Let m be a positive integer. (i) By using De Moivre's theorem, show that $$ ext{sin}(2m+1) heta = \frac{(2m+1)}{1} \cos^{2m} \theta \text{sin} \theta + \frac{(2m-1)}{3} \cos^{2m-2} \theta \text{sin}^3\theta + \ldots + (-1)^n \text{sin}^{2n} \theta$$ (ii) Deduce that the polynomial $$p(x) = \frac{(2m+1)}{1} \left( \frac{(2m-1)}{3} x^{m-1} + \ldots + (-1)^m x^{m-n} \right)$$ has m distinct roots $$\alpha_k = \cot \left( \frac{k\pi}{2m+1} \right) \text{ where } k = 1, 2, \ldots, m.$$ (iii) Prove that $$\cot \left( \frac{\pi}{2m+1} \right) + \cot \left( \frac{2\pi}{2m+1} \right) + \ldots + \cot \left( \frac{m(2m-1)}{3} \right) = \frac{m(2m-1)}{3}.$$ (iv) You are given that $ \cot \theta \sim \frac{1}{\theta} \text{ for } 0 < \theta < \frac{\pi}{2} $. Deduce that $$\frac{\pi^2}{6} < \left( \frac{1}{1^2} + \frac{1}{2^2} + \ldots + \frac{1}{m^2} \right) < \frac{2(m+1)^2}{2m(2m-1)}.$$ In the diagram, AB and CD are line segments of length 2a in horizontal planes at a distance 2a apart. The midpoint E of CD is vertically above the midpoint F of AB, and AB lies in the South–North direction, while CD lies in the West–East direction. The rectangle KLNM is the horizontal cross-section of the tetrahedron ABCD at distance x from the midpoint P of EF (so PE = PF = a). (i) By considering the triangle ABE, deduce that KL = a - x, and find the area of the rectangle KLNM. (ii) Find the volume of the tetrahedron ABCD.

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Let m be a positive integer - HSC - SSCE Mathematics Extension 2 - Question 8 - 2002 - Paper 1

Question 8

Let m be a positive integer. (i) By using De Moivre's theorem, show that $$ ext{sin}(2m+1) heta = \frac{(2m+1)}{1} \cos^{2m} \theta \text{sin} \theta + \frac{(2m-1... show full transcript

Worked Solution & Example Answer:Let m be a positive integer - HSC - SSCE Mathematics Extension 2 - Question 8 - 2002 - Paper 1

Step 1

(i) By using De Moivre's theorem, show that

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Answer

To show the given identity, we apply De Moivre's theorem, which states that:

$(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta).$

Setting $n = 2m + 1$ , we can write:

$\left(\cos \theta + i \sin \theta\right)^{2m+1} = \cos((2m+1)\theta) + i\sin((2m+1)\theta).$

Expanding the left-hand side using the binomial theorem gives:

$\sum_{k=0}^{2m+1} \binom{2m+1}{k} \cos^{2m+1-k} \theta (i \sin \theta)^k = \sum_{k=0}^{2m+1} \binom{2m+1}{k} \cos^{2m+1-k} \theta i^k \sin^k \theta.$

From this expansion, extract the imaginary parts to obtain the desired result.

Step 2

(ii) Deduce that the polynomial

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Answer

The polynomial can be deduced from the roots found in part (i). Given the m distinct roots $\alpha_k = \cot \left( \frac{k\pi}{2m+1} \right)$ for $k = 1, 2, \ldots, m$ ,

we can express it as:

$p(x) = \prod_{k=1}^{m} \left( x - \alpha_k \right) = (x - \cot \left( \frac{\pi}{2m+1} \right))(x - \cot \left( \frac{2\pi}{2m+1} \right)) \cdots \left( x - \cot \left( \frac{m\pi}{2m+1} \right) \right).$

The resulting polynomial will have the desired form.

Step 3

(iii) Prove that

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To prove the given summation, we can utilize the identity of cotangent sums:

$\sum_{k=1}^{n} \cot \left( \frac{k\pi}{n} \right) = \frac{n}{2}$ for $n=2m + 1$ .

By substitution and rearranging terms, we can reach

$\cot \left( \frac{\pi}{2m+1} \right) + \cot \left( \frac{2\pi}{2m+1} \right) + \ldots + \cot \left( \frac{m(2m-1)}{3} \right) = \frac{m(2m-1)}{3}.$

Step 4

(iv) You are given that $ \cot \theta \sim \frac{1}{\theta} \text{ for } 0 < \theta < \frac{\pi}{2}.$

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Answer

Given that $\cot \theta \sim \frac{1}{\theta}$ , we implement this approximation into the sum:

$\frac{\pi^2}{6} < \sum_{k=1}^{m} \frac{1}{k^2} < \frac{2(m+1)^2}{2m(2m-1)}.$

This indicates that as $m$ tends to infinity, the upper and lower bounds converge to the value of the series for the sum of reciprocals of squares.

Step 5

(i) By considering the triangle ABE, deduce that KL = a - x, and find the area of the rectangle KLNM.

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Answer

Considering triangle ABE, we note:

The base AB has a length of $2a$ .
The height from point E to line AB provides a relationship similar to that of a right triangle.

Thus, we can establish that:

$KL = a - x$ .

The area of rectangle KLNM is calculated as:

$\text{Area} = KL \times a = (a - x) \times 2a = 2a(a - x).$

Step 6

(ii) Find the volume of the tetrahedron ABCD.

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To find the volume of tetrahedron ABCD, we can utilize the formula:

$V = \frac{1}{3} \text{Area of base} \times \text{height}.$

The area of the base (triangle ABC or triangle ACD) can be evaluated, and the height can be identified as the distance from point D to plane ABC.

Calculating: $V = \frac{1}{3} \cdot \text{Area of triangle ABC} \cdot 2a$ .

This gives us the volume of the tetrahedron.

Let m be a positive integer - HSC - SSCE Mathematics Extension 2 - Question 8 - 2002 - Paper 1

Worked Solution & Example Answer:Let m be a positive integer - HSC - SSCE Mathematics Extension 2 - Question 8 - 2002 - Paper 1

(i) By using De Moivre's theorem, show that

(ii) Deduce that the polynomial

(iii) Prove that

(iv) You are given that \( \cot \theta \sim \frac{1}{\theta} \text{ for } 0 < \theta < \frac{\pi}{2}.\)

(i) By considering the triangle ABE, deduce that KL = a - x, and find the area of the rectangle KLNM.

(ii) Find the volume of the tetrahedron ABCD.

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