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    1. hellow
      • 3 years ago
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      k even, i.e. k=2n for some integer n, then :\[\sin(\pi/2+k \pi)= \sin(\pi/2 + 2n \pi )= 1\]k odd, i.e. k=2n-1 for some integer n:\[\sin(\pi/2+k \pi)= \sin(\pi/2 + (2n-1) \pi )= -1\] (Remember this from trig? You can see this on a graph of y=sin(theta)) We find the y-coordinates of the points by substituting the x-coordinates into sin(x)+sin(y)=1/2, and using the values calculated above. It turns out only some of the x-values will give real answers for y, so only these x-coordinates are kept. We are then left with the x-coordinates and y-coordinates that work - the points we wanted.

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    spraguer (Moderator)
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