4
$\begingroup$

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.
Link to desmos

$\endgroup$
3
  • 5
    $\begingroup$ 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. $\endgroup$
    – Blue
    Feb 11 at 12:33
  • 1
    $\begingroup$ 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 $\frac{\pi}{4} \approx 0.785398$. $\endgroup$
    – Dan
    Feb 15 at 17:35
  • $\begingroup$ $\arcsin \left(\dfrac{2}{\sqrt 5} \right)>\arcsin \left(\dfrac{1}{\sqrt 2} \right) = \dfrac{\pi}{4}$. So in your equation RHS>$\dfrac{\pi}{2}$ whereas LHS <$\dfrac{\pi}{2}$ $\endgroup$ Feb 27 at 5:57

3 Answers 3

3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 $\dfrac{1}{x^2+2x+2}-\dfrac{1}{x^2+1}$ $\endgroup$
    – Piquito
    Feb 11 at 15:27
2
$\begingroup$

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 $\,\arctan x=\arctan (x+1)-\frac{\pi}4 <\frac{\pi}2-\frac{\pi}4=\frac{\pi}4\,$.

$\endgroup$
1
$\begingroup$

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 $\pi/2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .