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

What’s the answer for the solution to odd functions?-example-1

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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
User Smokku
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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.
User Mlschechter
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