1. The problem asks to explain two terms and prove a trigonometric identity involving the resultant of two equal forces. 2. First, explain the terms: i) Moment of a force: It is the turning effect produced by a force acting at a distance from a pivot point, calculated as the product of the force magnitude and the perpendicular distance to the pivot. ii) Statics: It is the branch of mechanics that deals with bodies at rest or in equilibrium, analyzing forces to ensure the sum of forces and moments equals zero. 3. Now to prove: When two equal forces are inclined at angle 2\alpha, their resultant is twice as great as when they are inclined at angle 2\beta, show that $$2\cos \beta = \cos \alpha$$. 4. Let each force be $F$. The magnitude of their resultant when inclined at angle $2\theta$ is given by the law of cosines for vectors: $$R = \sqrt{F^2 + F^2 + 2FF\cos(2\theta)} = F \sqrt{2 + 2\cos(2\theta)}$$ 5. Simplify inside the square root using trigonometric identity $\cos(2\theta)=2\cos^2\theta -1$: $$R = F \sqrt{2 + 2(2\cos^2 \theta -1)} = F \sqrt{2 + 4\cos^2 \theta - 2} = F \sqrt{4\cos^2 \theta} = 2F \cos \theta$$ 6. According to the problem, the resultant at angle $2\alpha$ is twice the resultant at angle $2\beta$: $$R_{2\alpha} = 2 R_{2\beta}$$ 7. Substitute the expressions: $$2F \cos \alpha = 2 \times (2F \cos \beta)$$ Simplify: $$2F \cos \alpha = 4F \cos \beta$$ 8. Divide both sides by $2F$ (assuming $F \neq 0$): $$\cos \alpha = 2 \cos \beta$$ 9. Rearranged, it becomes: $$2 \cos \beta = \cos \alpha$$ 10. Hence, the required identity is proved. Final answer: $$2 \cos \beta = \cos \alpha$$