How would I sum this $\sum_{k = 0}^{\infty} \frac{(-1)^k \cos(x + 2ky)}{(2k+1)!}$?

Question

I'm studying for an admission test. One of the problem asked in a previous exam I'm looking at asks to find the sum of the series $$\sum_{k = 0}^{\infty} \frac{(-1)^k \cos(x + 2ky)}{(2k+1)!}.$$

This is not related to anything I have studied very recently and would appreciate if someone can help me with this. Thank you.

lab bhattacharjee · Accepted Answer · 2017-07-25T13:31:23.387

5

$$(-1)^k\dfrac{\cos(x+2ky)}{(2k+1)!}$$

is the Real part of

$$=\dfrac{i^{2k}e^{i(x+2ky)}}{(2k+1)!}=\dfrac{e^{ix}}{ie^{iy}}\cdot\dfrac{(ie^{iy})^{2k+1}}{(2k+1)!}$$

Now $$2\sum_{k=0}^\infty\dfrac{a^{2k+1}}{(2k+1)!}=e^a-e^{-a}$$

How to prove Euler's formula: $e^{it}=\cos t +i\sin t$? should be handy

edited Jul 25 '17 at 13:31

answered Jul 25 '17 at 13:29

lab bhattacharjee

274,582

That was fast. Thanks so much. – 100110 Jul 25 '17 at 13:30
sir i really appericiate your answer but still it is labouring to separate the real – Noor Aslam Jul 25 '17 at 14:18

How would I sum this $\sum_{k = 0}^{\infty} \frac{(-1)^k \cos(x + 2ky)}{(2k+1)!}$?

1 Answers1