Trigonometry in Geometry

Question

One of my friends gave me this problem. Initially I thought that this would be easy but i tried this for two days unable to do this in a proper manner.

There is a circle with an angle A located at the center of the circle

Prove that: $\sin A < A < \tan A$

Answer 1

The area of $\triangle ABC$ is $\frac{1}{2}\sin(x)$. The area of the colored wedge is $\frac{1}{2}x$, and the area of $\triangle ABD$ is $\frac{1}{2}\tan(x)$. By inclusion, we get $$ \frac{1}{2}\tan(x)\ge\frac{1}{2}x\ge\frac{1}{2}\sin(x)\tag{1} $$

Answer 2

The shortest way between two points is the straight line. This establishes $$\sin \dfrac A2<\dfrac A2.$$

Then the area of the triangle formed with the tangent to the circle is larger than the area of the sector, hence

$$\frac A2<\tan\frac A2.$$

Replace $\dfrac A2$ by $A$.

Answer 3

On $(0,\pi/2)$ we have $\cos t < 1 < \sec^2 t.$ Integrating each term from $t=0$ to $t=A$ gives your desired inequality as long as $0<A<\pi/2.$ [inequality false for at least some $A$ not in this range.]

Of course this proof is usually done via a diagram; see a calc book.