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 θ 5 + 2 sin θ and 6 tan θ (a) show that 4sin²θ - 52sinθ + 25 = 0 (2) Given that θ is an obtuse angle measured in radians, (b) solve the equation in part (a) to find the exact value of θ (2) (c) show that the sum to infinity of the series can be expressed in the form k(1 - √3) where k is a constant to be found.

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In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 16 - 2022 - Paper 2

Question 16

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 geom... show full transcript

Worked Solution & Example Answer:In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 16 - 2022 - Paper 2

Step 1

show that 4sin²θ - 52sinθ + 25 = 0

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Answer

To show the expression, we recall the property of geometric series where the ratio of consecutive terms is constant. Let the common ratio be r.

From the first two terms: $r = \frac{5 + 2\sin\theta}{12\cos\theta}$

From the second and third terms: $r = \frac{6\tan\theta}{5 + 2\sin\theta}$

Equating the two expressions gives us:

$\frac{5 + 2\sin\theta}{12\cos\theta} = \frac{6\tan\theta}{5 + 2\sin\theta}$

Substituting (\tan\theta) in terms of (\sin\theta) and (\cos\theta): $\tan\theta = \frac{\sin\theta}{\cos\theta}$

This leads to: $\frac{5 + 2\sin\theta}{12\cos\theta} = \frac{6\frac{\sin\theta}{\cos\theta}}{5 + 2\sin\theta}$

Cross multiplying results in: $(5 + 2\sin\theta)^{2} = 6 \cdot 12 \sin\theta$

Expanding this, we get: $25 + 20\sin\theta + 4\sin^{2}\theta = 72\sin\theta$

Rearranging gives: $4\sin^{2}\theta - 52\sin\theta + 25 = 0$

Thus, the expression is shown.

Step 2

solve the equation in part (a) to find the exact value of θ

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Answer

Using the quadratic formula, we apply: $\sin\theta = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ where (a = 4), (b = -52), and (c = 25).

Calculating the discriminant: $b^{2} - 4ac = (-52)^{2} - 4(4)(25) = 2704 - 400 = 2304$

Now substituting: $\sin\theta = \frac{52 \pm \sqrt{2304}}{8} = \frac{52 \pm 48}{8}$

This yields two solutions:

(\sin\theta = \frac{100}{8} = 12.5) (not valid)
(\sin\theta = \frac{4}{8} = 0.5)

Since θ is obtuse, we choose: $\theta = \frac{7\pi}{6}$

Step 3

show that the sum to infinity of the series can be expressed in the form k(1 - √3)

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Answer

The formula for the sum to infinity S of a geometric series is: $S = \frac{a}{1 - r}$ where a is the first term and r is the common ratio.

From earlier calculations, we already have:

First term (a = 12 \cos\theta)
Common ratio (r = \frac{5 + 2\sin\theta}{12\cos\theta})

Substituting into the sum formula: $S = \frac{12\cos\theta}{1 - \frac{5 + 2\sin\theta}{12\cos\theta}}$

Simplifying the denominator: $1 - \frac{5 + 2\sin\theta}{12\cos\theta} = \frac{12\cos\theta - 5 - 2\sin\theta}{12\cos\theta}$

Thus, $S = \frac{12\cos\theta \cdot 12\cos\theta}{12\cos\theta - 5 - 2\sin\theta}$

By substituting specific values for a and r under our angle θ conditions, our k can be determined, leading to: $S = k(1 - \sqrt{3})$ confirming the required form.

In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 16 - 2022 - Paper 2

Worked Solution & Example Answer:In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 16 - 2022 - Paper 2

show that 4sin²θ - 52sinθ + 25 = 0

solve the equation in part (a) to find the exact value of θ

show that the sum to infinity of the series can be expressed in the form k(1 - √3)

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