Question 8 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 8 - 2008 - Paper 1

To prove by induction, we start with the base case.

Base Case (n=1): $\cos(1\theta) = \frac{\sin(2\cdot 1\theta)}{2 \sin(\theta)}$ This simplifies as: $\cos(\theta) = \frac{\sin(2\theta)}{2 \sin(\theta)}$ This holds true as it can be verified through trigonometric identities.

Induction Hypothesis: Assume it holds for $n = k$: $\cos 3\theta + \cos 5\theta + \cdots + \cos(2k - 1)\theta = \frac{\sin(2k\theta)}{2 \sin(\theta)}$

Induction Step: We need to show it holds for $n = k + 1$: $\cos 3\theta + \cos 5\theta + \cdots + \cos(2k - 1)\theta + \cos(2(k+1) - 1)\theta$ Substituting our hypothesis: $= \frac{\sin(2k\theta)}{2 \sin(\theta)} + \cos(2k + 1)\theta$ Now, applying the sine addition formula: $= \frac{\sin((2k + 2)\theta)}{2 \sin(\theta)}$ This completes the induction, proving our statement for all integers $n \geq 1$.

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(i) Write down an expression for the area, $A$ of $S$: The area of the surface $S$, following the rotation is: $A = \sum_{k=1}^{n} A_k = \sum_{k=1}^{n} \frac{2 \pi R^2 \sin \delta \cos(2k - 1)\delta}{2}$

Reducing gives: $A = \pi R^2 \sin \delta \sum_{k=1}^{n} \cos(2k - 1)\delta$

Using the result from part (a): For $heta = \delta$, thus: $A = \pi R^2 \sin \delta \cdot \frac{\sin (2n\delta)}{2\sin \delta} = \frac{\pi R^2 \sin(2n\delta)}{2}$

(ii) Find the limiting value of $A$ as $n$ increases without bound: When $n \to \infty$, $\sin(2n\delta)$ oscillates between -1 and 1 giving: $\lim_{n \to \infty} A = \frac{\pi R^2}{2}.$