# Trigonometry equation with arctan

Solve the following equation: $\arctan x + \arctan (x^2-1) = \frac{3\pi}{4}$.

## What I did

Let $\arctan x = \alpha, \arctan(x^2-1) = \beta$, $\qquad\alpha+\beta = \frac{3\pi}{4}$

$\tan(\alpha+\beta) = \tan(\frac{3\pi}{4}) = -1$

$$\frac{\tan\alpha + tan\beta}{1-\tan\alpha\tan\beta} = \frac{x+x^2-1}{1-x(x^2-1)} = -1$$

\begin{align} x^2+x-1 &= -(1-(x^3-x)) = -1+x^3-x \\ \iff x^2 + x &= x^3-x \\ \iff x(x+1) &= x(x^2-1) \qquad\implies \boxed{x_1 = 0}\\ \implies x+1 &= x^2-1 \\ \iff x^2-x-2 &= 0 \\ \end{align}

$\therefore x_1 = 0,\quad x_2 = 2,\quad x_3 = -1$

However, the equation only works for $x=2$.

### I wonder

Did I do this in an efficient manner? Is there any easy way to find $x$ where there's no fake solutions?

Your three answers all look good to me. You did this in a pretty efficient manner, although I would've just jumped straight to the identity $$\arctan(A)+\arctan(B) = \arctan\left(\frac{A+B}{1-AB} \right)$$ without making the substitution $\arctan(x) = \alpha$, etc.

Anyway, you are confident that the equation works for $x=2$ (which it does.) But for $x=0$ note that a calculator will tell us that $$\arctan(0)+\arctan(0^2-1) = 0+\arctan(-1) = \frac{-\pi}{4}$$ However, this is because $$\tan\left( \frac{3\pi}{4}\right) =\tan\left( \frac{-\pi}{4} \right)$$ where there is a discrepancy with the calculator due to the fact that Cosine is negative and Sine is positive in the quadrant where $\frac{3\pi}{4}$ lies, and Cosine is positive while Sine is negative in the quadrant where $\frac{-\pi}{4}$ lies. Long story short, the calculator doesn't know the difference in the calculation and defaults to $\frac{-\pi}{4}$. The same exact thing happens for $x = -1$. All three of your answers are right though.

• Hmm, you're saying all three answers are correct; however, shouldn't $x=2$ be the only solution? I'm picturing the unit circle, where $3\pi/4$ is a different angle than $-pi/4$ even though they're as large. But you're confident they're all correct solutions? – B. Lee Oct 12 '14 at 19:23
• I believe all three are correct. $$\tan\left( \frac{3\pi}{4}\right) =\frac{\sin\left( \frac{3\pi}{4}\right)}{\cos\left( \frac{3\pi}{4}\right)} = \frac{\frac{\sqrt{2}}{2}}{\frac{-\sqrt{2}}{2}} =-1= \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=\frac{\sin\left( \frac{-\pi}{4}\right)}{\cos\left( \frac{-\pi}{4}\right)} =\tan\left( \frac{-\pi}{4} \right)$$ Negative one can commute between numerator and denominator – graydad Oct 12 '14 at 19:40
• I think I need to clarify; we're looking for a specific angle in the original equation, therefore the solutions to $\tan(-\frac{pi}{4})$ are not correct, I believe. Even though $\tan (-\frac{pi}{4}) = \tan\frac{3\pi}{4}$, $\frac{-\pi}{4} \not= \frac{3\pi}{4}$ – B. Lee Oct 12 '14 at 19:48
• Now I'm not sure if you agree or disagree with me. But your algebra above is flawless, so from a purely logical point of view, given that $$\arctan x + \arctan (x^2-1) = \frac{3\pi}{4} \iff \frac{x+x^2-1}{1-x(x^2-1)} = -1$$ then the three solutions you found for $x$ on the RHS have to work for the quantity on the LHS. – graydad Oct 12 '14 at 20:19
• No problem haha. I'm trying to convince you that the problem lies with calculators; they will lie to you in this case! – graydad Oct 12 '14 at 20:37

$$\arctan x+\arctan(x^2-1)=\begin{cases} \arctan\left(\dfrac{x+x^2-1}{1-x(x^2-1)}\right) &\mbox{if } x(x^2-1)<1 \\\pi+ \arctan\left(\dfrac{x+x^2-1}{1-x(x^2-1)}\right) & \mbox{if } x(x^2-1)> 1. \end{cases}$$

Now, $-\dfrac\pi2\le\arctan(z)\le\dfrac\pi2$

$\implies-\dfrac\pi2\le\arctan x+\arctan(x^2-1)\le\dfrac\pi2$ if $x(x^2-1)<1$ which is true if $x=0,-1$

Then $\arctan x+\arctan(x^2-1)\ne\dfrac{3\pi}4$