By using the substitution $t = \tan\left( \frac{\theta}{2} \right)$, show that $ \frac{1 - \cos\theta}{\sin\theta} = \tan\left( \frac{\theta}{2} \right) $. Let $ f(x) = 2\cos^{-1}(x) $. (i) Sketch the graph of $ y = f(x) $, indicating clearly the coordinates of the endpoints of the graph. (ii) State the range of $ f(x) $. The polynomial $ P(x) = x^2 + ax + b $ has a zero at $ x = 2 $. When $ P(x) $ is divided by $ x + 1 $, the remainder is 18. Find the values of $ a $ and $ b $. A skydiver jumps from a hot air balloon which is 2000 metres above the ground. The velocity, $ v $ metres per second, at which she is falling t seconds after jumping is given by $ v = 50(1 - e^{-0.2t}) $. (i) Find her acceleration ten seconds after she jumps. Give your answer correct to one decimal place. (ii) Find the distance that she has fallen in the first ten seconds. Give your answer correct to the nearest metre.

SSCE HSC Mathematics Extension 1 Vectors in two dimensions: By using the substitution $t = \

By using the substitution $t = \tan\left( \frac{\theta}{2} \right)$, show that $ \frac{1 - \cos\theta}{\sin\theta} = \tan\left( \frac{\theta}{2} \right) $ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2007 - Paper 1

Question 2

$By-using-the-substitution-$t-=-\tan\left(-\frac{\theta}{2}-\right)$,-show-that-$-\frac{1---\cos\theta}{\sin\theta}-=-\tan\left(-\frac{\theta}{2}-\right)-$-HSC-SSCE Mathematics Extension 1-Question 2-2007-Paper 1.png$

By using the substitution $t = \tan\left( \frac{\theta}{2} \right)$, show that $ \frac{1 - \cos\theta}{\sin\theta} = \tan\left( \frac{\theta}{2} \right) $. Let \(... show full transcript

Worked Solution & Example Answer:By using the substitution $t = \tan\left( \frac{\theta}{2} \right)$, show that $ \frac{1 - \cos\theta}{\sin\theta} = \tan\left( \frac{\theta}{2} \right) $ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2007 - Paper 1

Step 1

By using the substitution $t = \tan\left( \frac{\theta}{2} \right)$, show that $ \frac{1 - \cos\theta}{\sin\theta} = \tan\left( \frac{\theta}{2} \right) $.

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Answer

To prove this, we can use the half-angle identities:

The half-angle identity for sine is given by: $\sin\theta = 2\sin\left( \frac{\theta}{2} \right)\cos\left( \frac{\theta}{2} \right)$
The half-angle identity for cosine is given by: $1 - \cos\theta = 2\sin^2\left( \frac{\theta}{2} \right)$

By substituting into the fraction, we have: $\frac{1 - \cos\theta}{\sin\theta} = \frac{2\sin^2\left( \frac{\theta}{2} \right)}{2\sin\left( \frac{\theta}{2} \right)\cos\left( \frac{\theta}{2} \right)} = \frac{\sin\left( \frac{\theta}{2} \right)}{\cos\left( \frac{\theta}{2} \right)} = \tan\left( \frac{\theta}{2} \right)$

Step 2

Let $ f(x) = 2\cos^{-1}(x) $. (i) Sketch the graph of $ y = f(x) $, indicating clearly the coordinates of the endpoints of the graph.

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To sketch the graph of ( f(x) = 2\cos^{-1}(x) ):

Identify the domain of ( f(x) ): the function ( \cos^{-1}(x) ) is defined for ( x \in [-1, 1] ).
The endpoints of the graph are:
- At ( x = -1 ), ( f(-1) = 2\cos^{-1}(-1) = 2\times \pi = 2\pi ).
- At ( x = 1 ), ( f(1) = 2\cos^{-1}(1) = 0 ).
Thus, the coordinates of the endpoints are ((-1, 2\pi)) and ((1, 0)).
The graph is a decreasing curve starting from ((1, 0)) to ((-1, 2\pi)).

Step 3

(ii) State the range of $ f(x) $.

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The range of ( f(x) = 2\cos^{-1}(x) ) is:

As ( x ) moves from -1 to 1, ( f(x) ) decreases from ( 2\pi ) to 0.
Therefore, the range is ( [0, 2\pi] ).

