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2. (a) By using the substitution $t = \tan\frac{\theta}{2}$, show that \( \frac{1 - \cos\theta}{\sin\theta} = \tan\frac{\theta}{2} \) - HSC - SSCE Mathematics Extension 1 - Question 2 - 2007 - Paper 1
Question 2
2. (a) By using the substitution $t = \tan\frac{\theta}{2}$, show that \( \frac{1 - \cos\theta}{\sin\theta} = \tan\frac{\theta}{2} \). (b) Let $f(x) = 2\cos^{-1} ... show full transcript
Worked Solution & Example Answer:2. (a) By using the substitution $t = \tan\frac{\theta}{2}$, show that \( \frac{1 - \cos\theta}{\sin\theta} = \tan\frac{\theta}{2} \) - HSC - SSCE Mathematics Extension 1 - Question 2 - 2007 - Paper 1
Step 1
Show that \( \frac{1 - \cos\theta}{\sin\theta} = \tan\frac{\theta}{2} \)
Answer
Using the substitution ( t = \tan\frac{\theta}{2} ), we have: [ \sin\theta = \frac{2t}{1+t^2}, \quad \cos\theta = \frac{1-t^2}{1+t^2} ] Substituting these into ( \frac{1 - \cos\theta}{\sin\theta} ): [ \frac{1 - \frac{1 - t^2}{1 + t^2}}{\frac{2t}{1 + t^2}} = \frac{t^2}{2t} = \frac{t}{2} = \tan\frac{\theta}{2}. ]
Step 2
Sketch the graph of $y = f(x)$, indicating clearly the coordinates of the endpoints of the graph.
Answer
To sketch the graph of ( y = f(x) = 2\cos^{-1} x ):
- Identify the domain: ( -1 \leq x \leq 1 ).
- The endpoints are at ( (1, 0) ) and ( (-1, 2\pi) ) since ( \cos^{-1}(x) ) ranges from 0 to ( \pi ).
- Graph the function, a decreasing curve from ( (1, 0) ) to ( (-1, 2\pi) ).
Step 3
Step 4
Find the values of $a$ and $b$.
Answer
Given ( P(2) = 0 ):
[
P(2) = 2^2 + 2a + b = 0 \rightarrow 4 + 2a + b = 0 \rightarrow b = -4 - 2a.
]
From the Remainder Theorem, when ( P(x) ) is divided by ( x + 1 ):
[
P(-1) = (-1)^2 + (-1)a + b = 1 - a + b = 18 \rightarrow 1 - a - 4 - 2a = 18 \rightarrow -3 - 3a = 18 \rightarrow a = -7.
]
Substituting back:
[
b = -4 - 2(-7) = -4 + 14 = 10.
]
Thus, ( a = -7 ) and ( b = 10 ).
Step 5
Find her acceleration ten seconds after she jumps. Give your answer correct to one decimal place.
Answer
The velocity function is ( v = 50(1 - e^{-2t}) ). To find acceleration, differentiate:
[
a(t) = \frac{dv}{dt} = 50 \cdot 2e^{-2t} = 100e^{-2t}.
]
At ( t = 10 ):
[
a(10) = 100e^{-20} \approx 0.0000454 \text{ (very small)} \approx 0.0.
]
Step 6
Find the distance that she has fallen in the first ten seconds.
Answer
To find the distance ( s ) fallen, integrate the velocity function: [ s(t) = \int_{0}^{t} v dt = \int_{0}^{10} 50(1 - e^{-2t}) dt. ] Calculating: [ s(10) = [50t + 25e^{-2t}]_{0}^{10} = 50(10) - 25e^{-20} - (0 -25) = 500 + 25 \approx 525 ext{ metres.} \