BlackJack
Vertigo!
I don't get this particular q...
(given, from previous parts.)
c=1+cos@.cos@+cos^2@.cos2@+...cos^(n-1)@.cos(n-1)@
s=cos@.sin@+cos^2@.sin2@+...cos^(n-1)@.sin(n-1)@
c+is=(1-cos^n@.e^[in@])/(1-cos@.e^[i@])
(1-cos@.e^[i@])(1-cos@.e^-1@)=sin^2@
Hence, show
c=1+(cos^n@.sin(n-1)@)/sin@
and find a similar expression for s.
I'm trying to find a way to express the real part of c + is after I slot the fourth equation into the third, but I can't rearrange the third one to suit it.
Any hints? It's for an assignment, so no full answers right now plz.
(hmmm... btw, e^[i@]==cos@ + isin@... and e^[-i@]==cos@ - isin@)
(given, from previous parts.)
c=1+cos@.cos@+cos^2@.cos2@+...cos^(n-1)@.cos(n-1)@
s=cos@.sin@+cos^2@.sin2@+...cos^(n-1)@.sin(n-1)@
c+is=(1-cos^n@.e^[in@])/(1-cos@.e^[i@])
(1-cos@.e^[i@])(1-cos@.e^-1@)=sin^2@
Hence, show
c=1+(cos^n@.sin(n-1)@)/sin@
and find a similar expression for s.
I'm trying to find a way to express the real part of c + is after I slot the fourth equation into the third, but I can't rearrange the third one to suit it.
Any hints? It's for an assignment, so no full answers right now plz.
(hmmm... btw, e^[i@]==cos@ + isin@... and e^[-i@]==cos@ - isin@)
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