If $0 \lt x \lt \frac{\pi}{2}$, is $\cos(x) \le \frac{\sin(x)}{x} \le \frac{1}{\cos(x)}$?

Question

How can I find the following product using elementary trigonometry?

Suppose $0 \lt x \lt \frac{\pi}{2}$ is an angle measured in radians. Use the trigonometric circle and show that $\cos(x) \le \frac{\sin(x)}{x} \le \frac{1}{\cos(x)}$.

I have been trying to solve this question. I can't figure out whether or not the solution requires a trigonometric circle or if it can be done using another method.

Answer 1

Draw a segment $OAB$ of the unit circle such that $x=\angle AOB$ is an acute angle. Let $C$ be the foot of the perpendicular from $B$ to $OA$. Let $D$ be the point on $OB$ extended such that $AD$ is perpendicular to $OA$.

Now calculate and compare the three areas of the triangle $OBC$, the circular segment $OAB$ and the triangle $OAD$, then do a little algebra.

Can you take it from here?

Answer 2

since x, sinx and cosx are all positive in (0, pi/2), we can write the ineq as cosxsinx <= x or 2cosxsinx <= 2x or sin2x <= 2x or siny <= y where y= 2x lies in (0, pi). however, siny <= y is extremely easy to prove (can also be seen from graph).