# If i^2=-1 then the sum of cos45+icos135+…i^n*cos(45+90n)+…i^40*cos3645=?

Jul 21, 2018

cos45+icos135+…i^n*cos(45+90n)+…i^40*cos3645=?

Sum of First 4 terms

${\sum}_{r = 1}^{r = 4} {S}_{r}$

$= \cos 45 + i \cos 135 + {i}^{2} \cos \left(45 + 90 \times 2\right) + {i}^{3} \cos \left(45 + 90 \times 3\right)$

$= \cos 45 - i \cos 45 + \cos 45 - i \cos 45$

$= \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} + \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} = \sqrt{2} \left(1 - i\right)$

Sum of Second 4 terms

${\sum}_{r = 5}^{r = 9} {S}_{r}$

$= {i}^{4} \cos \left(45 + 90 \times 4\right) + {i}^{5} \cos \left(45 + 90 \times 5\right) + {i}^{6} \cos \left(45 + 90 \times 6\right) + {i}^{7} \cos \left(45 + 90 \times 7\right)$

$= \cos 45 - i \cos 45 + \cos 45 - i \cos 45$

$= \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} + \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} = \sqrt{2} \left(1 - i\right)$

Considering this way we get 10 groups of 4terms having same sum.

So our sum

$S = {\sum}_{r = 1}^{r = 40} {S}_{r} + {i}^{40} \cdot \cos 3645$

$= 10 \times \sqrt{2} \left(1 - i\right) + \cos \left(45 + 90 \times 40\right)$

$= 10 \sqrt{2} \left(1 - i\right) + \cos 45$

$= 10 \sqrt{2} \left(1 - i\right) + \frac{1}{\sqrt{2}}$