# Solve the equation $\arcsin\bigg(\dfrac{x+1}{\sqrt{x^2+2x+2}}\bigg)-\arcsin\bigg(\dfrac{x}{\sqrt{x^2+1}}\bigg)=\dfrac{\pi}{4}$

Solve the equation $$\arcsin\bigg(\dfrac{x+1}{\sqrt{x^2+2x+2}}\bigg)-\arcsin\bigg(\dfrac{x}{\sqrt{x^2+1}}\bigg)=\dfrac{\pi}{4}$$

My solution: I converted this equation in terms of $$\arctan$$ and applied tangent to both sides, and I got my answer as $$x=-1,0$$.

But then one of my friends said that $$x=2$$ satisfies too the above equation, and the reason he gave is as follows: $$\arcsin\bigg(\dfrac{3}{\sqrt{10}}\bigg)=\dfrac{\pi}{4}+\arcsin\bigg(\dfrac{2}{\sqrt{5}}\bigg)$$ and he applied sinus to both sides to obtain $$\dfrac{3}{\sqrt{10}}=\dfrac{3}{\sqrt{10}}\:$$.

Now I don't have any explanation for him. Can anyone here explain the reason behind this situation?

I plotted it in desmos, and I am getting $$x=-1,0$$ only.

• You can check numerically that $\arcsin(3/\sqrt{10})\neq \pi/4 + \arcsin(2/\sqrt{5})$; the left-hand side is an acute angle, and the right-hand side is obtuse. Now, the left- and right-hand values happen to be supplements of each other, so they do have the same sine; that doesn't make them equal. (Just like knowing $\cos 0=\cos 2\pi$ doesn't allow you to conclude that $0=2\pi$.) This is a common point of confusion.
– Blue
Feb 11 at 12:33
• Plugging in $x = 2$ to the LHS of the original equation gives $\arcsin\bigg(\dfrac{3}{\sqrt{10}}\bigg)-\arcsin\bigg(\dfrac{2}{\sqrt{5}}\bigg) \approx 0.141897$, which is not equal to $\frac{\pi}{4} \approx 0.785398$.
– Dan
Feb 15 at 17:35
• $\arcsin \left(\dfrac{2}{\sqrt 5} \right)>\arcsin \left(\dfrac{1}{\sqrt 2} \right) = \dfrac{\pi}{4}$. So in your equation RHS>$\dfrac{\pi}{2}$ whereas LHS <$\dfrac{\pi}{2}$ Feb 27 at 5:57

COMMENT.-Note that one of the two terms is equal to $$0$$ when $$x=0$$ and when $$x=-1$$. If the other term fits, then you do have solved directly the question. In fact this gives $$\arcsin\left(\frac{1}{\sqrt2}\right)-\arcsin(0)=\frac{\pi}{4}\\\arcsin(0)-\arcsin\left(-\frac{1}{\sqrt2}\right)=\frac{\pi}{4}$$ which is obviously true.

• However, it was necessary to show that $x=1$ and $x=0$ are the only solutions, which is easily shown with the derivative equal to $\dfrac{1}{x^2+2x+2}-\dfrac{1}{x^2+1}$ Feb 11 at 15:27

The solution you have obtained is correct and complete. And fully backed by the desmos plot.

Your friend's objection contains a wrong calculation, since we have non-equality $$1.249 \:\approx\:\arcsin\bigg(\frac3{\sqrt{10}}\bigg) \;\neq\; \frac\pi 4+\arcsin\bigg(\frac 2{\sqrt{5}}\bigg) \:\approx\: 1.893\,,$$ see also Blue's comment. Thus, $$\,x\!=\!2\,$$ does not satisfy the given equation.

The given equation can be rewritten as $$\arctan(\,x+1\,)\:=\:\arctan x+\frac\pi 4\,,\tag{1}$$ using the trigonometric identity $$\,\sin q =\dfrac{\tan q}{\sqrt{(\tan q)^2 + 1}}$$ $$\quad\stackrel{\tan q\,=\,x}{\iff}\quad$$ $$\arctan x =\arcsin\bigg(\dfrac{x}{\sqrt{x^2 + 1}}\bigg)\,$$.

Equation $$(1)$$ may be solved quickly by applying "$$\tan$$" and using the angle sum identity: \begin{align}x+1 & \:=\:\tan\Big(\arctan x+\frac{\pi}4\,\Big) \:=\: \frac{x+1}{1-x} \\[1.ex] \iff\;(x+1)\cdot x & \:=\: 0 \end{align} Please note $$\,1\!-\!x >0\:$$ in the denominator because $$\,\arctan x=\arctan (x+1)-\frac{\pi}4 <\frac{\pi}2-\frac{\pi}4=\frac{\pi}4\,$$.

The reason behind this situation is that $$\sin(\pi/2 - \theta) = \sin(\pi/2 + \theta)$$, even though $$\pi/2 - \theta \neq \pi/2 + \theta$$. Given this, $$\sin(\arcsin(\frac{3}{\sqrt{10}})) = \sin(\frac{\pi}{4} + \arcsin(\frac{2}{\sqrt{5}})),$$

but

$$\arcsin(\frac{3}{\sqrt{10}})\neq \frac{\pi}{4} + \arcsin(\frac{2}{\sqrt{5}})$$

The right-hand side is greater than $$\pi/2$$ and the left-hand side is less than $$\pi/2$$.