Both Sin[x]==0 && Cos[x]==0 as a solution

Question

If I do

FullSimplify[Reduce[Sin[p1] == 0 && Cos[p1] == 0, Reals]]

I get

 Cos[p1] == 0 && Sin[p1] == 0

while I would expect False. why? is there a way to have mathematica compute that both sin and cos cannot be 0?

Answer 1

As noted, specifying the variable of interest within Reduce[] yields the expected result:

Reduce[Sin[p1] == 0 && Cos[p1] == 0, p1]
   False

Answer 2

Because sin(0)=0 and cos(0)=1 (and cos(π/2)=0 while sin(π/2)=1), it is clear that the result is false, as expected.