# I'm not comfortable with solving `tan^(-1)[tan((4pi)/5)] ` . How do I solve this if I cannot find the value on the unit circle?

ishpiro | College Teacher | (Level 1) Educator

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In order to find the inverse tangent of a tangent, you need to use the definition of inverse function:

If a function f is a rule that makes a value of x into a value of y, then the inverse function `f^(-1) ` makes the value of y back into the value of x.

For example, if y = f(x) = sin(x) (sine function) is a rule that takes an angle and creates its sine, then inverse sine function would be

` f^(-1)(x) = sin^(-1)(x) ` , the function that takes a number and creates an angle, in radians, that has a sine equal x.

In other words, the inverse function "undoes" what the function does. So the inverse function of a function, or vice versa, function of an inverse function, would result in x:

`f^(-1)(f(x)) = f(f^(-1)(x)) = x `

So, generally, inverse tangent of the tangent would be the angle itself. However, the example of ` tan^(-1)[tan((4pi)/5)] ` is a bit trickier. Note that angles `(4pi)/5 ` and ` -pi/5` have the same tangent, because the difference between these two angles is `pi `, and tangent is periodic with the period of `pi ` . So, when you take the inverse tangent of this tangent, would it result in `(4pi)/5 ` or `-pi/5 ` ?

Since tangent is a periodic function, it is not invertible. It is, however, invertible on the interval `[-pi/2, pi/2] ` , where for every value of x there is only one value of the tangent, which makes the function invertible (see the graph.)

For the inverse tangent, the restricted domain of the tangent, `[-pi/2, pi/2] ` , becomes the range. So any output of the inverse tangent must lie within that range. The angle `(4pi)/5 ` is greater than ` pi/2 ` , which makes it outside of the range, but the angle `-pi/5 ` is inside `[-pi/2, pi/2] ` .

Therefore,

` tan^(-1)[tan((4pi)/5)] = -pi/5 ` .

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to find out that `tan^(-1)(tan a) = a` , hence, replacing` (4pi)/5` for a, yields:

`tan^(-1)(tan ((4pi)/5))= (4pi)/5`

You may evaluate (`4pi)/5` , such that:

`(4pi)/5 = (5pi)/5 - pi/5`

Reducing by 5 the fraction (5pi)/5, yields

`(4pi)/5 = pi - pi/5`

Since `pi/5 < pi/4` , hence, the angle `pi - pi/5` is located in the second quadrant, between pi/2 and pi.

You should also know that converting radians in degrees, yields:

`pi = 180^o => pi - pi/5 = 180^o - (180^o)/5`

`pi - pi/5 = 180^o - 36^o`

`(4pi)/5 = pi - pi/5 = 144^o`

Hence, evaluating `tan^(-1)(tan ((4pi)/5))` yields `tan^(-1)(tan ((4pi)/5)) = tan^(-1)(tan 144^o) = 144^o.`