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Show that the function f: R->R given by ...

Show that the function `f: R->R` given by `f(x)=cosx` for all `x in R` , is neither one-one nor onto.

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Given function `f(x)=cosx`
Now, `f(π/2​)=cosπ/2​=0`
`⇒ f(−π/2​)=cosπ/2​=0`
`⇒ f(π/2​)=f(−π/2​)`
But `π/2​!=−π/2`​
So, `f(x)` is not one-one
Now, f(x)=cosx is not onto as
there is no preimage for an real number.
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