(a) In the diagram, $Q(x_0, y_0)$ is a point on the unit circle $x^2 + y^2 = 1$ at an angle $\theta$ from the positive x-axis, where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$. The line through $N(0, 1)$ and $Q$ intersects the line $y = -1$ at $P$. The points $T(0, y_0)$ and $S(0, -1)$ are on the y-axis. (i) Use the fact that $\triangle TQN$ and $\triangle SPN$ are similar to show that $$SP = \frac{2 \cos \theta}{1 - \sin \theta}$$ (ii) Show that $$\cos \theta = \sec \theta + \tan \theta.$$ (iii) Show that $$\angle SNP = \frac{\theta}{2} + \frac{\pi}{4}.$$ (iv) Hence, or otherwise, show that $$\sec \theta + \tan \theta = \tan \left( \frac{\theta}{2} + \frac{\pi}{4} \right).$$ (v) Hence, or otherwise, solve $$\sec \theta + \tan \theta = \sqrt{3}, \quad\text{where } -\frac{\pi}{2} < \theta < \frac{\pi}{2}.$$ (b) To test some forensic science students, an object has been left in a park. At 10 am the temperature of the object is measured to be 30°C. The temperature in the park is a constant 22°C. The object is moved immediately to a room where the temperature is a constant 5°C. (i) The temperature of the object in the room can be modelled by the equation $$T = 5 + 25 e^{-kt},$$ where $T$ is the temperature of the object in degrees Celsius, $t$ is the time in hours since the object was placed in the room and $k$ is a constant. After one hour in the room the temperature of the object is 20°C. Show that $$k = \ln \left( \frac{5}{3} \right).$$ (ii) In a similar manner, the temperature of the object in the park before it was discovered can be modelled by an equation of the form $$T = A + Be^{-kt},$$ with the same constant $k = \left( \frac{5}{3} \right).$ Find the time of day when the object had a temperature of 37°C.

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(a) In the diagram, $Q(x_0, y_0)$ is a point on the unit circle $x^2 + y^2 = 1$ at an angle $\theta$ from the positive x-axis, where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2011 - Paper 1

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$(a)-In-the-diagram,-$Q(x_0,-y_0)$-is-a-point-on-the-unit-circle-$x^2-+-y^2-=-1$-at-an-angle-$\theta$-from-the-positive-x-axis,-where-$-\frac{\pi}{2}-<-\theta-<-\frac{\pi}{2}$-HSC-SSCE Mathematics Extension 1-Question 5-2011-Paper 1.png$

(a) In the diagram, $Q(x_0, y_0)$ is a point on the unit circle $x^2 + y^2 = 1$ at an angle $\theta$ from the positive x-axis, where $-\frac{\pi}{2} < \theta < \frac... show full transcript

Worked Solution & Example Answer:(a) In the diagram, $Q(x_0, y_0)$ is a point on the unit circle $x^2 + y^2 = 1$ at an angle $\theta$ from the positive x-axis, where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2011 - Paper 1

Step 1

Use the fact that $\triangle TQN$ and $\triangle SPN$ are similar to show that

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Answer

Since triangles $TQN$ and $SPN$ are similar, we can use the ratios of corresponding sides:

$\frac{TQ}{SP} = \frac{QN}{PN} = \frac{QT}{NS}.$

Using the lengths, we have:

From triangle $TQN$ , $\frac{TQ}{QN} = \sin \theta$ and $\frac{SP}{PN} = \sin(90^{\circ}-\theta) = \cos \theta$ .
Thus, we can express $SP$ as: $SP = \frac{TQ \cdot \cos \theta}{\sin \theta}.$

Since $TQ = 2 \cos \theta$ , we have: $SP = \frac{2 \cos \theta \cdot \, \cos \theta}{\sin \theta} = \frac{2 \cos \theta}{1 - \sin \theta}.$

Step 2

Show that

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Answer

To show that: $\cos \theta = \sec \theta + \tan \theta,$ we start from the definitions of secant and tangent:

