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# Trig functions

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Find all numbers in the range of f(x) = arctan x + arctan (1 - x).

May 14, 2022

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Ok, so the function is not in the simplest form yet, let's simplify it.

Note that $$\arctan a + \arctan b = \arctan \left(\dfrac{a + b}{1 - ab}\right)$$, so we have:

$$\arctan x + \arctan(1 - x) = \arctan\left(\dfrac1{1 - x(1 - x)}\right) = \arctan\left(\dfrac1{x^2 -x +1}\right)$$

What is the range of $$\dfrac1{x^2 - x + 1}$$? We can find that by completing the squares:

$$x^2 - x + 1 = \left(x - \dfrac12\right)^2 + \dfrac34\\$$, so the range of x^2 - x + 1 is $$\left[\dfrac34, \infty\right)$$.

Therefore, it means that the range of $$\dfrac1{x^2 - x + 1}$$ is $$\left(0, \dfrac43\right]$$.

(If you don't understand interval notations, that means $$\dfrac34 \leq x^2 - x + 1$$ and $$0<\dfrac1{x^2 -x + 1} \leq \dfrac43$$.)

Note that arctan is a strictly increasing function. As $$0<\dfrac1{x^2 -x + 1} \leq \dfrac43$$, we have $$\arctan 0 < \arctan\left(\dfrac1{x^2 - x + 1}\right) \leq \arctan\left(\dfrac43\right)$$, so $$0 < \arctan\left(\dfrac1{x^2 - x + 1}\right) \leq \arctan\left(\dfrac43\right)$$

The range of the function is $$\left(0, \arctan\left(\dfrac43\right)\right]$$.

May 14, 2022