Step 4

The polynomial $ P(x) = x^2 + ax + b $ has a zero at $ x = 2 $. When $ P(x) $ is divided by $ x + 1 $, the remainder is 18. Find the values of $ a $ and $ b $.

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Answer

Given that ( P(2) = 0 ):

Substitute ( x = 2 ): $P(2) = 2^2 + 2a + b = 0 \Rightarrow 4 + 2a + b = 0 \Rightarrow 2a + b = -4 \quad \text{(1)}$
Use the Remainder Theorem for ( P(-1) = 18 ): $P(-1) = (-1)^2 + a(-1) + b = 18 \Rightarrow 1 - a + b = 18 \Rightarrow -a + b = 17 \quad \text{(2)}$
Solve equations (1) and (2): From (1), ( b = -4 - 2a ) and substitute into (2): $-a + (-4 - 2a) = 17 \Rightarrow -3a - 4 = 17 \Rightarrow -3a = 21 \Rightarrow a = -7$ Substitute ( a o -7 ): ( b = -4 - 2(-7) = -4 + 14 = 10 )

Thus, ( a = -7 ) and ( b = 10 ).

Step 5

A skydiver jumps from a hot air balloon which is 2000 metres above the ground. The velocity, $ v $ metres per second, at which she is falling t seconds after jumping is given by $ v = 50(1 - e^{-0.2t}) $. (i) Find her acceleration ten seconds after she jumps. Give your answer correct to one decimal place.

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Answer

To find acceleration, we need to differentiate the velocity function:

Differentiate ( v ): $\frac{dv}{dt} = 50 \cdot 0.2e^{-0.2t} = 10e^{-0.2t}$
Evaluate at ( t = 10 ): $\frac{dv}{dt}\bigg|_{t=10} = 10e^{-0.2 \times 10} = 10e^{-2} \approx 10(0.1353) \approx 1.353 \text{ m/s}^2$

So, the acceleration ten seconds after she jumps is approximately ( 1.4 \text{ m/s}^2 ).

Step 6

(ii) Find the distance that she has fallen in the first ten seconds. Give your answer correct to the nearest metre.

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Answer

The distance fallen can be calculated by integrating the velocity:

Find the position function, ( s(t) ): $s(t) = \int v(t) dt = \int 50(1 - e^{-0.2t}) dt$ This results in: $s(t) = 50t + 250e^{-0.2t} + C$
As she starts falling from ( s(0) = 0 ), we can find ( C ): At ( t = 0 ), ( s(0) = 0 \Rightarrow 0 + 250 + C = 0 \Rightarrow C = -250 )
The position function thus becomes: $s(t) = 50t + 250e^{-0.2t} - 250$
Evaluating at ( t = 10 ): $s(10) = 50(10) + 250e^{-2} - 250 = 500 + 250(0.1353) - 250 \approx 500 + 33.825 - 250 \approx 283.825$

Thus, rounding to the nearest metre, the distance fallen in the first ten seconds is approximately ( 284 , ext{m} ).

By using the substitution $t = \tan\left( \frac{\theta}{2} \right)$, show that \( \frac{1 - \cos\theta}{\sin\theta} = \tan\left( \frac{\theta}{2} \right) \) - HSC - SSCE Mathematics Extension 1 - Question 2 - 2007 - Paper 1

Worked Solution & Example Answer:By using the substitution $t = \tan\left( \frac{\theta}{2} \right)$, show that \( \frac{1 - \cos\theta}{\sin\theta} = \tan\left( \frac{\theta}{2} \right) \) - HSC - SSCE Mathematics Extension 1 - Question 2 - 2007 - Paper 1

By using the substitution $t = \tan\left( \frac{\theta}{2} \right)$, show that \( \frac{1 - \cos\theta}{\sin\theta} = \tan\left( \frac{\theta}{2} \right) \).

Let \( f(x) = 2\cos^{-1}(x) \). (i) Sketch the graph of \( y = f(x) \), indicating clearly the coordinates of the endpoints of the graph.

(ii) State the range of \( f(x) \).

The polynomial \( P(x) = x^2 + ax + b \) has a zero at \( x = 2 \). When \( P(x) \) is divided by \( x + 1 \), the remainder is 18. Find the values of \( a \) and \( b \).

(ii) Find the distance that she has fallen in the first ten seconds. Give your answer correct to the nearest metre.

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