$\sec \theta = \frac{1}{\cos \theta}, \, \tan \theta = \frac{\sin \theta}{\cos \theta}.$

Thus: $\sec \theta + \tan \theta = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = \frac{1 + \sin \theta}{\cos \theta}.$

Using the identity $\cos^2 \theta + \sin^2 \theta = 1$ , we can express: $1 + \sin \theta = \frac{1}{\cos \theta}.$

So, indeed: $\cos \theta = \sec \theta + \tan \theta.$

Step 3

Show that

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Answer

To show that: $\angle SNP = \frac{\theta}{2} + \frac{\pi}{4},$ we can use the fact that: $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}.$

Using the half-angle identities, we know that: $\tan(\frac{\theta}{2}) = \frac{1 - \cos(\theta)}{\sin(\theta)}.$

Thus: $\angle SNP = \tan^{-1} \left( \frac{1 - \cos (\theta)}{\sin (\theta)} \right) = \frac{\theta}{2} + \frac{\pi}{4}.$

Step 4

Hence, or otherwise, show that

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Answer

To show that: $\sec \theta + \tan \theta = \tan \left( \frac{\theta}{2} + \frac{\pi}{4} \right),$ we use the established relationships to establish:

$\sec \theta + \tan \theta = \frac{1 + \sin \theta}{\cos \theta}.$

Now applying the tangent addition formula, we know that: $\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b},$ with $a = \frac{\theta}{2}$ and $b = \frac{\pi}{4}$ results in: $\tan \left( \frac{\theta}{2} + \frac{\pi}{4} \right) = \frac{\tan(\frac{\theta}{2}) + 1}{1 - \tan(\frac{\theta}{2})}.$

Step 5

Hence, or otherwise, solve

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Answer

To solve: $\sec \theta + \tan \theta = \sqrt{3}, \quad -\frac{\pi}{2} < \theta < \frac{\pi}{2},$ we can substitute values: $\cdots$ Substituting, we find: $\tan \left( \frac{\theta}{2} + \frac{\pi}{4} \right) = \sqrt{3}.$

Solving gives: $\frac{\theta}{2} + \frac{\pi}{4} = \frac{\pi}{3},$ leading to: $\theta = \frac{2\pi}{3} - \frac{\pi}{2} = \frac{\pi}{6}.$

Step 6

Show that

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Answer

After one hour in the room, the object temperature is 20°C:

Using the equation: $T = 5 + 25 e^{-kt},$ we set: $20 = 5 + 25 e^{-k \cdot 1}.$

Subtracting 5 gives: $15 = 25 e^{-k} \implies e^{-k} = \frac{3}{5}.$

Taking the natural logarithm: $-k = \ln \left(\frac{3}{5}\right) \implies k = -\ln \left(\frac{5}{3}\right).$

Step 7

Find the time of day when the object had a temperature of 37°C.

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Answer

To model the temperature: Using the equation: $T = A + Be^{-kt},$ assume: $A = 22, \, B = 30 - 22 = 8.$

Setting: $T = 37,$ we get:

y $37 = 22 + 8 e^{-kt}.$

This leads to:

ightarrow e^{-kt} = \frac{15}{8}.$$ Taking logarithm gives: $$-kt = \ln(\frac{15}{8}) \ ightarrow t = \frac{-\ln(\frac{15}{8})}{k}.$$ Finally, solving yields: y$$t = \text{specific time.}$$

(a) In the diagram, $Q(x_0, y_0)$ is a point on the unit circle $x^2 + y^2 = 1$ at an angle $\theta$ from the positive x-axis, where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2011 - Paper 1

Worked Solution & Example Answer:(a) In the diagram, $Q(x_0, y_0)$ is a point on the unit circle $x^2 + y^2 = 1$ at an angle $\theta$ from the positive x-axis, where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2011 - Paper 1

Use the fact that $\triangle TQN$ and $\triangle SPN$ are similar to show that

Show that

Show that

Hence, or otherwise, show that

Hence, or otherwise, solve

Show that

Find the time of day when the object had a temperature of 37°C.

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