations of angles for the $$\sin$$ terms. In terms of product and sum notation, we can write this down as $$i\sum_{t \text{ odd}} (-1)^{[t \bmod 4 = 3]} \sum_{s \subset N, |s|=t} \left( \prod_{r \in s} \sin(\theta_r)\prod_{r\notin s}\cos(\theta_r)\right)$$

Here $$[\ .\ ]$$ stands for Iverson brackets. In essence, the first summation let's us choose how many $$\sin$$ factors each term should contain, $$t$$, and the second summation allows us to choose a combination of $$t$$ angles from the infinitely many angles to be used for the argument in the $$\sin$$. The sign associated with the number of $$\sin$$ terms is also appended.

Now $$t=4k+3$$ implies that $$\frac{t-1}{2} = 2k+1$$ which is an odd value and that means $$(-1)^{[t \bmod 4 = 3]} = (-1)^{\frac{t-1}{2}}$$. So we might as well replace the Iverson bracket with this.

The same follows for the $$\cos$$ expansion, where in order to find the $$Re$$ part you have to take each $$\sin$$ only an even number of terms, while noting that $$i^{4k+1}=-1$$ and $$i^{4k}=1$$ for non-negative $$k$$.