# Why did this extraneous root creep into the solution?

I was solving this equation and proceeded as follows:

$$\arcsin (1-x) - 2\arcsin (x) = \frac{π}{2}$$

$$\implies \arcsin(1-x) = \frac{π}{2} + 2\arcsin (x)$$ $$\implies \sin (\arcsin (1-x)) = \sin \left( \frac {π}{2} + 2\arcsin (x)\right)$$ $$\implies (1-x) = \cos \left(2\arcsin(x)\right)$$ $$\require{cancel}\implies \cancel{1}-x =\cancel{1}- 2 \left(\sin\arcsin (x)\right)^2$$ $$\implies -x = -2x^2$$ $$\therefore x = 0$$ Or $$x=\frac{1}{2}$$

However, $x=\frac{1}{2}$ doesn't satisfy the original equation.

I understand that extraneous roots do creep in while solving inverse trig problems, but I wonder why it crept in here.

If possible (if it doesn't make the question too broad), I'd like to know the general causes for the occurrence of extraneous roots in inverse trig equations, too.

• At your second implication sign. Commented Mar 16, 2018 at 8:43
• @LordSharktheUnknown since the sine function is not one-one? Commented Mar 16, 2018 at 8:44
• Exaclty when you applied a non-injective function to both sides of the equation( $\sin$ which, of couse, has been not one-to-one since always). Truth to be told, I don't understand why so many people are taught exotic and ineffective notions such as "extraneous solutions", rather than a proper method to turn an equation into an equivalent system of equations+inequations upon application of the usual four non-injective functions.
– user228113
Commented Mar 16, 2018 at 8:47
• @G.Sassatelli Um, the usual four non-injective functions? I'm a stranger to them. Where can I learn of them (maybe a link to some website may help)? I've never heard of your method, but it sounds interesting. Commented Mar 16, 2018 at 8:51
• Your "And" should be "Or" Commented Mar 16, 2018 at 8:52

When $x=\frac12$, you have $\arcsin(x)=\frac{\pi}{6}$ and $\arcsin(1-x)=\frac{\pi}{6}$

so $\sin (\arcsin (1-x))=\sin \left( \frac {π}{6} \right)$ and $\sin \left( \frac {π}{2} + 2\arcsin (x)\right) =\sin \left( \frac {5π}{6} \right)$

showing your third line would be a correct equality when $x=\frac12$ since $\sin \left( \frac {π}{6} \right)=\sin \left( \frac {5π}{6} \right) =\frac12$

but your second line would not be a correct equality when $x=\frac12$ since $\frac {π}{6}\not = \frac {5π}{6}$

and it is this use of $\sin$ which creates an equality which was not in the original expression.

$\arcsin (1-x) - 2\arcsin (x) = \frac{π}{2} \Rightarrow x=0 \text{ or } x=\frac12$ is a correct statement, but checking shows $x=0 \Rightarrow \arcsin (1-x) - 2\arcsin (x) = \frac{π}{2}$ while $x=\frac12 \not \Rightarrow \arcsin (1-x) - 2\arcsin (x) = \frac{π}{2}$

• Thank you for the helpful explanation! By the way, would it be better to verify the solutions at the very end (to eliminate extraneous roots) or to keep track of possible extraneous roots as one solves the problem? Commented Mar 16, 2018 at 11:40

$$-2\arcsin(x)=\dfrac\pi2-\arcsin(1-x)=\arccos(1-x)$$

Now using Principal values of $\arccos$,

$0\le-2\arcsin(x)\le\pi\iff0\ge\arcsin(x)\ge-\dfrac\pi2\implies-1\le x\le0$

Now let $\arcsin x=y\implies x=\sin y$

and $\arcsin(1-\sin y)=\dfrac\pi2+2y$

$$1-\sin y=\sin\left(\dfrac\pi2+2y\right)=\cos2y=1-2\sin^2y$$

$$\implies \sin y=0,\dfrac12$$ but $\sin y=x\le0$

• Thanks for the answer! I guess one must always keep an eye on the principal values to root out extra solutions! (+1) Commented Mar 16, 2018 at 11:41

The sine function is not one-to-one. When you applied that function you also got

$\sin(\arcsin(1-x))=\sin(\frac{\pi}{2}-2\arcsin(x))$,

as this RHS equals the one you intended. Note the sign change on the right side.

Then $x=\frac{1}{2}$ satisfies

$\arcsin(1-x)+2\arcsin(x)=\frac{\pi}{2}$.

• Thanks for your answer! (+1) So extraneous roots creep in when one uses non-injective functions on both sides of an equation! Commented Mar 16, 2018 at 11:42
• And, many people do not recognize that the sine function is included in that description. How is the sine function non-injective when it doesn't have anything that's squared? Uh-oh, time to review our geometry text, specifically the section where they talk about how a line (with a fixed ordinate) intersects a circle. Commented Mar 16, 2018 at 12:35

# None of your implication steps are wrong

Let's review the concept of implication. In mathematics, an implication is represented by the symbol $\implies$ and when we write $A \implies B$, it means precisely that if A is true, then B is true.

Here is a quote of your second implication:

$$\arcsin(1-x) = \frac{\pi}{2} + 2\arcsin (x) \implies \sin (\arcsin (1-x)) = \sin \left( \frac {\pi}{2} + 2\arcsin (x)\right)$$

This is perfectly correct, since the form of that statement is $A = B \implies \sin(A) = \sin(B)$, and clearly if A = B, they are the same thing, and their sine will also be equal.

# Well then where is my mistake? Surely there is a mistake somewhere

Let's summarize what you've proved:

$$\arcsin (1-x) - 2\arcsin (x) = \frac{\pi}{2} \implies x = 0 \text{ or } x = \dfrac{1}{2}$$

This is correct! (bear with me, I'll get there).

Assume my car is blue. Then I tell you the following sentence: "My car is blue or my car is red." That sentence is correct!

See what's happening here?

When we solve equations, we should be using "equivalence" steps, not "implication" steps, because we want to find all the values of $x$ that satisfy the equation, no more no less.

I find it enlightening to look at it this way because we don't lose the benefit of formal math which IMO makes much easier to find mistakes. By understanding well how implications work, it is clear that there is no mistake in the sequence of implications you created.

Of course, the other answers are correct as well, but I'd like to provide this formal insight because I find it very useful.

If you replace the implication symbols ($\implies$) in your attempt by equivalence symbols ($\iff$), you'll have a mistake on your second step, for the reasons other people already mentioned.