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I have a problem with deriving the formula for the sum: $cos(\phi) + cos(3\phi) + \dots + cos((2n+1)\phi)=\dots $ Any hint of which series should be used here?

Mia
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  • https://math.stackexchange.com/q/17966/892252 Hopefully this will help you. It has both a trigonometrical and a complex numbers based solution in the answers. –  Feb 27 '21 at 13:22

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$$\underbrace{cos(\phi) + cos(3\phi)}_{3\phi-\phi=2\phi}+ \dots + cos((2n+1)\phi)$$ so multiply/divide by half of deffernce betwen angles $\frac{2\cos(\phi)}{2\cos(\phi)}$ then use product to sum formulas $$\frac{2\cos(\phi)}{2\cos(\phi)}(cos(\phi) + cos(3\phi) + \dots + cos((2n+1)\phi))=\\\frac{1}{2\cos(\phi)}(2\cos(\phi)\cos(\phi)+2\cos(\phi)\cos(3\phi)+2\cos(\phi)\cos(5\phi)+...+2\cos(\phi)cos((2n+1)\phi))= \frac{1}{2\cos(\phi)}\begin{cases}\cos(\phi-\phi))-\cos(\phi+\phi)\\\cos(\phi-3\phi))-\cos(\phi+3\phi)\\\cos(\phi-5\phi))-\cos(\phi+5\phi)\\\vdots\\\cos(\phi-(2n-1)\phi))-\cos(\phi+(2n+1)\phi)\end{cases} $$simplify $$=\frac{1}{2\cos(\phi)}\begin{cases}\cos(\phi-\phi)-\cos(\phi+(2n+1)\phi)\}\end{cases}$$

Khosrotash
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