anonymous
  • anonymous
Trig Identities Question: say we have sin(x+deltax) = sinx*cosdeltax + cosx*sindeltax and we want to introduce -sinx on both sides. So, how can I put -sinx on the left, and a -1 on the right like this: sin(x+deltax) - sinx = sinx*(cosdeltax-1) + cosx*sindeltax Professor Strang is doing this in a video lecture, and I want to understand what the basis for this is. Can you help me understand this?
Mathematics
chestercat
  • chestercat
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jim_thompson5910
  • jim_thompson5910
I'm going to use y instead of deltax to save time. sin(x+y) = sin(x)*cos(y) + cos(x)*sin(y) Now subtract sin(x) from both sides sin(x+y) - sin(x) = sin(x)*cos(y) + cos(x)*sin(y) - sin(x) Group like terms and factor sin(x+y) - sin(x) = (sin(x)*cos(y) - sin(x)) + cos(x)*sin(y) sin(x+y) - sin(x) = sin(x)*(cos(y) - 1) + cos(x)*sin(y) Now if you want, you can replace y with deltax again to get sin(x+deltax) - sin(x) = sin(x)*(cos(deltax) - 1) + cos(x)*sin(deltax)
anonymous
  • anonymous
you are thinking way too hard. it is not trig, it is algebra
anonymous
  • anonymous
what jimthompson wrote. it is not a trig identity at all

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anonymous
  • anonymous
Thanks, I seem to have lost my mind due to cos and sin being on the scene.

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