Finding The Inverse Tangent Based On The Definition Of Tangent Derived From Euler's Formula

In the section https://brilliant.org/wiki/eulers-formula/#trigonometric-applications, tan(x) is defined as (e^(i * x) - e^(-i * x)) / i * (e^(i * x) + e^(-i * x)). I have been trying to find the inverse tan(x) from this definition, but have been unable to do so. I first wrote the problem as y = (e^(i * x) - e^(-i * x)) / i * (e^(i * x) + e^(-i * x)), y being the tan(x), and then substituted all of the x for y and y for x. I now had the problem x = (e^(i * y) - e^(-i * y)) / i * (e^(i * y) + e^(-i * y)). I multiplied both sides by i and got x * i = (e^(i * y) - e^(-i * y)) / (e^(i * y) + e^(-i * y)), and then, knowing that a^(-b) = 1 / a^(b), rewrote the problem as x * i = (e^(i * y) - (1 / e^(i * y))) / (e^(i * y) + (1 / e^(i * y))), and combined terms from there to get x * i = ((e^(2 * i * y) - 1) / (e^(i * y))) / ((e^(2* i * y) + 1) / (e^(i * y))). I then divided the numerator's fraction by the denominator's fraction to get x * i = (e^(2 * i * y) - 1) / (e^(2 * i * y) + 1). From here, I do not know how to proceed. My goal is to isolate the y. Any suggestions on how I might do this would be greatly appreciated.

#Algebra

Note by Michael Vonica
3 years, 11 months ago

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Comments

Let y = e^(ix) first, then rearrange the equation to get a quadratic expression in terms of y.

Pi Han Goh - 3 years, 11 months ago

Thanks Pi Han Goh. That lead me to my answer.

Michael Vonica - 3 years, 11 months ago
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