Mathematical Induction

[Emmanuel]

Prove by mathematical induction that ½+cos(a)+…+cos(n-1)(a)=(sin(a/2)(2n-1))/(2sin(a/2))

 

[ksdhart]

I'm sorry, but I'm finding it quite difficult to understand what the problem you've been tasked with is, exactly... I believe, if properly formatted, the problem should look like this:

\(\displaystyle \displaystyle \frac{1}{2}+cos\left(1a\right)+cos\left(2a\right)+cos\left(3a\right)+...+cos\left(\left(n-1\right)a\right)= \frac{sin\left(\left(\frac{a}{2}\right)\left(2n-1\right)\right)}{2sin\left(\frac{a}{2}\right)}\)

Can you please confirm if the above is the correct problem, so that we're all operating on the same page. When you reply back, if you could also please include any efforts you've made and work you've done on this problem (even if you know it's wrong) that would be helpful as well. Thank you.

 

[Emmanuel]

you are correct in your question i have only been able to prove the statement for n=1

 

[Emmanuel]

re:math induction

your math statement is correct. I have only been able to prove the statement for n=1

 

[Prove It]

Prove by mathematical induction that ½+cos(a)+…+cos(n-1)(a)=(sin(a/2)(2n-1))/(2sin(a/2))

I would personally lean more towards using a direct proof with complex series instead. Notice that \(\displaystyle \displaystyle \begin{align*} \mathrm{e}^{\mathrm{i}\,a} \equiv \cos{ \left( a \right) } + \mathrm{i}\sin{ \left( a \right) } \end{align*}\), that means we could write

\(\displaystyle \displaystyle \begin{align*} \sum_{k = 1}^{n-1} \left[ \left( \mathrm{e}^{\mathrm{i}\,a} \right)^k \right] &= \sum_{k = 1}^{n - 1} \cos{ \left( k\,a \right) } + \mathrm{i}\sum_{k = 1} \sin{ \left( k\,a \right) } \end{align*}\)

Notice that the LHS is a geometric series, with common ratio \(\displaystyle \displaystyle \begin{align*} \mathrm{e}^{\mathrm{i}\,a} \end{align*}\), thus its closed form solution is

\(\displaystyle \displaystyle \begin{align*} \sum_{k = 1}^{n-1} \left[ \left( \mathrm{e}^{\mathrm{i}\,a} \right) ^k \right] &= \frac{\mathrm{e}^{\mathrm{i}\,a} \, \left( 1 - \mathrm{e}^{\left( n - 1 \right) \, \mathrm{i}\,\mathrm{a} } \right)}{1 - \mathrm{e}^{\mathrm{i}\,a} } \\ &= \frac{\left[ \cos{ \left( a \right) } + \mathrm{i}\sin{ \left( a \right) } \right] \left\{ 1 - \cos{ \left[ \left( n - 1 \right) \, a \right] } - \mathrm{i}\sin{ \left[ \left( n - 1 \right) \, a \right] } \right\} }{1 - \cos{ \left( a \right) } - \mathrm{i}\sin{ \left( a \right) } } \end{align*}\)

with some simplification you would be able to write this in terms of its real and imaginary parts. I'd start by simplifying by multiplying the top and bottom by the bottom's conjugate. The real part is the series you are trying to find (at least once you add 1/2 to it). You might also need to apply some trigonometric identities.

 

[Steven G]

you are correct in your question i have only been able to prove the statement for n=1

So what did you do next? Can you show us your work. Even if you know it's wrong, that is ok. This way we can point you in the correct direction.