# Re: O.K. I'll rephrase...

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Posted by Joel on October 22, 2002 at 22:03:05:

In Reply to: O.K. I'll rephrase... posted by T.Gracken on October 22, 2002 at 21:13:09:

: : : : : : R={(x,y)| 0<=x<=(2Pi), 0<=y<=(2Pi), [Sin(x+y)Cosy]>=(Sin x) }

: : : : : sin(x+y)cos(y) >= sin(x)
: : : : : (sin(x)cos(y)+cos(x)sin(y))*cos(y) >= sin(x)
: : : : : sin(x)cos^2(y) + cos(x)cos(y)sin(y) >= sin(x)
: : : : : cos(x)cos(y)sin(y) >= sin(x) - sin(x)cos^2(y)
: : : : : (1/2)cos(x)cos(2y) >= sin(x)(1-cos^2(y)) ignore this one
: : : : : (1/2)cos(x)cos(2y) >= sin(x) - sin(x)cos^2(y) ignore this one (here we are again :)

: : : I don't see how you got the next line [below]. Shouldn't it be (1/2)cos(x)sin(2y) >=...?
: : no change on left side; on right side substituting: cos^2(y)=(1+cos(2y))/2

: I don't have a problem with the right side. How did you get the factor "cos(2y)" on the left???

Ah, yes...

So now it's:

:
: : : : : (1/2)cos(x)sin(2y) >= sin(x) - sin(x)(1+cos(2y)/2
: : : : : (1/2)cos(x)sin(2y) >= sin(x) - (1/2)sin(x) - (1/2)sin(x)cos(2y)
: : : : : cos(x)sin(2y) + sin(x)cos(2y) >= sin(x)
: : : : : sin(x+2y) >= sin(x)

which doesn't appear to be very helpful

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