# What is the period of y = cot 2 د€x?

I know period = 2د€/B2د€ / B.

The equation is y=cot(2د€x)y = \cot (2 د€x).

I think period is one of these three things:

1)2د€/2د€=11)\;2د€ / 2د€ = 1

2)2د€/2=د€2) \;2د€ / 2 = د€

3)2د€/د€=23) \;2د€ / د€ = 2

None of these answers are correct. It has to be either 1/2,1/4,د€/41/2, 1/4, د€/4, or د€/2د€/2. What am I doing wrong?

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Try π/2π\pi/2\pi.
Oct 21 at 1:58

@AdamHughes Thank you! Why is 2د€ in the denominator and not the numerator to solve the equation? I thought period = 2د€/B?
– user7045242
Oct 21 at 2:00

Do you mind saying what BB is, or is that something I’m supposed to know?
– bof
Oct 21 at 2:00

The default period for cotx\cot x is π\pi, if you augment it by aa, i.e. do cot(ax)\cot(ax) the new period is always π/a\pi/a. In this case you just happen to have a=2πa=2\pi. Your use of 2π/a2\pi/a is from sinax\sin ax or cosax\cos ax where the default period is 2π2\pi, but it’s always the default over the multiplier.