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sin2u=2sinucosu for all u, because this is a trig identity. If 2sinucosu=-2sinucosu, then 2sin2u=0, making 2u=n(pi), where n is an odd integer, and u=(pi)n/2. sin((pi)n/2)=1 or -1, both odd numbers, the only odd numbers that sine can be equal to. So it is true that if sinu is odd, the only odd value defines u as (pi)n/2, and 2u=(pi)n, the sine of which is zero for all n, particularly odd n. Since +0=-0=0, 2sinucosu is the same as -2sinucosu, when u=(pi)n/2 for n odd.


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