simple but yet confusing calculus even odd function problem?

Question

i am having trouble with this...

1 - cos x and x + cosx

ok for the first one i know you substitute the f(-x) but what is
cos(-x) and sin(-x) [even though its not on there i want to know how to figure it out. are these functions even or odd. can someone try to show me a step by step procedure please. i just dont understand the cos(-x) and sin(-x). how do we know whats x?

Answer 1

Recall some trigonometric identities:
1) sin(x) = cos(pi/2-x)
2) sin(A+B) =sin A cos B + cos A sin B
3) cos(x) = sin(pi/2-x)
4) cos(A + B) = cosA cosB - sinA sinB

So using 1)
sin(-x)=cos(pi/2-(-x)) = cos(pi/2+x)
now using 4) on cos(pi/2+x)
sin(-x)=
cos(pi/2)cos(x) -sin(pi/2)sin(x)
=-sin(x)

Remember this identity I just proved.. 5) sin(-x) = -sin(x)

Using 3)
cos(-x)=sin(pi/2-(-x)) = sin(pi/2+x)
Using 2) on sin(pi/2+x)
cos(-x)=
sin(pi/2)cos(x) +cos(pi/2)sin(x)
= cos(x)

Remember this identity I just proved ... 6) cos(-x) = cos(x)

Answer 2

Since (1 - cos(x)) = (1 - cos(-x)), that function is even. But (x + cos(x)) is not equal to (-x + cos (-x)), so that function is not even. Since (x + cos(x)) is not equal to -(-x + cos(-x)), it is not odd either.

Answer 3

thinking (a million - cos(x)) = (a million - cos(-x)), that function is even. yet (x + cos(x)) isn't equivalent to (-x + cos (-x)), so as that function isn't even. thinking (x + cos(x)) isn't equivalent to -(-x + cos(-x)), it really is not any more spectacular both.