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Thread: prove that if sin(a/2) ≠ 0 then...

  1. #1
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    prove that if sin(a/2) ≠ 0 then...

    prove that if sin(a/2) ≠ 0 then

    cosa + cos2a +...+cos(n+1)a = (sin((n+1)a/2)*cos((n+1)a/2)/sin(a/2)

    FInd this sum also in the case sin(a/2) = 0

    Find a similar formula for sina + sin 2a +...+ sin(n+1)a
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  2. #2
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    $\displaystyle cosa + cos2a +...+cos(n+1)a=S$

    Multiply it by $\displaystyle sin(\frac{a}{2})$. When you have sums like this - multiply by $\displaystyle sin(\frac{r}{2})$, r is ratio. Is signs alternate multiply the sum by $\displaystyle cos(\frac{r}{2})$.
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  3. #3
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    Not sure what you mean.
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  4. #4
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    $\displaystyle cosa + cos2a +...+cos(n+1)a=S \Rightarrow (cosa + cos2a +...+cos(n+1)a)sin(\frac{a}{2})=S\cdot sin(\frac{a}{2})$
    $\displaystyle sin a \cdot cos b=\frac{1}{2}(sin (a+b)+sin(a-b))$

    "Is signs alternate multiply the sum by $\displaystyle cos(\frac{a}{2})$." for example: $\displaystyle S=cos a - cos 2a + cos 3a - cos 4a+...+(-1)^n cos(n+1)a$
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  5. #5
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    still not sure what you mean
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  6. #6
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    Okay, what do you think I mean?
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