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Mathematics 14 Online
OpenStudy (anonymous):

A simple functions one, : A real valued function,f(x), satisfies the equation: f(x - y) = f(x)f(y) - f(a-x)f(a+y) where a is a consant and f(0) = 1, Then f(2a-x) is?

OpenStudy (anonymous):

You have to find it in terms of f(x).

OpenStudy (experimentx):

is it -f(x)??

OpenStudy (anonymous):

Yes.:) Method?

OpenStudy (experimentx):

put x=y then put x=0, that will yield f(a) = 0 if we put, 2a-x as a - (x-a), we will get -f(x)

OpenStudy (experimentx):

Ah ,,, putting x=y=0 would be shortcut

OpenStudy (anonymous):

After x=y=0, f(a)=0 Put.x = a, and y = a -x works as well. However, the quickest way to do this is by comparing it to arc(cos(x-y))

OpenStudy (experimentx):

arc(cos(x-y)) or arccos(x-y) ??

OpenStudy (anonymous):

cos(inverse) (x-y)

OpenStudy (anonymous):

Not used to the arc notation.

OpenStudy (experimentx):

and a would be pi/2??

OpenStudy (experimentx):

i see .. nice technique

OpenStudy (anonymous):

Yeahhhhh. :)

OpenStudy (experimentx):

-f(a-x)f(a+x) had me confused ... though i would have never guessed

OpenStudy (anonymous):

Still. Nice way yours too! :)

OpenStudy (experimentx):

:)

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