## Thursday, May 7, 2015

### Intermediate Value Theorem to Prove Inequality Problem

In my previous post, I talked about how to use calculus method to prove that $\sin 10^{\circ} > \dfrac{1}{6}$ (Never Be Complacent).

Or more precisely, the Intermediate Value Theorem as our weapon to prove it.

We will use the triple angle identity for $\sin 3x=3\sin x-4\sin^3 x$ and the value for $\sin 30^{\circ}=\dfrac{1}{2}$ when $x=10^{\circ}$ in our solution:

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

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

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

$4\sin^3 (10^{\circ})-3\sin (10^{\circ})+\dfrac{1}{2}=0$

Here, if we treat $\sin (10^{\circ})=r$, the equation above becomes

$4r^3-3r+\dfrac{1}{2}=0$---(1)

Suppose we want to find the solutions (i.e. the value for $r=\sin (10^{\circ})$ when $0<r<0.5$ since $0<\sin 10^{\circ}<\sin 30^{\circ}$ ) for equation (1).

We know this cubic equation doesn't factor nicely, so, we won't aim to find for the exact solution but to look for the interval where that specific solution lies, in this case, we will use the Intermediate Value Theorem.

Intermediate Value Theorem states that:

If $f$ is continuous on the closed and finite interval $[a,\,b]$, and if $K$ is any number between $f(a)$ and $f(b)$, then there exists (at least one) number $c$ where $c\in[a,\,b]$ such that $K=f(c)$.

Corollary:

If $f$ is continuous on the closed and finite interval $[a,\,b]$, and if $f(a)$ and $f(b)$ have opposite sign $(f(a)f(b)<0)$, then there exists (at least one) number $c$ where $c\in[a,\,b]$ such that $f(c)=0$ and so $c$ is the root of $f$.

We could exploit this fact to our problem, but we must think of two "wise" values as $a$ and $b$ so that we could have $f(a)<0$ and $f(b)>0$, and we must relate $a$ and $b$ with $\dfrac{1}{6}$, or to be more exact, if $a<b$, then $a<\dfrac{1}{6}<b$.

We also need to define $f(r)=4r^3-3r+\dfrac{1}{2}$.

We let $a=\dfrac{1}{7}$ and $b=\dfrac{1}{5}$.

$f(a)=f\left(\dfrac{1}{7}\right)=4\left(\dfrac{1}{7}\right)^3-3\left(\dfrac{1}{7}\right)+\dfrac{1}{2}>0$

$f(b)=f\left(\dfrac{1}{5}\right)=4\left(\dfrac{1}{5}\right)^3-3\left(\dfrac{1}{5}\right)+\dfrac{1}{2}<0$

Since $f(r)$ is continuous, by the Intermediate Value Theorem, there is a number $c$ between $a=\dfrac{1}{7}$ and $b=\dfrac{1}{5}$ such that $f(c)=0$. In other words, $c$ is the root of $f(r)$.

In our case, $c>\dfrac{1}{7}$ and our $c$ represents the solution for the equation $4r^3-3r+\dfrac{1}{2}=0$.

Therefore, we can conclude by now that

$\sin 10^{\circ}>\dfrac{1}{7}> \dfrac{1}{10}$ and we are hence done.