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Question

Solve for the following : $$\frac{\pi}{4}=\frac{e^x-e^{-x}}{2}+\arctan(x+1)$$

Clearly I would mutliply both sides by 2 but then I would get the following :

$$\frac{\pi}{2}=e^x-e^{-x}+\arctan(x+1)$$

I was wondering how I would simplify from here. Would i convert $\arctan$ into $\frac{\arcsin}{\arccos}?$

preferably done without using hyperbolic trig functions

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    $e^x - e^{-x} = 2\sinh(x)$. – 2017-01-24
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    You have multiplied by 2 and got the same equation. Are you sure there is no typo? If the LHS is $\frac {\pi }{4} $, then $0$ is a solution. – 2017-01-24
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    @rohan indeed it is 0 but i dont know how to arrive at that conclusion. and also i fixed it – 2017-01-24
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    Note! $\arctan \neq \frac{\arcsin}{\arccos}$ ! Not necessary for this problem, but it will lead to mistakes in the future. Try evaluating each side for a few different values with a calculator... – 2017-01-24
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    @JohnRawls In general, with these kind of equations you need to use methods that successively find better approximations to the roots of the function. There are different methods (e.g. Newton's Method) but for this case you can clearly see that $0$ is a solution. By taking the first derivative you can see that it always increases, therefore the only solution is $0$. Note: when you encounter equations of this kind, it is a good start to find a possible interval that might contain the zero, then use the Intermediate Value Theorem to prove it. From there you use Newton's Method to find the root. – 2017-01-24

2 Answers 2

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You can guess that $0$ is a solution, and because the derivative of $e^x-e^{-x}+2\arctan(x+1)$ is always positive, we know that $0$ is the only solution.

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Let $f(x)=\frac{e^x-e^{-x}}{2}+\arctan(x+1)$.

Since $f'(x)=\frac{e^x+e^{-x}}{2}+\frac{1}{x^2+2x+2}>0$,

we get that $f$ is an increasing function.

Thus, our equation has one root maximum and since $0$ is a root, we get an answer: $\{0\}$.

I fixed a typo.