Jump to content
Sign in to follow this  

Proving Using Mathematical Induction

Recommended Posts

This is the Lagrange's trig identity. I think the key to prove by induction is to convert sin (n+3/2)x into sin ((n+1) + 1/2) x = sin (n+1)x cos (x/2) + cos (n+1)x * sin (x/2), but to also do the same for sin (n+1/2)x as sin ((n+1) - 1/2)x
Proof f(n)--> f(n+1)

 sin (n+1/2)x / (2 sin (x/2)) + cos (n+1)x ... use sin (a - b), where a = (n + 1)x, b = x/2

= [sin (n+1)x cos (x/2) - cos (n+1)x sin (x/2) + 2 cos(n+1)x sin (x/2)] / (2 sin (x/2)) 

= [sin (n+1)x cos(x/2) + cos(n+1)x sin (x/2)] / (2 sin (x/2))
= sin (n+3/2) x / (2 sin (x/2))
QED

EDIT:
If you are trying to prove f(1), then maybe tan (x / 2) = sin x / (1 + cos x) is helpful.

Edited by kw0573

Share this post


Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Sign in to follow this  

×
×
  • Create New...