Why is this trigonometry function neither odd nor even?

[adelin]

f(x)=1+sinx
what am I doing wrong here?
1+sin(-x)= 1-sin(x)

 

[PhysicsandSuch]

Nothing wrong--just remember the definition of an odd or even function that you just used on the sin(x) to calculate your parity transform, and think about whether the function f(x) falls into either of those catagories.

 

[lendav_rott]

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.

 

[UltrafastPED]

Every function can be separated into an even part and an odd part ... f(x) = f_Even(x) + f_Odd(x).

See http://en.wikipedia.org/wiki/Even_and_odd_functions#Some_facts

 

[willem2]

..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.

The sum of an odd function and an even function is never odd or even, unless one of the funtions is the zero funtion.

let e(x) be an even function and o(x) be an odd function.

if f(x) = e(x) + o(x) is an even function, than we have

f(-x) = e(-x) + o(-x) = e(x) - o(x) = f(x) = e(x) + o(x). the e(x) terms cancel and you get 2o(x) = 0.

if it is an odd function than

f(-x) = e(-x) + o(-x) = e(x) - o(x) = -f(x) = -e(x) - o(x). the o(x) terms cancel and you get 2e(x) = 0