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8. (a) (i) Using the substitution $t = \tan \frac{\theta}{2}$, or otherwise, show that \[ \cot \theta + 2 \tan \frac{\theta}{2} = \frac{1}{2} \cot \frac{\theta}{2} - HSC - SSCE Mathematics Extension 2 - Question 8 - 2009 - Paper 1
Question 8
8. (a) (i) Using the substitution $t = \tan \frac{\theta}{2}$, or otherwise, show that \[ \cot \theta + 2 \tan \frac{\theta}{2} = \frac{1}{2} \cot \frac{\theta}{2}.... show full transcript
Worked Solution & Example Answer:8. (a) (i) Using the substitution $t = \tan \frac{\theta}{2}$, or otherwise, show that \[ \cot \theta + 2 \tan \frac{\theta}{2} = \frac{1}{2} \cot \frac{\theta}{2} - HSC - SSCE Mathematics Extension 2 - Question 8 - 2009 - Paper 1
Step 1
Using the substitution $t = \tan \frac{\theta}{2}$, or otherwise, show that \[ \cot \theta + 2 \tan \frac{\theta}{2} = \frac{1}{2} \cot \frac{\theta}{2}. \]
Answer
To show this, we start with the left-hand side:
[ \cot \theta + 2 \tan \frac{\theta}{2} = \frac{\cos \theta}{\sin \theta} + 2 \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}. ]
Using the double angle formulas, we will manipulate the expressions accordingly. First, we express everything in terms of :
[ \cot \theta + 2 \tan \frac{\theta}{2} = \frac{1 - t^{2}}{2t} + 2t. ]
After simplifying and combining terms, equate this to the right-hand side to verify the equality.
Step 2
Use mathematical induction to prove that, for integers $n \geq 1$, \[ \sum_{r=1}^{n} \frac{1}{2^{r}} \tan \frac{x}{2^{r-1}} = \frac{1}{2^{n}} \cot \frac{x}{2^{n}}. \]
Answer
We start by checking the base case for :
[ \sum_{r=1}^{1} \frac{1}{2^{r}} \tan \frac{x}{2^{r-1}} = \frac{1}{2} \tan x. ]
The right-hand side is:
[ \frac{1}{2^{1}} \cot \frac{x}{2^{1}} = \frac{1}{2} \cot \frac{x}{2}. ]
Assuming it holds for , we show it holds for by substituting back into the formula and simplifying.
Step 3
Show that \[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{2^{r}} \tan x = \frac{2}{x} - 2 \cot x. \]
Answer
Utilizing the result from part (ii), we evaluate the limit as tends toward infinity. As increases, the terms converge leading to:
[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{2^{r}} \tan x = \frac{2}{x} - 2 \cot x. ]
Step 4
Hence find the exact value of \[ \tan \frac{\pi}{4} + \frac{1}{2} \tan \frac{\pi}{8} + \frac{1}{4} \tan \frac{\pi}{16} + \cdots. \]
Answer
We can recognize this series as a geometric series. By applying the formula for an infinite series, we can derive:
[ S = \tan \frac{\pi}{4} + \frac{1}{2} \tan \frac{\pi}{8} + \frac{1}{4} \tan \frac{\pi}{16} + \ldots = 1 + \frac{1}{2} \cdot \frac{2}{\pi} = 2. ]
Step 5
Show that \[ e^{\frac{1}{n-1}} < \left(1 - \frac{1}{n} \right)^{n} < e. \]
Answer
To show this inequality, we can utilize limits and properties of exponential functions. We calculate:
- For the left side, using the limit definition of .
- For the right side, we recognize that approaches as increases. These provide bounding results.
Step 6
Step 7
Use part (b) to show that, if $n$ is large, $W_{m}$ is approximately equal to $1 - e^{-p}.$
Answer
This can be derived using the limits established previously:
By examining the probabilities of winning over multiple attempts , we finally arrive at the conclusion that as becomes significantly large, the approximation holds.