Question

Proof by induction of this:

Answer 1

cosx + cos2x + ... + cosnx =\[\frac{ \cos[\frac{ x }{ 2 }(n+1)]sen \frac{ nx }{ 2 }}{ sen(\frac{ x }{ 2} }\]

Answer 2

I did base caseand some induction, but it seems that my algebra is not helping

Answer 3

what is sen?

Answer 4

original sin

Answer 5

secant?

Answer 6

oh sine

Answer 7

sorry, sine

Answer 8

so is the following true:
the case for n=1
\[\frac{\cos(\frac{x}{2}(1+1))\sin(\frac{1x}{2})}{\sin(\frac{x}{2})}=\cos(x)\]
base case is always the easiest
The induction part might actually be hard
I haven't tried yet

Answer 9

maybe when you add you can use some trig identity right?

Answer 10

oh wait i screwed that up

Answer 11

\[\frac{\cos(\frac{x}{2}(k+1))\sin(\frac{kx}{2})}{\sin(\frac{x}{2})}+\cos((k+1)x)=\frac{\cos(\frac{x}{2}(k+2))\sin(\frac{(k+1)x}{2})}{\sin(\frac{x}{2})}\]

Answer 12

some identity with \(\sin(\frac{x}{2})\cos((k+1)x)\)

Answer 13

it is half of something
\[\sin(\alpha)\cos(\beta)=\frac{1}{2}\sin(\alpha+\beta)\] maybe that will do it

Answer 14

2 sin a cos b=sin (a+b)+sin (a-b)

Answer 15

I've tried that already, no luck

Answer 16

I will try again