Question

Prove that
\[\frac{1}{2}+\cos x+\cos2x+\cdots+\cos nx=\frac{\sin\left(n+\dfrac{x}{2}\right)}{2\sin\dfrac{x}{2}}\]
(If anyone discovers that this actually isn't an identity, please let me know...)

Answer 1

can we use induction?

Answer 2

I'm not certain, but I can try here if no one knows, please if you know interrupt me

Answer 3

Any way you see fit! I haven't tried it myself, so I can't say if it'll work.

Answer 4

So we need to show it is true for every n, then prove for n+1.

Answer 5

I amthinking we can use identities

Answer 6

so what if we work backwards

Answer 7

If we look at sin(A+B)=? then go through, it may work

Answer 8

ok maybe check this out, I may be out of my league http://math.stackexchange.com/questions/520719/prove-frac12-cosx-cos2x-dots-cosnx-frac-sinn-frac1

Answer 9

I feel like proving those hints may not be simple though

Answer 10

for inductionn n:
for n=1, w have:
\[\frac{ 1 }{ 2 }=\frac{ 1 }{ 2 }\]
now I suppose that your expression is true for n-1, then for n:
\[\frac{ 1 }{ 2 }+\cos x+...+\cos ((n-1)x)+\cos (nx)=\frac{ \sin(n-1/2)x }{ 2 \sin x }+\cos nx=\]
\[=\frac{ \sin(n-1/2)x+2 \cos nx \sin (x/2) }{ 2 \sin (x/2) }\]
we get our thesis,if we note that
\[\sin \left( n-\frac{ 1 }{ 2 } \right)x=\sin nx \cos (x/2)\cos nx \sin (x/2)\]

Answer 11

oops for n=0

Answer 12

but the question is how do we get that, how can we modify our original into that

Answer 13

(I use strong induction ususally, but they are logically equivalent)

Answer 14

oops i have made an error of typo:
\[\sin \left( n-\frac{ 1 }{ 2 } \right)x=\sin nx \cos(x/2)-\cos nx \sin(x/2)\]

Answer 15

we have to subtitute the identity above into the last step of the application of induction principle

Answer 16

after that substitution, we get the subsequent expression at numerator:
\[\sin nx \cos(x/2)-\cos nx \sin(x/2)+2 \cos nx \sin(x/2)=\]
\[=\sin nx \cos (x/2)+\cos nx \sin (x/2)=\sin \left( n+\frac{ 1 }{ 2 } \right)x\]

Answer 17

wow! never seen an induction version for this before looks very interesting xD
Here is an interesting problem that can be solved easily using this identity
http://openstudy.com/study#/updates/53ec6330e4b0f30a87d43077

Answer 18

Alternative :
\[\frac{1}{2}+\sum\limits_{k=1}^n \cos(kx) = \frac{1}{2} + \mathfrak{Real}\sum\limits_{k=1}^n e^{ikx} = \cdots \]

Notice it is a geometric series with common ratio of \(e^{ix}\)
so we can evaluate using partial sum formula and take the real part

Answer 19

Yup I was shooting for the complex exponential form and using induction on that. Nice work @Michele_Laino !

Answer 20

Thank you! @SithsAndGiggles