# Mary L Boas Ch1, Section 15, Problem 29

$$\frac{F}{W} = \frac{T \, \sin \theta}{T \, \cos \theta} = \tan \theta = \theta \; + \; \frac{\theta^3}{3} \; + \; \frac{2\theta^5}{15} \; + ...$$;

How to solve part b? Solution in a book is $$\frac{x}{l} \; + \; \frac{x^3}{2l^3} \; + \; \frac{3x^5}{8l^5}$$

It's clear that $$\sin(\theta) = \frac{x}{l}$$ Considering $$\theta \ll 0$$, we have $$\theta = \frac{x}{l}$$