Using Taylor even when Maclaurin converges everywhere
Just because the series representation converges everywhere doesn't mean it converges rapidly everywhere. The graph below illustrates this. Note that if the series is shifted by pi/2 then we can use the memorized Maclaurin series for the cosine function but with shifted x values and write out the representation almost as quickly as if we used the also-memorized Maclaurin series for the sine.