Question: User: Trigonometric Equations - Year 2 Core
3.
In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
Given that the first three terms of a geometric series are
12 cos \theta 5 + 2 sin \theta and 6 tan \theta
(a) show that
4 sin² \theta - 52 sin \theta + 25 = 0
Given that \theta is an obtuse angle measured in radians,
(b) solve the equation in part (a) to find the exact value of \theta
(c) show that the sum to infinity of the series can be expressed in the form
k(1 - \sqrt{3})
where k is a constant to be found
(3)
(2)
(5)
1. **Problem Statement:**
Given the first three terms of a geometric series: $$12 \cos \theta,$$ $$5 + 2 \sin \theta,$$ and $$6 \tan \theta,$$ show that $$4 \sin^2 \theta - 52 \sin \theta + 25 = 0,$$ find the exact value of $$\theta$$ (obtuse angle in radians), and express the sum to infinity in the form $$k(1 - \sqrt{3})$$ where $$k$$ is a constant.
2. **Step (a): Show the quadratic in $$\sin \theta$$**
- Since the terms form a geometric series, the ratio between consecutive terms is constant:
$$r = \frac{5 + 2 \sin \theta}{12 \cos \theta} = \frac{6 \tan \theta}{5 + 2 \sin \theta}$$
- Cross-multiplied:
$$(5 + 2 \sin \theta)^2 = 12 \cos \theta \times 6 \tan \theta$$
- Simplify right side:
$$12 \cos \theta \times 6 \tan \theta = 72 \cos \theta \times \frac{\sin \theta}{\cos \theta} = 72 \sin \theta$$
- So:
$$(5 + 2 \sin \theta)^2 = 72 \sin \theta$$
- Expand left side:
$$25 + 20 \sin \theta + 4 \sin^2 \theta = 72 \sin \theta$$
- Rearrange:
$$4 \sin^2 \theta + 20 \sin \theta + 25 = 72 \sin \theta$$
- Bring all terms to one side:
$$4 \sin^2 \theta + 20 \sin \theta + 25 - 72 \sin \theta = 0$$
- Simplify:
$$4 \sin^2 \theta - 52 \sin \theta + 25 = 0$$
3. **Step (b): Solve the quadratic equation for $$\sin \theta$$**
- The quadratic is:
$$4 \sin^2 \theta - 52 \sin \theta + 25 = 0$$
- Use quadratic formula:
$$\sin \theta = \frac{52 \pm \sqrt{(-52)^2 - 4 \times 4 \times 25}}{2 \times 4}$$
- Calculate discriminant:
$$(-52)^2 - 4 \times 4 \times 25 = 2704 - 400 = 2304$$
- Square root:
$$\sqrt{2304} = 48$$
- So:
$$\sin \theta = \frac{52 \pm 48}{8}$$
- Two solutions:
$$\sin \theta = \frac{52 + 48}{8} = \frac{100}{8} = 12.5$$ (not possible since $$|\sin \theta| \leq 1$$)
$$\sin \theta = \frac{52 - 48}{8} = \frac{4}{8} = 0.5$$
- Since $$\theta$$ is obtuse (between $$\pi/2$$ and $$\pi$$), $$\sin \theta = 0.5$$ corresponds to:
$$\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$ radians.
4. **Step (c): Express sum to infinity in form $$k(1 - \sqrt{3})$$**
- The common ratio $$r$$ is:
$$r = \frac{5 + 2 \sin \theta}{12 \cos \theta}$$
- Substitute $$\sin \theta = 0.5$$ and $$\theta = \frac{5\pi}{6}$$:
$$\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$$
- Calculate numerator:
$$5 + 2 \times 0.5 = 5 + 1 = 6$$
- Calculate denominator:
$$12 \times \left(-\frac{\sqrt{3}}{2}\right) = -6 \sqrt{3}$$
- So:
$$r = \frac{6}{-6 \sqrt{3}} = -\frac{1}{\sqrt{3}}$$
- Sum to infinity formula for geometric series with $$|r| < 1$$:
$$S_\infty = \frac{a}{1 - r}$$
- First term $$a = 12 \cos \theta = 12 \times \left(-\frac{\sqrt{3}}{2}\right) = -6 \sqrt{3}$$
- Substitute values:
$$S_\infty = \frac{-6 \sqrt{3}}{1 - \left(-\frac{1}{\sqrt{3}}\right)} = \frac{-6 \sqrt{3}}{1 + \frac{1}{\sqrt{3}}}$$
- Simplify denominator:
$$1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}}$$
- So:
$$S_\infty = -6 \sqrt{3} \times \frac{\sqrt{3}}{\sqrt{3} + 1} = \frac{-6 \times 3}{\sqrt{3} + 1} = \frac{-18}{\sqrt{3} + 1}$$
- Rationalize denominator:
$$S_\infty = \frac{-18}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{-18 (\sqrt{3} - 1)}{3 - 1} = \frac{-18 (\sqrt{3} - 1)}{2} = -9 (\sqrt{3} - 1)$$
- Rewrite:
$$S_\infty = -9 \sqrt{3} + 9 = 9 (1 - \sqrt{3})$$
- Hence, $$k = 9$$.
**Final answers:**
- (a) $$4 \sin^2 \theta - 52 \sin \theta + 25 = 0$$
- (b) $$\theta = \frac{5\pi}{6}$$ radians
- (c) $$S_\infty = 9 (1 - \sqrt{3})$$