f is a real valued function and not identically zero satisfying f(x+y)+f(x-y) = 2f(x)f(y) for all x,y belonging to R then it is what type of function even or odd ? How?

Question

Deepak Kumar Shringi · Answer

as we know for even function
f(x)=f(-x) and for odd function f(x)=-f(-x)
ifPut y=0

=> f(x)+f(x)=2f(x).f(0)

=> f(0)=1

Now put x=0 and y=x

=> f(x)+f(-x)=2f(0)f(x)

=> f(x)+f(-x)=2f(x)

=> f(-x)=f(x)

So, even function.

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f is a real valued function and not identically zero satisfying f(x+y)+f(x-y) = 2f(x)f(y) for all x,y belonging to R then it is what type of function even or odd ? How?

f is a real valued function and not identically zero satisfying f(x+y)+f(x-y) = 2f(x)f(y) for all x,y belonging to R then it is what type of function even or odd ? How?

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f is a real valued function and not identically zero satisfying f(x+y)+f(x-y) = 2f(x)f(y) for all x,y belonging to R then it is what type of function even or odd ? How?

f is a real valued function and not identically zero satisfying f(x+y)+f(x-y) = 2f(x)f(y) for all x,y belonging to R then it is what type of function even or odd ? How?

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