Posted by Helen on October 22, 2002 at 20:47:06:
In Reply to: sorry, lots of missing tags. try this one posted by T.Gracken on October 22, 2002 at 10:13:56:
Answer: 2xPi^2.
Method:
Sin(x+y)Cos(y) >= Sin(x)
Sin(x)Cos(y)Cos(y) + Cos(x)Sin(y)Cos(y) - Sin(x) >= 0
Sin(x)(Cos(y)Cos(y) - 1) + Cos(x)Sin(y)Cos(y) >= 0
-Sin(x)Sin(y)Sin(y) + Cos(x)Sin(y)Cos(y) >= 0
Sin(y)(Cos(x)Cos(y) - Sin(x)Sin(y)) >= 0
Sin(y)Cos(x+y) >= 0
Sin(y) and Cos(x+y) have the same sign.
'Cos' is 2Pi periodic and is positive half the time and negative half the time
(ignoring zeros, which have no effect). So, for each value of y, exactly half
of the x-values between 0 and 2Pi give Cos(x+y) to have the same sign as
Sin(y).
Hence, exactly half of the (x,y) values in the region 0<=x<=2Pi, 0<=y<=2Pi
(whose area is 4xPi^2) satisfy the condition, giving the area of the region R
as 2xPi^2.