To solve this problem, we are asked to find approximate values for the given trigonometric expressions when $\theta$ is small. In these cases, we can use small-angle approximations for sine, cosine, and tangent. These approximations are:

• $\sin \theta \approx \theta$
• $\cos \theta \approx 1 - \frac{\theta^2}{2}$
• $\tan \theta \approx \theta$

Let's solve each part using these approximations:

Part (a): $\frac{1 - \cos 2\theta}{\theta \sin \theta}$

1. For small $\theta$, we use the approximation for $\cos 2\theta$: $\cos 2\theta \approx 1 - 2\theta^2$ So, $1 - \cos 2\theta \approx 1 - (1 - 2\theta^2) = 2\theta^2$

2. Also, $\sin \theta \approx \theta$. Therefore, the expression becomes: $\frac{1 - \cos 2\theta}{\theta \sin \theta} \approx \frac{2\theta^2}{\theta \cdot \theta} = 2$

Thus, for small $\theta$, the expression in part (a) approximately equals 2.

Part (b): $\frac{\cos 6\theta - 1}{\sin 3\theta \tan \theta}$

1. For small $\theta$, $\cos 6\theta \approx 1 - 18\theta^2$. Thus: $\cos 6\theta - 1 \approx -18\theta^2$

2. $\sin 3\theta \approx 3\theta$ and $\tan \theta \approx \theta$, so: $\sin 3\theta \tan \theta \approx 3\theta^2$

3. Substituting into the expression, we get: $\frac{\cos 6\theta - 1}{\sin 3\theta \tan \theta} \approx \frac{-18\theta^2}{3\theta^2} = -6$

Thus, for small $\theta$, the expression in part (b) approximately equals $-6$.

Part (c): $\frac{\theta \tan \theta}{1 - \cos \theta}$

1. $\tan \theta \approx \theta$, so $\theta \tan \theta \approx \theta^2$.

2. Using $\cos \theta \approx 1 - \frac{\theta^2}{2}$, we have: $1 - \cos \theta \approx 1 - \left(1 - \frac{\theta^2}{2}\right) = \frac{\theta^2}{2}$

3. Substituting, we get: $\frac{\theta \tan \theta}{1 - \cos \theta} \approx \frac{\theta^2}{\frac{\theta^2}{2}} = 2$

Thus, for small $\theta$, the expression in part (c) approximately equals 2.

Summary of approximate values:

• (a) $\frac{1 - \cos 2\theta}{\theta \sin \theta} \approx 2$
• (b) $\frac{\cos 6\theta - 1}{\sin 3\theta \tan \theta} \approx -6$
• (c) $\frac{\theta \tan \theta}{1 - \cos \theta} \approx 2$

Let me know if you need more details or clarifications!

Here are 5 additional related questions for practice:

1. What is the small-angle approximation for $\sin 2\theta$ and $\tan 2\theta$?
2. How does the approximation for $\cos \theta$ change if higher-order terms are included?
3. Why is $\theta$ assumed to be small in these approximations?
4. Can you derive the small-angle approximation for $\sin(3\theta)$ using a series expansion?
5. How does the accuracy of small-angle approximations vary as $\theta$ increases?

Tip:

Small-angle approximations are extremely useful in physics, especially in mechanics and wave theory, where angles are often small enough to simplify trigonometric functions for easier computation.