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My attempt:

First I converted this expression to sum notation:

$\sin(1^\circ) + \sin(3^\circ) + \sin(5^\circ) + ... + \sin(175^\circ) + \sin(177^\circ) + \sin(179^\circ)$ = $\sum_{n=1}^{90}\sin(2n-1)^\circ$

Next, I attempted to use Euler's formula for the sum, since I needed this huge expression to be simplified in exponential form:

$\sum_{n=1}^{90}\sin(2n-1)^\circ$ = $\operatorname{Im}(\sum_{n=1}^{90}cis(2n-1)^\circ)$

$\operatorname{Im}(\sum_{n=1}^{90}cis(2n-1)^\circ)$ = $\operatorname{Im}(\sum_{n=1}^{90}e^{i(2n-1)^\circ})$

$\operatorname{Im}(\sum_{n=1}^{90}e^{i(2n-1)^\circ})$ = $\operatorname{Im}(e^{i} + e^{3i} + e^{5i} + ... + e^{175i} + e^{177i} + e^{179i})$

Next, I used the sum of the finite geometric series formula on this expression:

$\operatorname{Im}(e^{i} + e^{3i} + e^{5i} + ... + e^{175i} + e^{177i} + e^{179i})$ = $\operatorname{Im}(\dfrac{e^i(1-e^{180i})}{1-e^{2i}})$

$\operatorname{Im}(\dfrac{e^i(1-e^{180i})}{1-e^{2i}})$ = $\operatorname{Im}(\dfrac{2e^i}{1-e^{2i}})$

Now I'm stuck in here;

Albus Dumbledore
  • 409

4 Answers4

2

Without complex numbers. We have $$\cos (x-y)=\cos x \cos y+\sin x \sin y,$$ $$\cos (x+y)=\cos x \cos y-\sin x \sin y.$$ Subtracting and dividing by $2$ we have $$\frac {1}{2}[\;\cos (x-y)-\cos (x+y)\;]=\sin x \sin y.$$ $$\text {Therefore }\quad \sum_{k=1}^{89}\sin (2 k-1)^\circ=\frac {1}{\sin 1^\circ}\sum_{k=1}^{89}\sin (2 k-1)^\circ\sin 1^\circ=$$ $$=\frac {1}{\sin 1^\circ}\sum_{k=1}^{89}\frac {1}{2}[\;\cos (2 k-2)^\circ-\cos (2k)^\circ\;]=\frac {1}{\sin 1^\circ}$$ because the summation telescopes to $\cos 0^\circ-\cos 180^\circ$, which is $2$.

Quuxplusone
  • 601
DanielWainfleet
  • 57,985
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Observe $\sin (p\,^{\circ})=\sin\left(\frac{p\pi}{180}\right)$, hence \begin{align} \sum_{k=1}^{90}\sin\left((2k-1)^{\circ}\right)&=\sum_{k=1}^{90}\sin\left(\frac{(2k-1)\pi}{180}\right)\\[5pt] &=\Im\left(\sum_{k=1}^{90}e^{i\cdot\frac{(2k-1)\pi}{180}}\right)\\[5pt] &=\Im\left(e^{i\frac{\pi}{180}}\cdot\sum_{k=0}^{90}\left(e^{i\cdot\frac{\pi}{89}}\right)^k\right)\\[5pt] &=\Im\left(e^{i\frac{\pi}{180}}\cdot\frac{1-e^{i\cdot\frac{\pi}{90}\cdot 90}}{1-e^{i\cdot\frac{\pi}{90}}}\right)\\[5pt] &=\Im\left(\frac{2e^{i\frac{\pi}{180}}}{1-e^{i\cdot\frac{\pi}{90}}}\right)\qquad\text{since }e^{i\pi}=-1 \end{align}

Ángel Mario Gallegos
  • 19,882
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What you have done is not entirely kosher. Euler's identity $e^{ix} = cos\, x + i\, sin\, x$ is true when x is in radians. However, it does work. There exists an $a$ such that $a^{i\theta} = cos\, \theta + i\, sin\, \theta$ with $\theta$ in degrees.

$\operatorname{Im}(\dfrac{2a^i}{1-a^{2i}})$

Multiply by the complex conjugate of the denominator. $(1-a^{-2i})$

$\operatorname{Im}(\dfrac{2(a^i-a^{-i})}{2-a^{2i}-a^{-2i}})$

$a^{i\theta}+a^{-i\theta} = 2\, cos \theta$; $a^{i\theta}-a^{-i\theta} = 2i\, sin \theta$

$\operatorname{Im}(\dfrac{4i\, sin\, 1}{2-2\,cos\,2})$

$\dfrac{2\, sin\, 1}{1-cos\,2}$

Doug M
  • 57,877
1

You have a sum of odd powers of $z=c_1+is_1$,

$$\sum_0^{n-1} z^{2k+1}=z\sum_0^{n-1}(z^2)^k=z\frac{1-z^{2n}}{1-z^2}.$$

Then multiplying by the conjugate,

$$z\frac{1-z^{2n}}{1-z^2}=\frac{(1-z^{2n})(z-z\bar z^2)}{(1-z^2)(1-\bar z^2)}=\frac{(1-c_{2n}-is_{2n})2is_1}{1-2c_2+1}.$$

The imaginary part is

$$\frac{(1-c_{2n})s_1}{1-c_2}=\frac{(1-\cos(180°))\sin(1°)}{1-\cos(2°)}.$$