Deduce non-type template parameter return value

Hi,

I've written a function template for applying a free function to every
element of a std::vector:

#include <vector>

template<class T, T F(T)>
std::vector<T> map(const std::vector<T>& from)
{
    std::vector<T> to;
    to.reserve(from.size());
    for (typename std::vector<T>::const_iterator iter = from.begin();
iter != from.end(); ++iter) {
        to.push_back(F(*iter));
    }
    return to;
}

This can be used as follows:

#include <cmath>

int main()
{
    std::vector<float> angles;
    angles.push_back(1);
    angles.push_back(2);
    angles.push_back(3);
    std::vector<float> sines = map<float, std::sin>(angles);
}

As you can see, std::sin is passed as a non-type template parameter.
However, I'd like to get rid of the the first parameter (float in the
example) so the call would become map<std::sin>(angles). Is there a
way to do this in C++03 or C++11?

Please note that I don't want to pass function pointers at run-time;
the mapping function (here, std::sin) is known at compile-time so I
want to take advantage of that.

Thanks,
Chris