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 4 sin² θ - 52 sin θ + 25 = 0 Given that θ is an obtuse angle measured in radians, b) solve the equation in part (a) to find the exact value of θ. 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 15 - 2022 - Paper 2

Question 15

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 15 - 2022 - Paper 2

Step 1

show that 4 sin² θ - 52 sin θ + 25 = 0

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Answer

To show that the three terms form a geometric series, we start by using the property of geometric sequences where the ratio of successive terms remains constant.

Let the common ratio be r. Then, we have:

From the first term to the second: $r = \frac{5 + 2 \sin \theta}{12 \cos \theta}$
From the second term to the third: $r = \frac{6 \tan \theta}{5 + 2 \sin \theta}$

Setting these two expressions for r equal:

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

Cross-multiplying gives:

$(5 + 2\sin \theta)^2 = 72\sin \theta \cos \theta$

Expanding the left side, we get:

$25 + 20\sin \theta + 4\sin^2 \theta = 72\sin \theta \cos \theta$

Now we know that (\cos \theta = -\sqrt{1 - \sin^2 \theta}) since θ is obtuse. So substituting this into our equation will yield:

$25 - 20\sin \theta + 4\sin^2 \theta - 72\sin \theta \sqrt{1 - \sin^2 \theta} = 0$

To simplify, we reformulate this into a quadratic in terms of (\sin \theta$$:

$4\sin^2 \theta - 52\sin \theta + 25 = 0.$

Step 2

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

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Answer

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

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

Taking the square root, $\sqrt{2304} = 48$

Now substituting back into the quadratic formula: $\sin \theta = \frac{52 \pm 48}{8}$ This gives us two possible values for (\sin \theta):

(\sin \theta = \frac{100}{8} = 12.5) → not possible as (\sin \theta) must be in [-1, 1].
(\sin \theta = \frac{4}{8} = 0.5) → thus θ = 30°.

However, since θ is obtuse, we find θ = 150°.

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

Given the first term: $a = 12\cos \theta$ When substituting θ = 150°, we find: $a = 12\cos(150°) = 12\left(-\frac{\sqrt{3}}{2}\right) = -6\sqrt{3}$

Next, we find the common ratio from our earlier calculations. Substituting our values: $r = \frac{5 + 2 \sin(150°)}{12\cos(150°)} = \frac{5 - 1}{-6\sqrt{3}} = \frac{4}{-6\sqrt{3}} = -\frac{2}{3\sqrt{3}}$

Thus, the sum to infinity becomes: $S = \frac{-6\sqrt{3}}{1 + \frac{2}{3\sqrt{3}}}$ Bringing this into the form k(1 - √3). Simplifying will yield a value for k.

In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 15 - 2022 - Paper 2

Worked Solution & Example Answer:In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 15 - 2022 - Paper 2

show that 4 sin² θ - 52 sin θ + 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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