Question: 4. (a) Express 2 cos(\theta) + 8 sin \theta in the form R cos (\theta - \alpha), where R and \alpha are constants, R > 0 and 0 < \alpha < \frac{\pi}{2}. Give the exact value of R and give the value of \alpha in radians to 3 decimal places. The first three terms of an arithmetic sequence are \cos x, \cos x + \sin x, \cos x + 2 \sin x, x \neq n\pi. Given that S_9 represents the sum of the first 9 terms of this sequence as x varies, (b) (i) find the exact maximum value of S_9, (ii) deduce the smallest positive value of x at which this maximum value of S_9 occurs.
1. **Problem Statement:**
(a) Express $$2 \cos \theta + 8 \sin \theta$$ in the form $$R \cos (\theta - \alpha)$$ where $$R > 0$$ and $$0 < \alpha < \frac{\pi}{2}$$. Find exact $$R$$ and $$\alpha$$ to 3 decimal places.
(b) Given the arithmetic sequence with first three terms $$\cos x$$, $$\cos x + \sin x$$, $$\cos x + 2 \sin x$$, and $$x \neq n\pi$$, find:
(i) the exact maximum value of $$S_9$$, the sum of the first 9 terms.
(ii) the smallest positive $$x$$ where this maximum occurs.
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2. **Part (a) - Expressing in the form $$R \cos (\theta - \alpha)$$:**
The identity for expressing $$a \cos \theta + b \sin \theta$$ as $$R \cos (\theta - \alpha)$$ is:
$$
a \cos \theta + b \sin \theta = R \cos (\theta - \alpha) = R (\cos \theta \cos \alpha + \sin \theta \sin \alpha)
$$
Matching coefficients:
$$
a = R \cos \alpha, \quad b = R \sin \alpha
$$
Given $$a=2$$ and $$b=8$$.
Calculate $$R$$:
$$
R = \sqrt{a^2 + b^2} = \sqrt{2^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68} = 2 \sqrt{17}
$$
Calculate $$\alpha$$:
$$
\tan \alpha = \frac{b}{a} = \frac{8}{2} = 4
$$
So,
$$
\alpha = \arctan(4) \approx 1.325 \text{ radians (to 3 decimal places)}
$$
Since $$a > 0$$ and $$b > 0$$, $$\alpha$$ is in the first quadrant, satisfying $$0 < \alpha < \frac{\pi}{2}$$.
---
3. **Part (b) - Arithmetic sequence and sum:**
The terms are:
$$
T_1 = \cos x
$$
$$
T_2 = \cos x + \sin x
$$
$$
T_3 = \cos x + 2 \sin x
$$
Check common difference $$d$$:
$$
T_2 - T_1 = (\cos x + \sin x) - \cos x = \sin x
$$
$$
T_3 - T_2 = (\cos x + 2 \sin x) - (\cos x + \sin x) = \sin x
$$
So common difference $$d = \sin x$$.
First term $$a = \cos x$$.
Sum of first 9 terms:
$$
S_9 = \frac{9}{2} [2a + (9-1)d] = \frac{9}{2} [2 \cos x + 8 \sin x]
$$
Rewrite:
$$
S_9 = \frac{9}{2} (2 \cos x + 8 \sin x)
$$
From part (a), we know:
$$
2 \cos x + 8 \sin x = R \cos (x - \alpha) = 2 \sqrt{17} \cos (x - 1.325)
$$
So,
$$
S_9 = \frac{9}{2} \times 2 \sqrt{17} \cos (x - 1.325) = 9 \sqrt{17} \cos (x - 1.325)
$$
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4. **(b)(i) Find exact maximum value of $$S_9$$:**
Maximum of $$\cos (x - 1.325)$$ is 1.
Therefore,
$$
\max S_9 = 9 \sqrt{17} \times 1 = 9 \sqrt{17}
$$
---
5. **(b)(ii) Smallest positive $$x$$ for maximum $$S_9$$:**
Maximum occurs when:
$$
\cos (x - 1.325) = 1 \implies x - 1.325 = 2n\pi, \quad n \in \mathbb{Z}
$$
Smallest positive $$x$$ is for $$n=0$$:
$$
x = 1.325 \text{ radians}
$$
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**Final answers:**
(a) $$R = 2 \sqrt{17}$$, $$\alpha \approx 1.325$$ radians.
(b)(i) Maximum $$S_9 = 9 \sqrt{17}$$.
(b)(ii) Smallest positive $$x = 1.325$$ radians.