To prove: 1. \(\displaystyle{\sum_{{{k}={1}}}^{{n}}}{\sin{{\left({2}{k}-{1}\right)}}}\theta={\frac{{{{\sin}^{{2}}{n}}\theta}}{{{\sin{\theta}}}}}\) 2. \(\displaystyle{\sum_{{{k}={1}}}^{{n}}}{{\sin}^{{2}}{\left({2}{k}-{1}\right)}}\theta={\frac{{{n}}}{{{2}}}}-{\frac{{{\sin{{4}}}{n}\theta}}{{{4}{\sin{{2}}}\theta}}}\)

Reagan Gomez

Reagan Gomez

Answered question

2022-04-09

To prove:
1. k=1nsin(2k1)θ=sin2nθsinθ
2. k=1nsin2(2k1)θ=n2sin4nθ4sin2θ

Answer & Explanation

srasloavfv

srasloavfv

Beginner2022-04-10Added 6 answers

From cos(a+b)=cosacosbsinasinb we get
sinasinb=cos(ba)cos(b+a)2
Giving appropriate values to a, b, ontain a telescopical on the left.
Let a=x, b=(2k1)x. Then
sinxsin(2k1)x. Then
sinxk=1nsin(2k1)x=1cos2nx2
so
k=1nsin(2k1)x=sin2nxsinx
tabuevniru8op

tabuevniru8op

Beginner2022-04-11Added 14 answers

For the second one, note that
sin2(2k1)θ=1cos(2k1)2θ2
so it suffices to find
k=1ncos(2k1)x
And again, we can use
sin(a+b)sin(ab)2=sinbcosa
to get
sinxcos(2k1)x=sin2kxsin(2k2)x2
k=1ncos(2k1)x=sin2nx2sinx
so
k=1nsin2(2k1)x=n2sin4nx4sin2x

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