Compute $(\sin4^\circ)^2 +(\sin8^\circ)^2+(\sin12^\circ)^2+\cdots+(\sin176^\circ)^2$ Angle of sine is in degrees, can anyone show me an easy soln to this? This was question was given to us for 1minute without calcu. 
I know that $\sin4^\circ=\sin176^\circ$, $\sin8^\circ=\sin172^\circ$.....
But dont know how to proceed
 A: We may add $\sin^2(0)=0$, too. 
In that way, one is summing $180/4=45$ terms of the form $\sin^2(x)$ where $x$ is uniformly and rather densely distributed over the angles (because the periodicity is $180$ degrees), so one basically calculates $45$ times the average value of $\sin^2(x)$. The average value of $\sin^2(x)$ is $1/2$ so the result is approximately $45/2=22.5$.
One may explicitly verify that this result $22.5$ is exact by using 
$$\sin^2(x) = \frac{1-\cos(2x)}{2}$$
Now, sum the right hand side for $x$ going from 0 degrees to 176 degrees, with the step 4 degrees. The first term gives us $45/2=22.5$.
The second term, the sum of $-\cos(2x)/2$ for $x$ from 0 degrees to 176 degrees with the step 4 degrees, is zero. It's because the argument $2x$ of the cosine goes from 0 to 352 degrees, with the step of 8 degrees. 
The cosine $\cos A$ is the real part of $\exp(iA)$. But the sum of $\exp(iA)$ over $A$ from 0 to 352 degrees with the step of 8 degrees vanishes because these complex numbers $\exp(iA)$ are uniformly distributed along the unit circle. By the ${\mathbb Z}_{45}$ symmetry of the polygon with 45 vertices, the sum is zero. Schematically,
$$\sum \cos(2x) = {\rm Re}\sum \exp(2ix),\\
\sum\exp(2ix) = \exp(i\cdot 8^\circ) \sum\exp(2ix)=0 $$
The sum of the phases has to vanish because it is equal to itself times a number different from one (the phase that just cyclically permutes the 45 terms).
So the result is $22.5$.
A: A simple trigo way as you noted continuing from there we can write it as $$2(\sin^2(4)\ldots\sin^2(88))$$ then we can use $$2\sin(\theta)\sin(\theta)=\cos(0)-\cos(2\theta)$$ thus we get $$22-(\cos(8)+\ldots +\cos(176))$$ now as terms are in ap ie angles we use A.P. formula for $\cos$ i.e. $$\frac{\sin(a+(n-1)b/2)}{\sin(b/2)}\cdot \cos((c+d)/2)$$ where a is first term,b is common difference and c,d are first,last angles respectively which yields result as $-0.500000$ thus sum is $22-(-0.500000) = 22.50000$
