May 2019

Now, solve $\sin(x)=x$ .

By inspection, $x=0$ .

Is this the only solution?

Visually, it is difficult to tell (see Fig. 1 and Fig. 2)

Fig. 1

Fig. 2

However, we can prove that $0$ is the only solution:

Fig. 3

If $0 < x \leq 1 < \frac{\pi}{2}$ then from Fig. 3, we have

Area of triangle OAB < Area of circular sector OAB.

That is,

$\frac{1}{2}\cdot 1 \cdot \sin(x)< \frac{1}{2}\cdot 1 \cdot x$ .

Hence,

$\forall 0< x \leq 1, \sin(x) < x\quad\quad\quad(1)$

If $x>1$ then

$\sin(x) \leq 1 \overset{x>1}{\implies} \sin(x) \leq 1< x \implies \forall x>1, \sin(x) <x\quad\quad\quad(2)$

Put (1) and (2) together, we have

$\forall x > 0, \sin(x) < x\quad\quad\quad(3)$

If $x < 0$ then

$-x > 0 \overset{(3)}\implies \sin(-x) < -x \overset{\sin(-x)=-\sin(x)}\implies -\sin(x) < -x \implies \sin(x) > x$

i.e.,

$\forall x < 0, \sin(x) > x\quad\quad\quad(4)$

Therefore by (3) and (4), we conclude that

only when $x = 0, \sin(x)=x$ .

Month: May 2019