Bijectivity of sinx on the interval [−π,π] to [−1,1]how can I prove that sinx is a bijective map from the domain [−π,π] to the co-domain [−1,1]. I had no problem proving that using the graphical representation of sinx, but rigorously could not.

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bicicletagyp · Accepted Answer

sin⁡ is not a bijective map from [−π,π] to [−1,1]. If it was, then ∀a,b∈[−π,π], sin⁡a=sin⁡b⇒a=b. However notice that sin⁡(−π)=sin⁡(π)=0 and −π≠π. Therefore we have a contradiction.Restrict yourself to [−π2,π2] and then you have bijectivity.

fodlonwyrgxc · Accepted Answer

Let f(x)=x2 from [−π,π] to [−π2,π2] is a bijection. Let g(x)=sin⁡x from [−π2,π2] to [−1,1]  is a bijection. Taking composition of these two function.The required function is h(x)=sin⁡(x2)

Bijectivity of sinx on the interval \(\displaystyle{\left[-\pi,\pi\right]}\)

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