wnvl
Jan14-12, 06:52 AM
1. The problem statement, all variables and given/known data
Prove the following equation
2^{n-1}\prod_{k=0}^{n-1}\sin(x+\frac{k\pi}{n}) = \sin(nx)
2. Relevant equations
3. The attempt at a solution
Below you find my unsuccessfull attempt using complex numbers.
When you convert it to complex numbers the equality can be rewritten as
2^{n-1}\prod_{k=0}^{n-1}\frac{e^{i(x+\frac{k\pi}{n})}-e^{-i(x+\frac{k\pi}{n})}}{2i} = \frac{e^{inx}-e^{-inx}}{2i}
i^{-n+1}\prod_{k=0}^{n-1}e^{i(x+\frac{k\pi}{n})}-e^{-i(x+\frac{k\pi}{n})} = e^{inx}-e^{-inx}
e^{i\frac{(-n+1)\pi}{2}}\prod_{k=0}^{n-1}e^{i(x+\frac{k\pi}{n})}-e^{-i(x+\frac{k\pi}{n})} = e^{inx}-e^{-inx}
The LHS of this equation contains terms in e^{inx}, e^{i(n-1)x}, e^{i(n-2)x}, ..., e^{-inx}.
Calculate the coefficient for each term.
\underline{e^{inx}}
e^{i\frac{(-n+1)\pi}{2}}\frac{(-n+1)\pi}{2}\prod_{k=0}^{n-1}e^{i(x+\frac{k\pi}{n})}=e^{i\frac{(-n+1)\pi}{2}}e^{inx+i\sum_{k=0}^{n-1}\frac{k\pi}{n}}=e^{inx}
\underline{e^{-inx}}
e^{i\frac{(-n+1)\pi}{2}}\prod_{k=0}^{n-1}-e^{-i(x+\frac{k\pi}{n})}=(-1)^ne^{i(-n+1)\pi}e^{-inx}=(-1)^{n}e^{-inx} = -e^{-inx}
Now we still have to prove that for -n<k<n the coefficient of {e^{ikx}} equals 0 to conclude the proof. But I don't know how to do this.
All suggestions, ideas are welcome. Proofs not using complex numbers will be appreciated as well.
Prove the following equation
2^{n-1}\prod_{k=0}^{n-1}\sin(x+\frac{k\pi}{n}) = \sin(nx)
2. Relevant equations
3. The attempt at a solution
Below you find my unsuccessfull attempt using complex numbers.
When you convert it to complex numbers the equality can be rewritten as
2^{n-1}\prod_{k=0}^{n-1}\frac{e^{i(x+\frac{k\pi}{n})}-e^{-i(x+\frac{k\pi}{n})}}{2i} = \frac{e^{inx}-e^{-inx}}{2i}
i^{-n+1}\prod_{k=0}^{n-1}e^{i(x+\frac{k\pi}{n})}-e^{-i(x+\frac{k\pi}{n})} = e^{inx}-e^{-inx}
e^{i\frac{(-n+1)\pi}{2}}\prod_{k=0}^{n-1}e^{i(x+\frac{k\pi}{n})}-e^{-i(x+\frac{k\pi}{n})} = e^{inx}-e^{-inx}
The LHS of this equation contains terms in e^{inx}, e^{i(n-1)x}, e^{i(n-2)x}, ..., e^{-inx}.
Calculate the coefficient for each term.
\underline{e^{inx}}
e^{i\frac{(-n+1)\pi}{2}}\frac{(-n+1)\pi}{2}\prod_{k=0}^{n-1}e^{i(x+\frac{k\pi}{n})}=e^{i\frac{(-n+1)\pi}{2}}e^{inx+i\sum_{k=0}^{n-1}\frac{k\pi}{n}}=e^{inx}
\underline{e^{-inx}}
e^{i\frac{(-n+1)\pi}{2}}\prod_{k=0}^{n-1}-e^{-i(x+\frac{k\pi}{n})}=(-1)^ne^{i(-n+1)\pi}e^{-inx}=(-1)^{n}e^{-inx} = -e^{-inx}
Now we still have to prove that for -n<k<n the coefficient of {e^{ikx}} equals 0 to conclude the proof. But I don't know how to do this.
All suggestions, ideas are welcome. Proofs not using complex numbers will be appreciated as well.