Question

evaluate the series:
sin^2 0 + sin^2 1 + sin^2 2+ sin^2 3 .... sin^2 90

Answer 1

Disclaimer: This gives the wrong answer (off by 0.5), but the approach seems correct. I must have made some minor error in the computation. I'll look this over and let you know if I find it later.
\[\begin{align*}S&=\sum_{k=0}^{90}\sin^2k^\circ\\\\
&=\frac{1}{2}\sum_{k=0}^{90}(1-\cos2k^\circ)\\\\
&=\frac{90}{2}-\frac{1}{2}\sum_{k=0}^{90}\cos2k^\circ\\\\
&=45-\frac{1}{2}\sum_{k=0}^{90}\cos\frac{\pi k}{90}\\\\
&=45-\frac{1}{2}\text{Re}\left(\sum_{k=0}^{90}e^{(\pi/90)ik}\right)\\\\
&=45-\frac{1}{2}\text{Re}\left(\sum_{k=0}^{90}\left(e^{i\pi/90}\right)^k\right)\\\\
&=45-\frac{1}{2}\text{Re}\left(\frac{\left(e^{i\pi/90}\right)^{90+1}-1}{e^{i\pi/90}-1}\right)\\\\
&=45-\frac{1}{2}\text{Re}\left(\frac{\cos\dfrac{91\pi}{90}-1+i\sin\dfrac{91\pi}{90}}{\cos\dfrac{\pi}{90}-1+i\sin\dfrac{\pi}{90}}\right)\\\\
&=45-\frac{1}{2}\text{Re}\left(\frac{\cos\dfrac{91\pi}{90}-1+i\sin\dfrac{91\pi}{90}}{\cos\dfrac{\pi}{90}-1+i\sin\dfrac{\pi}{90}}\times\frac{\cos\dfrac{\pi}{90}-1-i\sin\dfrac{\pi}{90}}{\cos\dfrac{\pi}{90}-1-i\sin\dfrac{\pi}{90}}\right)\\\\
&=45-\frac{1}{2}\text{Re}\left(\frac{\left(\left(\cos\dfrac{91\pi}{90}-1\right)+i\sin\dfrac{91\pi}{90}\right)\left(\left(\cos\dfrac{\pi}{90}-1\right)-i\sin\dfrac{\pi}{90}\right)}{\cos^2\dfrac{\pi}{90}-2\cos\dfrac{\pi}{90}+1+\sin^2\dfrac{\pi}{90}}\right)\end{align*}\]
Simplifying the numerator:
\[\left(\cos\frac{91\pi}{90}-1\right)\left(\cos\frac{\pi}{90}-1\right)+i\left(\cos\frac{\pi}{90}-1\right)\sin\frac{91\pi}{90}\\
\quad\quad+i\left(\cos\frac{91\pi}{90}-1\right)\sin\frac{\pi}{90}+\sin\frac{91\pi}{90}\sin\frac{\pi}{90}\]
We're only interested in the real part, so we have
\[\begin{align*}
S&=45-\frac{1}{2}\left(\frac{\left(\cos\dfrac{91\pi}{90}-1\right)\left(\cos\dfrac{\pi}{90}-1\right)+\sin\dfrac{91\pi}{90}\sin\dfrac{\pi}{90}}{\cos^2\dfrac{\pi}{90}-2\cos\dfrac{\pi}{90}+1+\sin^2\dfrac{\pi}{90}}\right)\\\\
&=45-\frac{1}{4}\left(\frac{\cos\dfrac{91\pi}{90}\cos\dfrac{\pi}{90}+\sin\dfrac{91\pi}{90}\sin\dfrac{\pi}{90}-\cos\dfrac{91\pi}{90}-\cos\dfrac{\pi}{90}+1}{1-\cos\dfrac{\pi}{90}}\right)\\\\
&=45-\frac{1}{4}\left(\frac{\cos\dfrac{90\pi}{90}-\cos\dfrac{91\pi}{90}-\cos\dfrac{\pi}{90}+1}{1-\cos\dfrac{\pi}{90}}\right)\\\\
&=45+\frac{1}{4}\left(\frac{\cos\dfrac{91\pi}{90}+\cos\dfrac{\pi}{90}}{1-\cos\dfrac{\pi}{90}}\right)\\\\
&=45
\end{align*}\]
since \(\cos\dfrac{91\pi}{90}=-\cos\dfrac{\pi}{90}\).

Answer 2

Trying another approach... Notice that
\[S=\color{red}{\sin^20^\circ}+\color{blue}{\sin^21^\circ}+\cdots+\sin^245^\circ+\cdots+\color{blue}{\sin^289^\circ}+\color{red}{\sin^290^\circ}\]
We know that \(\sin^2x=1-\cos^2x\), and additionally that \(\cos (x-180^\circ)=\sin x\).

Consider the red terms:
\[\begin{align*}\sin^20^\circ+\sin^290^\circ&=\sin^20^\circ+1-\cos^290^\circ\\
&=\sin^20^\circ+1-\cos^2(180-90)^\circ\\
&=\sin^20^\circ+1-\sin^290^\circ\\
&=\frac{1}{2}\end{align*}\]
You have this sort of pattern occurring 45 times for each paired term, plus the middle \(\sin^245^\circ=\dfrac{1}{2}\). This means the sum must be \(45.5\).