Detailed Answer
a)
b)
\[ f(x) = 2\cos{\left(x + \frac{\pi}{2}\right)} + 1 = 0 \]
\[ 0 = -2\sin{(x)} + 1 \]
\[ \sin{(x)} = \frac{1}{2} \]
\[ x = \frac{\pi}{6} + 2k\pi\]
\[ x = \frac{5\pi}{6} + 2k\pi \]
Close
a)
\[ \text{Area} = \frac{1}{2}r^2\theta = \frac{1}{2}(10)^2(1.5) = 75\;cm^2 \]
b)
\[ \text{Area} = \frac{1}{2}r^2\sin{\theta} = \frac{1}{2}(10)^2\sin{1.5} = 50\;cm^2 \]
c)
\[ \text{Area} = 75 - 50 = 25\;cm^2 \]
d) We calculate the length of the arc and the line AB:
\[ \text{Arc} = 10 \cdot 2 = 20 \]
\[ AB = \sqrt{10^2 + 10^2 - 2 \cdot 10 \cdot 10 \cos{1.5}} = 13.63 \quad \textit{(Cosine rule)} \]
\[ \text{Total length} = 33.63 \]
a) (i) \( \pi \) is how long it takes to complete 1 revolution.
a) (ii) \( \frac{Max - Min}{2} = 4 \)
a) (iii) \( y = 6 \)
b) Therefore, the function can be expressed as:
\[ f(x) = 4\sin\left(2x\right) + 6 \]
Having no solution here means that the 2 graphs of \( f(x) \) do not intersect. We know that cosine can only take values between 1 and -1. However here the amplitude of it is 2, and it is shifted 1 unit up. Thus it can take \( y \) values between 1 and 3. So \( z \) can be any number outside of this range, and will not intersect our trigonometric graph.
a) \( 1 + \tan{x}^2 = 1 + \frac{\sin{x}^2}{\cos{x}^2} = \frac{\cos{x}^2 + \sin{x}^2}{\cos{x}^2} = \frac{1}{\cos{x}^2} \)
b) \( \frac{1 + 2\sin{(x)\cos{(x)}}}{\sin{(x)} + \cos{(x)}} = \frac{\cos{(x)}^2 + \sin{(x)}^2 + 2\sin{(x)\cos{(x)}}}{\sin{(x)} + \cos{(x)}} = \frac{(\sin{(x)} + \cos{(x)})^2}{\sin{(x)} + \cos{(x)}} = \sin{(x)} + \cos{(x)} \)
c) \[ \frac{\cos{(x)}}{1 - \sin{(x)}} - \tan(x) = \frac{\cos{(x)}}{1 - \sin{(x)}} - \frac{\sin{(x)}}{\cos{(x)}} \]
\[ = \frac{\cos{(x)}^2 - \sin{(x)}(1 - \sin{(x)})}{(1 - \sin{(x)})\cos{(x)}} = \frac{\cos{(x)}^2 + \sin{(x)}^2 - \sin{x}}{(1 - \sin{(x)})\cos{(x)}} \]
\[ = \frac{1 - \sin{x}}{(1 - \sin{(x)})\cos{(x)}} = \frac{1}{\cos(x)} \]
\[ 0 = \frac{\sin{(x)}}{\cos{(x)}} - 2\sin{(x)}\cos{(x)} \]
\[ 0 = \sin{(x)} - 2\sin{(x)}\cos{(x)}^2 \]
\[ 0 = \sin{(x)}(1 - 2\cos{(x)}^2)\]
\[ \sin{(x)} = 0 \quad \Rightarrow \quad x = \pi, 2\pi \]
\[ \cos{(x)} = \pm \frac{\sqrt{2}}{2} \quad \Rightarrow \quad x = \frac{5\pi}{4}, \frac{7\pi}{4} \]
a)
b)
\[ f(x) = 2\cos{\left(x + \frac{\pi}{2}\right)} + 1 = 0 \]
\[ 0 = -2\sin{(x)} + 1 \]
\[ \sin{(x)} = \frac{1}{2} \]
\[ x = \frac{\pi}{6} + 2k\pi\]
\[ x = \frac{5\pi}{6} + 2k\pi \]
We need to consider in which quadrants \(\theta\) can be. Since the tangent of it is negative, we know it must either be in the second or fourth quadrant. First, let's consider it is in the second quadrant.
This means that the side opposite the angle is 12, and the side adjacent is 5, but since this is also the \(x\) coordinate, it is negative. We can also calculate the hypotenuse of this triangle, which is 13. From this we can see that \(\sin{\theta} = \frac{12}{13}\) and \(\cos{\theta} = -\frac{5}{13}\).
We can use a similar approach for the fourth quadrant, only here the \(y\)-axis is negative, therefore the opposite side will be -12. Like this \(\sin{\theta} = -\frac{12}{13}\) and \(\cos{\theta} = \frac{5}{13}\).
a) The water height at the end of the day is:
\[ W(24) = 3\sin\left(\frac{\pi}{4} \cdot 24 + \frac{\pi}{2}\right) + 20 = 23\;\text{meters} \]
b) \( \frac{2\pi}{\frac{\pi}{4}} = 8\;\text{hours} \)
c) The sine function can only return \(\pm 1\) as a maximum or minimum. Hence, \(W(t)\) can be \(3(\pm 1) + 20\). Thus, the maximum water level is 23 meters when \(\sin\) is \(+1\), and the minimum is 17 meters when \(\sin\) is \(-1\).
d)
\[ W(t) = 3\sin\left(\frac{\pi}{4}t + \frac{\pi}{2}\right) + 20 = 17 \]
\[ t = 4\;\text{hours}, 12\;\text{hours}, 20\;\text{hours} \; \text{[GDC]} \]
e) The graph of \(W(t)\) is shown below:
a) (i) 52 weeks, as it repeats on yearly basis.
a) (ii) \( c = \frac{2\pi}{52} \)
b)
\[ H(5) = a \cos\left(\frac{2\pi}{52} \cdot 5\right) + b = 15.3 \]
\[ H(5) = 0.82a + b = 15.3 \]
\[ H(10) = a \cos\left(\frac{2\pi}{52} \cdot 10\right) + b = 13.42 \]
\[ H(10) = 0.35a + b = 13.42 \]
Solving this system of equations gives us \( a = 4 \) and \( b = 12 \) approximately.
c)
\[ H(55) = 4 \cos\left(\frac{2\pi}{52} \cdot 55\right) + 12 = 15.74 \]
d) The graph of \(H(t)\) is shown below:
e) The graph of \(y=10\) allows us to see that the water level is below this line approximately from week 17 to week 35, which is 18 weeks.
