sum to product formula

**algebra2** · April 17th 2008, 03:27 PM

use sum to product formula:

[sinx+siny=2sin(x+y)cos(x-y)] to solve [0,2pi).

sin3x + sinx = 0

-----------------------

this is what i did and got stuck on:

1. sin(3x+x)
2. (2sin(3x+x)/2)cos(3x-x)/2
3. 2sin(4x/2)cos(2x/2)
4. 2sin2xcosx

stuck there, what do i do next?

**topsquark** · April 17th 2008, 04:45 PM

Originally Posted by algebra2

use sum to product formula:

[sinx+siny=2sin(x+y)cos(x-y)] to solve [0,2pi).

sin3x + sinx = 0

Sorry, nothing of what you did was correct.

$sin(3x) = sin(2x + x) = sin(2x)~cos(x) + sin(x)~cos(2x)$

$= 2~sin(x)~cos^2(x) + sin(x)(1 - 2sin^2(x))$

$= 2~sin(x)(1 - sin^2(x)) + sin(x)(1 - 2sin^2(x))$

$= 4~sin^3(x) + 3~sin(x)$

So solve
$4~sin^3(x) + 3~sin(x) + sin(x) = 0$

$4~sin^3(x) + 4~sin(x) = 0$

-Dan

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