My question is as follows: If $i^{2}$ = -1, find the value of $$\cos45° + i\cos135° + \ ...\ + i^{n}\cos(45+90n)° + \ ...\ + i^{40}\cos3645°$$ without the aid of a calculator. In terms of my attempts at the problem, I have begun by noticing that $$\cos(180°+x) = \cos180°\cos x - \sin180°\sin x = -\cos x,$$ and so $$\cos45° = i^{2}\cos225° = i^{4}\cos315° = \ ... \ = i^{40}\cos3645°.$$ How do I simplify this summation with this information in mind? I do not have any idea how to do so efficiently. Many thanks.
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Both the cosines and the powers of $i$ recur with period $4$, so you only need to compute the first four terms. There are eleven of the first and ten of each subsequent. So $$\cos 45^\circ=\frac {\sqrt 2}2\\ i\cos 135^\circ=-i\frac {\sqrt 2}2\\ i^2\cos 225^\circ=\frac {\sqrt 2}2\\ i^3\cos 315^\circ=-i\frac {\sqrt 2}2$$ and in fact we have repetition with period $2$ because the $i^2$ cancels with the change of sign of the cosine, so you sum is $21 \frac {\sqrt 2}2-20i\frac {\sqrt 2}2$
Ross Millikan
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As the principal value of $i$ is $e^{i90^\circ}$ and $e^{i180^\circ}=-1,$ using Euler Formula for any integer $r,$
$$i^r\{\cos(90^\circ r+45^\circ)+i\sin(90^\circ r+45^\circ)\}=(e^{ i90^\circ})^r e^{i(90^\circ r+45^\circ)}=e^{i45^\circ}(-1)^r$$
$$i^r\{\cos(90^\circ r+45^\circ)-i\sin(90^\circ r+45^\circ)\}=e^{r i90^\circ}\cdot e^{-i(90^\circ r+45^\circ)}=-e^{-i45^\circ}$$
$$\implies2\sum_{r=0}^{2n}i^r\cos(90^\circ r+45^\circ)=e^{i45^\circ}(-1)^0+(2n+1)(-e^{-i45^\circ})=?$$
Now use Euler Formula to find $e^{\pm i45^\circ}=\dfrac{1\pm i}{\sqrt2}$
lab bhattacharjee
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See Article $#76$ of https://archive.org/details/treatiseonplanet00hobsuoft
– lab bhattacharjee Jul 17 '16 at 04:25