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# trig quickie watch

1. just wanna check a trig question

sinx+sin3x=sin2x

just give us a simple answer (only 1 needed)

i get 60 but can people show how they did this???
2. If you only need one answer, then how about x = 0?!

(cos(x) + i sin(x))^3 = cos(3x) + i sin(3x)
so, taking imaginary part, 3sin(x)cos^2(x) - sin^3(x) = sin(3x).

So sin(x) + sin(3x) = sin(2x) is equivalent to

sin(x) (1 - 2cos(x) + 3cos^2(x) - sin^2(x)) = 0
<=> sin(x) (-2cos(x) + 4cos^2(x)) = 0
<=> 2 sin(x) cos(x) (-1 + 2cos(x)) = 0
<=> sin(x) = 0 or cos(x) = 0 or cos(x) = 1/2.
3. (Original post by lgs98jonee)
just wanna check a trig question

sinx+sin3x=sin2x

just give us a simple answer (only 1 needed)

i get 60 but can people show how they did this???
sinx+sin3x-sin2x=0 =>2sin(2x)cos(x)-sin(2x)=0
4. sinx + sin3x = sin2x
=> 2sin2xcosx = sin2x
=> sin2x(2cosx - 1) =0
etc
5. (Original post by Jonny W)
If you only need one answer, then how about x = 0?!

(cos(x) + i sin(x))^3 = cos(3x) + i sin(3x)
so, taking imaginary part, 3sin(x)cos^2(x) - sin^3(x) = sin(3x).

So sin(x) + sin(3x) = sin(2x) is equivalent to

sin(x) (1 - 2cos(x) + 3cos^2(x) - sin^2(x)) = 0
<=> sin(x) (-2cos(x) + 4cos^2(x)) = 0
<=> 2 sin(x) cos(x) (-1 + 2cos(x)) = 0
<=> sin(x) = 0 or cos(x) = 0 or cos(x) = 1/2.
i think complex numbers is a bit beyond this!
6. expand to

sin x +sin2xcosx + cos2xsinx = 2cosxsinx
sin x + 2sinxcosxcosx + (2(cos^2)x -1)sin x = 2cosxsinx

sin x=0 is a solution

and

1 + 2(cos^2)x + 2(cos^2)x -1) = 2cos x

so

4(cos^2)x = 2cos x

so cos x = 0 is a solution and

cos x = 0.5 is a solution.
7. (Original post by Jonny W)
How did you prove sin(x) + sin(3x) = 2sin(2x)cos(x)?
Using the sum-product formula for sin A + sin B?
8. (Original post by Jonny W)
How did you prove sin(x) + sin(3x) = 2sin(2x)cos(x)?
sinA+sinB=2sin((A+B)/2)cos((A-B)/2). C'mon u should know that!
9. dont bother memorising it, its in the formula book lol

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