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

Question

By using the substitution $t = 	an \frac{	heta}{2}$, we can express the trigonometric identities in terms of $t$.

Starting with the identity for cosine:
$$\cos 	heta = \frac{1 - t^2}{1 + t^2}$$

We have:
$$1 - \cos 	heta = 1 - \frac{1 - t^2}{1 + t^2} = \frac{(1 + t^2) - (1 - t^2)}{1 + t^2} = \frac{2t^2}{1 + t^2}$$

Now for sine:
$$\sin 	heta = \frac{2t}{1 + t^2}$$

Thus,
$$\frac{1 - \cos 	heta}{\sin 	heta} = \frac{\frac{2t^2}{1 + t^2}}{\frac{2t}{1 + t^2}} = \frac{t}{1} = t$$

Hence, we have shown that:
$$\frac{1 - \cos 	heta}{\sin 	heta} = t$$

Let $f(x) = 2 \cos^{-1} x$.

(i) To sketch the graph of $y = f(x)$, we note that $f(x)$ is defined for $-1 \leq x \leq 1$. The endpoints of the graph occur at:
- When $x = -1$, $f(-1) = 2\pi$.
- When $x = 1$, $f(1) = 0$.

Thus, the coordinates are $(-1, 2\pi)$ and $(1, 0)$.

(ii) The range of $f(x)$ is $[0, 2\pi]$.

The polynomial $P(x) = x^2 + ax + b$ has a zero at $x = 2$. Thus, $P(2) = 0$:
$$P(2) = 2^2 + 2a + b = 0 \Rightarrow 4 + 2a + b = 0 \Rightarrow 2a + b = -4$$

When $P(x)$ is divided by $x + 1$, the remainder is 18:
$$P(-1) = (-1)^2 + a(-1) + b = 18 \Rightarrow 1 - a + b = 18 \Rightarrow -a + b = 17$$

Now we have a system of equations:
1. $2a + b = -4$
2. $-a + b = 17$

By solving these equations, we find:
Subtracting the first equation from the second:
$$(-a + b) - (2a + b) = 17 - (-4) \Rightarrow -3a = 21 \Rightarrow a = -7$$

Substituting $a$ back into the first equation:
$$2(-7) + b = -4 \Rightarrow -14 + b = -4 \Rightarrow b = 10$$

Thus, the values are $a = -7$ and $b = 10$.

For the skydive scenario:
The velocity function is given by $v = 50(1 - e^{-2t})$.

(i) To find the acceleration, we differentiate the velocity function:
$$a = \frac{dv}{dt} = 50(0 - e^{-2t})(-2) = 100e^{-2t}$$

At $t = 10$ seconds:
$$a(10) = 100e^{-20} \approx 0.0000454 	ext{ m/s}^2$$

(ii) To find the distance fallen, we integrate the velocity:
$$s = \int v dt = \int 50(1 - e^{-2t}) dt = 50(t + 0.5e^{-2t}) + C$$

As she starts from 2000 m:
$$s(0) = 2000 \Rightarrow C = 2000$$

Integrating from 0 to 10 seconds:
$$s = 50(10 + 0.5e^{-20}) + 2000 \approx 2284 	ext{ m}$$

By using the substitution $t = \tan \frac{\theta}{2}$, we can express the trigonometric identities in terms of $t$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2007 - Paper 1

Worked Solution & Example Answer:By using the substitution $t = \tan \frac{\theta}{2}$, we can express the trigonometric identities in terms of $t$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2007 - Paper 1

Show that $\frac{1 - \cos \theta}{\sin \theta} = t$

Sketch the graph of $y = f(x)$, indicating the endpoints

State the range of $f(x)$

Find the values of a and b

Find the acceleration ten seconds after she jumps

Find the distance fallen in the first ten seconds

Join the SSCE students using SimpleStudy...