Mathematical Induction (1 Viewer)

[serge]

hey people this question has been getting on my nerves

prove by induction

sinq + sin3q + sin5q+...+sin(2n-1)q= [(1-cos2nq) / 2sinq]

using the already proven identity that

[cosy-cos(y+2q)] / 2sinq= sin(y+q)

 

[KFunk]

hey people this question has been getting on my nerves

prove by induction

sinq + sin3q + sin5q+...+sin(2n-1)q= [(1-cos2nq) / 2sinq]

using the already proven identity that

[cosy-cos(y+2q)] / 2sinq= sin(y+q)

Using [cosy-cos(y+2q)] / 2sinq= sin(y+q) you get the identity:

[cosy-cos(y+2q)] = 2sinqsin(y+q)

You assume true for k so:
sinq + sin3q + sin5q+...+sin(2k-1)q= [(1-cos2kq) / 2sinq] then you add sin(2k+1)q to both sides...

LHS = sinq + sin3q + sin5q+...+sin(2k-1)q + sin(2k+1)q
RHS = (1-cos2kq)/2sinq + sin(2k +1)q
= [2sinqsin(2k+1)q +1 - cos2kq]/2sinq

Looking at the identity you know that 2sinqsin(2kq+q) = cos2kq - cos(2kq + 2q) so,

RHS = [cos2kq - cos(2kq + 2q) + 1 - cos2kq]/2sinq
= [1 - cos2(k+1)q]/2sinq

Hence S_k ==> S_k+1

 

[.ben]

well done KFunk were you in Mr. Hayes' class at school?

 

[KFunk]

If things had been right I should have been. They didn't really put me in a class. I had a couple periods a week with Mr Niven ontop of my normal 3U classes instead.

 

[.ben]

oh that's bad. was niven crap? should we be scared if we get him for 3/4U?