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
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HINT: Use $$\cos2A=1-2\sin^2A$$ and http://math.stackexchange.com/questions/117114/sum-cos-when-angles-are-in-arithmetic-progression – lab bhattacharjee Feb 04 '16 at 06:48
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Is the answer $-22.5$ – Archis Welankar Feb 04 '16 at 06:59
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Am i right copper hat – Archis Welankar Feb 04 '16 at 07:02
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1@ArchisWelankar: It can't be negative, it is the sum of non negative numbers (squares). – copper.hat Feb 04 '16 at 07:04
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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 $22.5$.
Luboš Motl
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Dear @ArchisWelankar, all the terms in the sum are clearly positive numbers (square of some other positive numbers) so their sum has to be positive, too. – Luboš Motl Feb 04 '16 at 07:04
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Just a non-essential addition. People normally use arguments in radians, so whenever I write $\cos A^\circ$ or $\exp(iA^\circ)$, I mean $\cos(\pi A/180)$ or $\exp(i\pi A/180)$. – Luboš Motl Feb 04 '16 at 07:30
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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 $22-(-0.500000) = 22.50000$
Archis Welankar
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LOL, what about the 0.000002 error? I assure you that the result 22.5 is exact and no such error arises. – Luboš Motl Feb 04 '16 at 07:31
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I don't know whether you understood what I wrote. The result is 22.50000000000000000000000, not 22.500002. – Luboš Motl Feb 04 '16 at 07:34
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3It depends for whom. For a mathematician, the difference between these two numbers usually does matter. – Luboš Motl Feb 04 '16 at 07:40
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2The task was to compute the sum, not to approximate it, and you haven't solved the task correctly. Alternatively, you may say that the difference between +1 and -1 doesn't matter, either. – Luboš Motl Feb 04 '16 at 07:42
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2@Luboš Motl: I cherished your blog, your answers at PSE; but now those comments..... awesome +1+1 . – Feb 04 '16 at 07:52