How do you show that tan (θ) has period π in terms of a unit circle?

1 Answer
Steve M
May 9, 2018

We seek to prove that #tan theta# has period #pi#.

Suppose that we denote the period of #tan theta# by #T# where #T in RR^+# then we have:

# tan(theta) = tan (theta+T) #

Using the tangent double angle formula, we can write:

# tan(theta) = (tantheta+tanT)/(1-tan theta tan T) #

Providing that #1-tan theta tan T != 0# then

# (1-tan theta tan T)tan theta = tan theta+tanT #

# :. tan theta-tan^2 theta tan T = tan theta+tanT #

# :. -tan^2 theta tan T = tanT #

# :. (tanT)(1+tan^2 theta) = 0#

We require that this equation holds #AA theta in RR# and so we must have:

# tanT = 0#

# :. T = arctan 0#

# :. T = 0, pi, 2pi, ...#

We want #T in RR^+#, and also the smallest possible value (otherwise we have a multiple of the periodicity)

Therefore, #T=pi \ \ \ #, QED

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