Show that:
$$\log(\sin(x)) = -\log(2) - \sum_{n \ge 1} \frac{\cos(2nx)}{n}$$
$$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$
$$\log(\sin(x)) = \log(e^{ix} - e^{-ix}) - \log(2i)$$
$$= \log(e^{ix} - e^{-ix}) - \log(2) - \frac{i \pi}{2}$$
What next?
Asked
Active
Viewed 349 times
0
-
See here and here and here for the full solution. – Matthew Cassell Feb 22 '15 at 09:49
1 Answers
1
$$e^{ix}-e^{-ix}=e^{ix}[1-e^{-2ix}]$$
$$\ln[e^{ix}-e^{-ix}]=ix+\ln[1-e^{-2ix}]$$
Now $\ln(1-x)=-\sum_{r=1}^\infty\dfrac{x^r}r$
Use Euler formula and equate the real and the imaginary parts
See also: Convergence for log 2
lab bhattacharjee
- 274,582
-
Wait so using: $$\sin^2(x) = \frac{1 - \cos(2x)}{2} = \frac{1 - e^{2ix}}{2}$$ – Lebes Feb 22 '15 at 10:48
-
@Lebes, No. We have $$\ln(1-e^{-2ix})=-\sum_{r=1}^{\infty}\frac{(e^{-2ix})^r}r$$ Now $(e^{-2ix})^r=e^{-i2rx}=\cos(2rx)-i\sin(2rx)$ – lab bhattacharjee Feb 22 '15 at 10:50
-
Okay so:
$$\log(1- e^{-2ix}) = -\sum_{r=1}^{\infty}\frac{\cos(2rx)}{r} + i\sum_{r=1}^{\infty} \frac{\sin(2rx)}{r}$$
But how to elimate $\sin$? @lab
– Lebes Feb 22 '15 at 11:14 -
@Lebes, The answer has "equate the real and the imaginary parts" – lab bhattacharjee Feb 22 '15 at 11:15
-
I see. let the expression by $H$. Then: $\Im H = \sum_{r=1}^{\infty} \frac{\sin(2rx)}{r}$. And on the LHS it is $0i$ hence:
$$\sum_{r=1}^{\infty} \frac{\sin(2rx)}{r}$$ Doesnt exist. So:
$$\log(1 - e^{-2ix}) = -\sum_{r=1}^{\infty} \frac{\cos(2rx)}{r}$$
And:
$$1 - e^{-2ix} = 2\sin^2(x)$$
So:
$$\log(\sin^2(x)) - \log(2) = -\sum_{r=1}^{\infty}...$$
That isnt correct though @lab
– Lebes Feb 22 '15 at 11:22