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Precalculus Mathematics Homework Help
Why is this trigonometry function neither odd nor even?
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[QUOTE="lendav_rott, post: 4504377, member: 224893"] What are the conditions of an odd or even function if it were even then f(x) = f(-x) or f(x) - f(-x) = 0 1+sinx =/= 1 + sin(-x) if it were odd, if you rotate it 180 deg about the origin (the 0,0 point) then the graph remains the same so -f(x) = f(-x) or my personal preference f(x) + f(-x) = 0 since it's easier to remember. 1+sinx + 1 + sin(-x) = 2 =/= 0 I don't know any mathematical explanation that explains why a function is neither odd nor even , but I have noticed that when you add a constant to any given function that does not have one (which usually makes them odd or even), it seems to do the trick. You must add a constant such that it moves the graph along the ordinate or y-axes in this case, which will make symmetry impossible, but.. ..then again if we look at f(x) = 1 + cosx we would see it is even, for 1 + cosx = 1 + cos(-x) From this I would assume adding a constant to an even function yields no change, but adding a constant to an odd function makes it neither odd nor even. I really don't know, perhaps someone can elaborate. [/QUOTE]
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Why is this trigonometry function neither odd nor even?
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