Question

Consider the cosine function \(\cos : \mathbb{R} \rightarrow \mathbb{R}\). Decide whether this function is injective and whether it is surjective. What if it had been defined as \(\cos : \mathbb{R} \rightarrow[-1,1] ?\)

Short answer

The cosine function \(\cos x : \mathbb{R} \rightarrow \mathbb{R}\) is neither injective nor surjective, because it repeats its values periodicaly and doesn't cover the whole set of real numbers. However, if it is redefined as \(\cos x : \mathbb{R} \rightarrow[-1,1]\), it is surjective, as all values in the interval [-1,1] are covered by the cosine function, but it still remains non-injective due to its periodic nature.

Step-by-step solution

Checking the Injectivity

\(\cos x\) is periodic, meaning it repeats its values as \(x\) advances by any integer multiple of \(2\pi\). Hence, multiple values of \(x\) lead to the same value of \(\cos x\), which implies that the cosine function is not injective in its given domain, \(\mathbb{R}\).

Checking the Surjectivity

The range of \(\cos x\) lies within [-1,1] only, regardless of the input \(x\) from all real numbers. Therefore, if we consider the codomain as \(\mathbb{R}\), not all elements in the codomain are mapped by the function, meaning it is not surjective. However, if we adjust the codomain to [-1,1], each of these elements is reached by \(\cos x\) at least once. Therefore, with this adjusted codomain, the function is surjective.