Removing $\cos \theta$ from the given equation?

Question

I have been stuck on a particular question. The question relates to the area of a triangle being equal to the area of a sector. As such, I came to the conclusion that $\sin\frac{\theta}{2} * \cos\frac{\theta}{2} = \frac{1}{2}*\theta $ . But I do not quite understand what to do next! Is there some concept to apply here, or did I simply mess up?

($\theta$ is in radians)

Answer 1

Use the identity:$$\sin\big(\frac{\theta}{2}\big)\cos\big(\frac{\theta}{2}\big)=\frac{1}{2}\sin(\theta)$$ to remove the cosine. Then set $$\frac{1}{2}\sin(\theta)=\frac{1}{2}\theta$$. Now it is much simpler. Can you finish?

Answer 2

Note that the equation is equivalent to

$$\sin \theta =\theta$$

whose solution is $\theta =0$.

Answer 3

The area of the triangle equals $\sin \frac \theta2 \cos \frac \theta2 = \frac 12 \sin \theta$ The area of the section of the circle is $\frac \theta 2$

From this geometric argument we can conclude that $|\sin \theta |\le |\theta|$ but when $\theta$ is small they are nearly equal.