Trigonometric inverse functions

  • Thread starter lizzie
  • Start date
  • #1
25
0

Main Question or Discussion Point

i want the most general solution for
sin6x=sin4x-sin2x
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
sin6x=sin4x-sin2x

sin6x-sin4x+sin2x=0

Then remember that

[tex]sinP+sinQ=2sin(\frac{P+Q}{2})cos(\frac{P-Q}{2})[/tex]
 
  • #3
25
0
thanks
 

Related Threads for: Trigonometric inverse functions

  • Last Post
Replies
1
Views
547
Replies
1
Views
5K
Replies
11
Views
3K
Replies
6
Views
2K
Replies
10
Views
1K
Top