Chapter 8.3, Problem 55AYU

To determine

To derive the fact that $\sin \theta <\theta$ , where $\theta >0$ is measured in radians.

Answer to Problem 55AYU

  $\boxed{\sin \theta <\theta }$,

Explanation of Solution

Given information:

Show that the length d of a chord of a circle of radius r is given by the formula

  $d=2r\sin \frac{\theta }{2}$

where $\theta$ is the central angle formed by the radii to the ends of the chord. See the figure. Use this result to derive the fact that $\sin \theta <\theta$ , where $\theta >0$ is measured in radians.

  

Calculation:

In the triangle $\Delta OAB$ , by Law of Cosines, we get

  ${\left(AB\right)}^2={\left(OA\right)}^2+{\left(OB\right)}^2-2\left(OA\right)\left(OB\right)\cos \theta$

  $d^2=r^2+r^2-2\left(r\right)\left(r\right)\cos \theta$

  $=2r^2\left(1-\cos \theta \right)$

  $=4r^2\left(\frac{1-\cos \theta }{2}\right)$

  $d=2r\sqrt{\frac{1-\cos \theta }{2}}$

  $=2r\sqrt{\frac{1-\left(1-2{\sin }^2\left(\frac{\theta }{2}\right)\right)}{2}}$

  $=2r\sin \left(\frac{\theta }{2}\right)\mathrm{.......}\left(1\right)$

This proves the first required result.

Now,

  $angl{e}_{in,radian}=\frac{arc}{length\left(radius\right)}$

  $\theta =\frac{arc-lengthAB}{OA}$

  $\theta =\frac{arc-lengthAB}{r}$

Arc length $AB=r\theta$ (angle measured in radian).....(2)

Since $lengthAB<arc-lengthAB$ ,we get

  $d<r\theta$ using (2)

  $2r\sin \left(\frac{\theta }{2}\right)<r\theta$ using (1)

  $\sin \left(\frac{\theta }{2}\right)<\frac{\theta }{2}$ ........(3)

Since $\theta$ is an interior angle of a triangle, we get

  $0<\theta <\pi$

  $0<\frac{\theta }{2}<\frac{\pi }{2}$

  $\cos \frac{\theta }{2}<1$

  $\cos \frac{\theta }{2},\sin \frac{\theta }{2}<\frac{\theta }{2}$ as $0<\theta <\pi \Rightarrow 0<\frac{\theta }{2}<\frac{\pi }{2}\Rightarrow 0<\sin \frac{\theta }{2}<1$

  $2\cos \frac{\theta }{2},\sin \frac{\theta }{2}<\theta$ using (3)

  $\sin \theta <\theta$

Hence, this proves the second required result.