How do you find the period for #y = -tan(x - pi/2)#?

1 Answer
Apr 19, 2015

The period is always #pi# in this case:

If you add a constant inside the brackets or multiply by a constant outside the period of a periodic function doesn't change, this is trivial:

if #f# has period #T#:
1)#f(x+T)=f(x) => f(x+T+a)=f((x+a)+T)=f(x+a)# so the period of #f(x+a)# is still #T# (you know it's minimal because it's minimal in #f#)
2) #f(x+T)=f(x) => af(x+T)=af(x)#, and still it's minimal because it's minimal in #f#

so the period is #pi#, the same as in #tan(x)#