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.
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?
Hints. (without using circles)
For the first inequality define $f(\theta) =\theta \cos(\theta) $ and $g(\theta) =\sin(\theta) $. Then $f(0)=g(0)$ and $f'(\theta) <g'(\theta) $ for $\theta\in (0,\pi/2)$. What can you conclude?
For the second inequality it is well known that $$\sin(\theta) <\theta$$ and it is easily proven with the use of Mean Value Theorem. After that use the fact that $\cos\theta <1$.
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).
