Sine/cosine series $$\frac{\sin²(1°) + \sin²(2°) + \sin²(3°) + .. + \sin²(90°)}{\cos²(1°) + \cos²(2°) + \cos²(3°) + .. + \cos²(90°)} = ?$$
I tried to use multiple identities but I couldn't simplify the expression. Where should I start? 
 A: HINT: For every argument $n^\circ$ we have $\sin^2n^\circ=\cos^2(90^\circ-n^\circ)$, where $n\in\{1,2,\dots, 89\}$.
A: Hint: Sketch the graph of $\sin^2$ and $\cos^2$ between $0$ and $90$ degrees. There is a symmetry between the numerator and the denominator. To formalize that, use $\sin n = \cos (90^o - n)$. The final answer is simple or would be if $\cos^2 0^o$ were also in the denominator.
A: Because $\sin(x)=\cos(90^\circ-x)$ and $\sin^2x+\cos^2x=1$, you can collapse most of the terms in the numerator and denominator separately, and your fraction simplifies to
$$ \frac{44+\sin^2(45^\circ)+\sin^2(90^\circ)}{44+\cos^2(45^\circ)+\cos^2(90^\circ)} = \frac{44+1/2+1}{44+1/2+0} $$
A: $$\begin{align}
&\frac{\sin^2 1+\sin^2 2+\sin^2 3+\dots+\sin^2 44+\sin^2 45+\sin^2 46+\dots+\sin^2 90}{\cos^2 1+\cos^2 2+\cos^2 3+\dots+\cos^2 44+\cos^2 45+\cos^2 46+\dots+\cos^2 90}\\
&=\frac{\sin^2 1+\dots+\sin^2 44+\sin^2 45+\sin^2 (90-44)+\dots+\sin^2 (90-1)+\sin^2 90}{\cos^2 1+\dots+\cos^2 44+\cos^2 45+\cos^2 (90-44)+\dots+\cos^2 (90-1)+\cos^2 90}\\
&=\frac{\sin^2 1+\dots+\sin^2 44+\frac{1}{2}+\cos^2 44+\dots+\cos^2 1+1}{\cos^2 1+\dots+\cos^2 44+\frac{1}{2}+\sin^2 44+\dots+\sin^2 1+0}\\
&=\frac{\overbrace{1+1+\dots+1}_{44\,\text{times}}+\frac{1}{2}+1}{\overbrace{1+1+\dots+1}_{44\,\text{times}}+\frac{1}{2}+0}\\
&=\frac{44+\frac{1}{2}+1}{44+\frac{1}{2}}\\
&=\frac{88+1+2}{88+1}=\frac{91}{89}
\end{align}$$
A: $$\frac{( sin²(1) + sin²(2) + sin²(3) + .. + sin²(90) )}{ ( cos²(1) + cos²(2) +cos²(3) + .. + cos²(90) )} = \\ \frac{2}{2}*\frac{( sin²(1) + sin²(2) + sin²(3) + .. + sin²(90) )}{ ( cos²(1) + cos²(2) +cos²(3) + .. + cos²(90°) )}$$ so try by $sin^2x=\frac{1-cos2x}{2} \\cos^2x=\frac{1-cos2x}{2}$
so 
$$\frac{1-cos2 +1-cos 4+1-cos 6+ ...+1-cos 180}{1+cos2 +1+cos 4+1+cos 6+ ...+1+cos 180}=\\ \frac{90-(cos 2+cos 4+cos 6+ ...+cos 180)}{90+(cos 2+cos 4+cos 6+ ...+cos 180)}$$ 
now see that $$(cos 2+cos 4+cos 6+ ...+cos 180)=\\(cos 2+cos 178 )+(cos 4 +cos 176) +...(cos 44 +cos 46) +cos 90 +cos 180 =0 +0 +0+0+...+cos 90 +cos 180 =0+0+(-1)$$ so  $$\frac{90-(cos 2+cos 4+cos 6+ ...+cos 180)}{90+(cos 2+cos 4+cos 6+ ...+cos 180)}=\frac{90-(-1)}{90+(-1)}=\frac{91}{89}$$
A: Hint: using
$$\sum_{k=1}^n\cos(kx)=\frac{\cos(\frac{n+1}{2}x)\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}$$
and $$ \sin^2x=\frac{1-\cos(2x)}{2}, \cos^2x=\frac{1+\cos(2x)}{2},$$
you can get the answer.
A: HINT:
$$\frac{\sum_{n=a}^{b} \sin^2(n)}{\sum_{n=a}^{b} \cos^2(n)}=$$
$$\frac{2-2a+2b-\csc(1)\sin(1-2a)-\csc(1)\sin(1+2b)}{2-2a+2b+\csc(1)\sin(1-2a)+\csc(1)\sin(1+2b)}$$

Now we know that $a=1$ and $b=90$:
$$\frac{\sum_{n=1}^{90} \sin^2(n)}{\sum_{n=1}^{90} \cos^2(n)}=$$
$$\frac{2-2\cdot 1+2\cdot 90-\csc(1)\sin(1-2\cdot 1)-\csc(1)\sin(1+2\cdot 90)}{2-2\cdot 1+2\cdot 90+\csc(1)\sin(1-2\cdot 1)+\csc(1)\sin(1+2\cdot 90)}=$$
$$\frac{180-\csc(1)\sin(-1)-\csc(1)\sin(181)}{180+\csc(1)\sin(-1)+\csc(1)\sin(181)}=$$
$$\frac{182}{178}=\frac{91}{89}$$
