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A  real valued function f\left ( x \right ) satisfies the functional equation  f\left ( x -y\right )=f(x)f(y)-f(a-x)f(a+y) where a  is a given constant and f(0)=1, \ f(2a-x)  is equal to

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f(x-y)=f(x) f(y)-f(a-x) f(a+y) \ldots($ i)
It is given that f(0)=1$ 
Substituting x=y=0 \ in \ (i) 
1=1-(f(a))^{2} 
Hence f(a)=0 
Now f(2 a-x)$ 
=f(a-(x-a)) \\ =f(a) f(x-a)-f(a-a) f(a+x-a)\\ =0-f(0) f(x)\\ =-f(x)\\

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