What is your Zodiac sine?

there are a total of 45 terms starting from $sin^{2}2$ to $sin^{2}90$ where the angles of the sines are in an increasing A.P. starting from $2$ with common difference of $2$ . notice that $sin88$ can be written as $cos(90-88)$ or $cos2$ similarly, $sin86, sin84, sin82$ .........upto $sin46$ can be written as $cos4,cos6, cos8$ ........upto $cos44$ (respectively) the given series:

$(sin^{2}2 + sin^{2}4 +..........sin^{2}88+ sin^{2}90)$ can be written as:

$(sin^{2}2+cos^{2}2)+(sin^{2}4+cos^{2}4)+............+(sin^{2}44+cos^{2}44)+sin^{2}90$ notice that, there are $22$ terms in the first brackets each of which can be written as $(sin^{2}x + cos^{2}x)$ ...which is, equal to $1$

$\therefore$ , the SUM: $(1+1+1+........(22times).......+1) + sin^{2}90$ which gives $22+1$ ....since $sin90=1$

$=23$

$\textbf{QED}$

First, notice that there are 45 terms

Let us try to evaluate the sum of the first 44 terms now:

Look, we have pairs like (sin 2°)(sin 2°) and (sin 88°)(sin 88°). sin 88° can be expressed as cos 2°

Thus, the pair becomes: (sin 2°)(sin 2°) + (cos 2°)(cos 2°) which is equal to 1 Now, for each of this 44 numbers, there are such complimentary pairs, like sin 6° and sin 84°.

So, since there are 44 terms, we can form 22 pairs and the sum of them each is 1. so, (sin2°)(sin2°)+(sin4°)(sin4°)+(sin6°)(sin6°)+(sin8°)(sin8°)+ ... + (sin 88°) (sin 88°) = 22

Now, lets add (sin 90°)(sin 90°) with it. Since sin 90° is 1, the total becomes 23

$\begin{aligned} S & = \sin^2 (2^\circ) + \sin^2 (4^\circ) + \sin^2 (6^\circ) + ... \sin^2 (88^\circ) + \sin^2 (90^\circ) \\ & = \sum_{n=1}^{44} \sin^2 (2n^\circ) + \sin^2 (90^\circ) \\ & = \sum_{n=1}^{22} \left(\sin^2 (2n^\circ) + \sin^2 (90^\circ - 2n^\circ) \right) + 1 \\ & = \sum_{n=1}^{22} \left(\sin^2 (2n^\circ) + \cos^2 (2n^\circ) \right) + 1 \\ & = \sum_{n=1}^{22} 1 + 1 \\ & = \boxed{23} \end{aligned}$