Wednesday, May 6, 2015

Prove that $\sin 10^{\circ}>\dfrac{1}{6}$.

Another heuristic problem solving technique that is so popular and widely use is called "Prove by Contradiction" technique.

Euclid's beautiful proof by contradiction to show there are infinitely many prime numbers is one of the most mind-boggling method to prove something is true that leave us speechless with absolute admiration.

According to wikipedia (Proof_by_contradiction):

In logic, proof by contradiction is a form of proof, and more specifically a form of indirect proof, that establishes the truth or validity of a proposition by showing that the proposition's being false would imply a contradiction. Proof by contradiction is also known as indirect proof, apagogical argument, proof by assuming the opposite, and reductio ad impossibilem. It is a particular kind of the more general form of argument known as reductio ad absurdum.

G. H. Hardy described proof by contradiction as "one of a mathematician's finest weapons", saying "It is a far finer gambit than any chess gambit: a chess player may offer the sacrifice of a pawn or even a piece, but a mathematician offers the game."

But not every problem can be tackled that way. If we proceed without the hunch when should we try this heuristic skill on the given math challenge problem, we might fret and feel helpless. For me, my hunch when do I think of using the method to prove something to be true by contradiction technique arises when all methods failed.

In this blog post, I will show one good and intriguing challenge problem that is designed for us to employ the prove by contradiction skill.

Prove that $\sin 10^{\circ}>\dfrac{1}{6}$.

We will use the method by contradiction to prove the actual value for $\sin 10^{\circ}$ is greater than $\dfrac{1}{6}$.

We first assume that $\sin 10^{\circ}<\dfrac{1}{6}$

But how do we prove that $\sin 10^{\circ}<\dfrac{1}{6}$? The triple angle formula for $\sin 3x$ should come into play since

1. We know the value for $\sin 30^{\circ}=\dfrac{1}{2}$,

2. We can relate $\sin 30^{\circ}$ and $\sin 10^{\circ}$ in one formula as $\sin 3x=3\sin x-4\sin^3 x$.

Putting all information in by letting $x=10^{\circ}$, and rearrange the equation to make $\sin 10^{\circ}$ as the subject, we have:

$\sin 3x=3\sin x-4\sin^3 x$

$\sin 3(10^{\circ})=3\sin 10^{\circ}-4\sin^3 10^{\circ}$

$\sin 30^{\circ}=3\sin 10^{\circ}-4\sin^3 10^{\circ}$

$\dfrac{1}{2}=3\sin 10^{\circ}-4\sin^3 10^{\circ}$

$\dfrac{1}{2}+4\sin^3 10^{\circ}=3\sin 10^{\circ}$

$\sin 10^{\circ}=\dfrac{\dfrac{1}{2}+4\sin^3 10^{\circ}}{3}=\dfrac{1}{6}+\dfrac{4\sin^3 10^{\circ}}{3}$

We have our assumption that $\sin 10^{\circ}<\dfrac{1}{6}$.

Therefore, we get:

$\dfrac{1}{6}+\dfrac{4\sin^3 10^{\circ}}{3}<\dfrac{1}{6}$

$\implies \dfrac{4\sin^3 10^{\circ}}{3}<0$ or $\sin 10^{\circ}<0$ which is truly impossible.

This result is contradicting our assumption, and the converse $\sin 10^{\circ}>\dfrac{1}{6}$ must be true and this completes the proof.

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