Do Trig Ever Age?

The average of the sequence of number is given by:

$\begin{aligned} \mu & = \frac 1{90} \sum_{k=1}^{90} 2k\sin (2k^\circ) \quad \quad \small \color{#3D99F6}{\text{As }\sin 180^\circ = 0} \\ & = \frac 1{90} \sum_{k=1}^{\color{#3D99F6}{89}} 2k\sin (2k^\circ) \quad \quad \small \color{#3D99F6}{\text{By }\sum_a^b f(k) = \sum_a^b f(a+b-k)} \\ & = \frac 1{180} \sum_{k=1}^{89} \left(2k\sin (2k^\circ) + (180-2k)\sin (180^\circ -2k^\circ) \right) \quad \quad \small \color{#3D99F6}{\text{Note: }\sin (180^\circ -x) = \sin x} \\ & = \sum_{k=1}^{89} \sin (2k^\circ) = \sum_{k=1}^{90-1} \sin \left(\frac {k\pi}{90} \right) = \cot \left(\frac {k\pi}{180} \right) = \boxed{\cot 1^\circ} \quad \quad \color{#3D99F6}{\text{(see Note.)}} \end{aligned}$

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Note:

Proving $\displaystyle \sum_{k=1}^{n-1} \sin \left(\frac {k \pi}n \right) = \cot \left(\frac {k \pi}{2n} \right)$

$\begin{aligned} \sum_{k=1}^{n-1} \sin \left(\frac {k \pi}n \right) & = \Im \left \{ \sum_{k=1}^{n-1} e^{\frac {k \pi}ni} \right \} = \Im \left \{ e^{\frac \pi n i} \left(\frac {1- e^{\frac {(n-1)\pi}n i}}{1- e^{\frac \pi n i}} \right) \right \} = \Im \left \{ \frac {e^{\frac \pi n i} - e^{\pi i}}{1- e^{\frac \pi n i}} \right \} \\ & = \Im \left \{ \frac {e^{\frac \pi n i} +1}{1- e^{\frac \pi n i}} \right \} = \Im \left \{ \frac {e^{\frac \pi{2n} i} + e^{-\frac \pi{2n} i}}{e^{-\frac \pi{2n} i} - e^{\frac \pi{2n} i}} \right \} = \Im \left \{ \frac {\cos \left(\frac \pi{2n}\right)}{-i \sin \left(\frac \pi{2n}\right)} \right \} \\ & = \Im \left \{i \cot \left(\frac \pi{2n}\right) \right \} = \boxed{\cot \left(\dfrac \pi{2n}\right)} \end{aligned}$

Consider $S\,=\,\cos(2x)+\cos(4x)+\cos(6x)+...+\cos(180x)$ .

Using, $\sum_{n=1}^k \,\cos \left( \alpha\,+\,(n-1)\cdot \beta \right) \,=\, \dfrac{ \sin \left(k \cdot \dfrac{\beta}{2} \right) }{ \sin \left(\dfrac{\beta}{2} \right)} \cdot \cos\left(\alpha\,+\,(n-1) \cdot \dfrac{\beta}{2} \right)$

$S\,=\,\dfrac{1}{2} \cdot \dfrac{\sin(181x)}{\sin(x)}\,-\,\dfrac{1}{2}$ .

So, $\dfrac{1}{2} \cdot \dfrac{\sin(181x)}{\sin(x)}\,-\,\dfrac{1}{2}\,=\, \cos(2x)+\cos(4x)+\cos(6x)+...+\cos(180x)$ .

Differentiating with respect to $x$ ,

$\dfrac{1}{2} \cdot \dfrac{181\cdot\sin(x)\cdot \cos(181x)\,-\,\cos(x) \cdot \sin(181x)}{\sin^{2}(x)}\,=\, -\left(2\sin(2x)+4\sin(4x)+6\sin(6x)+...+180\sin(180x) \right)$ .

Putting $x\,=\,1$ gives, $2\sin(2^{\circ})+4\sin(4^{\circ})+6\sin(6^{\circ})+...+180\sin(180^{\circ})\,=\,90\cdot \cot(1^{\circ})$ . Now, since, there're $90$ terms in the given sequence, therefore, its average is $\cot(1^{\circ})$ .