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Let \( \alpha = \cos \theta + i \sin \theta \), where \( 0 < \theta < 2\pi \) - HSC - SSCE Mathematics Extension 2 - Question 16 - 2017 - Paper 1
Question 16
Let \( \alpha = \cos \theta + i \sin \theta \), where \( 0 < \theta < 2\pi \). (i) Show that \( \alpha^k + \alpha^{-k} = 2 \cos k\theta \), for any integer \( k \).... show full transcript
Worked Solution & Example Answer:Let \( \alpha = \cos \theta + i \sin \theta \), where \( 0 < \theta < 2\pi \) - HSC - SSCE Mathematics Extension 2 - Question 16 - 2017 - Paper 1
Step 1
Step 2
By summing the series, prove that \( C = \frac{\alpha^{n} - \left(\alpha^{n+1} - \alpha^n\right)}{(1 - \alpha)(1 - \bar{\alpha})} \)
Answer
The series ( C ) can be rewritten:
[ C = 1 + \alpha + \alpha^2 + \cdots + \alpha^n = \frac{1 - \alpha^{n+1}}{1 - \alpha} ]
To show the required form, we note:
[ \frac{1 - \alpha^{n+1}}{1 - \alpha} = \frac{\alpha^n - (\alpha^{n+1} - \alpha^n)}{(1 - \alpha)(1 - \bar{\alpha})}. ]
Step 3
Deduce, from parts (i) and (ii), that \( 1 + 2\left(\cos \theta + \cos 2\theta + \cdots + \cos n\theta\right) = \frac{\cos n \theta - \cos (n+1)\theta}{1 - \cos \theta} \)
Answer
By substituting ( \alpha^k + \alpha^{-k} ) into the series form, we derive:
[ 1 + 2\left(\cos \theta + \cos 2\theta + \cdots + \cos n\theta\right) = \frac{\cos n\theta - \cos(n+1)\theta}{1 - \cos \theta} ].
Step 4
Show that \( \cos \frac{\pi}{n} + \cos \frac{2\pi}{n} + \cdots + \cos \frac{n\pi}{n} \) is independent of \( n \)
Answer
Using the result from part (iii) for a specific case where ( n ) approaches a limit, it can be shown that:
[ \sum_{k=1}^{n} \cos \frac{k\pi}{n} = \frac{n}{2} \quad \text{is constant for all n.} ]
Step 5
What are the possible values of \( a \) and \( b \)?
Answer
Given the eccentricity ( e = 2 ) for the hyperbola:
[ e = \frac{\sqrt{a^2 + b^2}}{a} \Rightarrow a = \frac{\sqrt{a^2 + b^2}}{2} \implies 4a^2 = a^2 + b^2 \implies 3a^2 = b^2 \implies b = \sqrt{3}a ].
Moreover, using the distance formula given: ( c = \sqrt{a^2 + b^2} \Rightarrow 2a = 1 \Rightarrow a = \frac{1}{2} \Rightarrow b = \frac{ ad3}{2} ).