how do i prove that f(x)=sin(x)*cos(x) is one-to-one in -pi/3

Question

how do i prove that f(x)=sin(x)*cos(x) is one-to-one in -pi/3<x<pi/3 ?

anonymous · Answer

sorry can you give me the definition of one-to-one?

anonymous · Answer

hi. i mean that the following implication holds:$$if f(x _{1})= f(x _{2}), then x _{1}=x _{2}$$or its logical equivalent:$$if x _{1}\neq x _{2}, then f(x _{1})\neq f(x _{2})$$

anonymous · Answer

the one-to-ont aka(Bijection http://en.wikipedia.org/wiki/Bijection) can be proved given the function is cotinuous in that interval, plus the slope is always either positive or negative in that particular interval

anonymous · Answer

you could also use the trigonometric identity $$\sin (x)\times \cos(x)=1\div2 \times \sin(2\times x)$$
and then compare $$\sin(2\times x1)=\sin(2\times x2)$$ using the same ideas

anonymous · Answer

ok, thank you very much; i'll try both methods.
by "one-to-one" i meant what wikipedia (and myself, in spanish (my first language)) calls an "injection"; a bijection is more than that, for it also requires the codomain to be the same as the image of f ("surjection"). i didn't know that the terms were ambiguous.