What’s the answer for the solution to odd functions?

The sine function is always odd.  That means you can elimiinate both cosine functions here.  Another requirement for an odd function is that the graph goes thru the origin (0,0).  This is true of Graph B.

We know that sine functions of the form
f(x)=a*sin(kx) are odd functions because
f(-x)=a*sin(-kx)=-a*sin(kx)=-f(x)
We also know that cosine functions of the form
g(x)=b*cos(kx) are even functions because
g(-x)=b*cos(-kx)=b*cos(kx)=g(x)

Out of the four choices, 
(a)F(x)=2sin(x/2+pi/2)=2cos(x/2)  => even function
(b)F(x)=2sin(x/2)=a*sin(kx)  where a=2, k=1/2 => odd function
(c)F(x)=2cos(2x) => even function
(d)F(x)=cos(x+pi)=-cos(x) = b*cos(kx) where b=-1, k=1, => even function