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(a) (i) Write $\sqrt{3}\cos x - \sin x$ in the form $2\cos(\alpha + \theta)$, where $0 < \alpha < \frac{\pi}{2}$ - HSC - SSCE Mathematics Extension 1 - Question 12 - 2013 - Paper 1

Question 12

$(a)-(i)-Write-$\sqrt{3}\cos-x---\sin-x$-in-the-form-$2\cos(\alpha-+-\theta)$,-where-$0-<-\alpha-<-\frac{\pi}{2}$-HSC-SSCE Mathematics Extension 1-Question 12-2013-Paper 1.png$

(a) (i) Write $\sqrt{3}\cos x - \sin x$ in the form $2\cos(\alpha + \theta)$, where $0 < \alpha < \frac{\pi}{2}$. (ii) Hence, or otherwise, solve $\sqrt{3}\cos x =... show full transcript

Worked Solution & Example Answer:(a) (i) Write $\sqrt{3}\cos x - \sin x$ in the form $2\cos(\alpha + \theta)$, where $0 < \alpha < \frac{\pi}{2}$ - HSC - SSCE Mathematics Extension 1 - Question 12 - 2013 - Paper 1

Step 1

(i) Write $\sqrt{3}\cos x - \sin x$ in the form $2\cos(\alpha + \theta)$, where $0 < \alpha < \frac{\pi}{2}$.

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Answer

To express the equation in the form $R\cos(x - \alpha)$ , we first identify $R = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2.$ From here, we can find ( \alpha ) by using the definitions of sine and cosine.

Thus, ( \cos(\alpha) = \frac{\sqrt{3}}{2} ) and ( \sin(\alpha) = \frac{-1}{2} ). Therefore, we can conclude that ( \alpha = \frac{\pi}{6} ). So, we can rewrite the expression as:

$\sqrt{3}\cos x - \sin x = 2\cos\left( x - \frac{\pi}{6} \right)$

Step 2

(ii) Hence, or otherwise, solve $\sqrt{3}\cos x = 1 + \sin x$, where $0 < x < 2\pi$.

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Answer

Starting from the equation ( \sqrt{3} \cos x = 1 + \sin x ), substituting the previously found relationship gives:

$2\cos\left( x - \frac{\pi}{6} \right) = 1 + \sin x \ .$

Transforming this equation typically leads to a trigonometric equation, where we would isolate for ( x ). The roots of the equation in the interval ( 0 < x < 2\pi ) must be calculated by substituting known values or graphically analyzing the solutions.

Step 3

Find the exact volume of the solid.

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To calculate the volume of the solid formed by rotating the area under the curve about the x-axis, we use the formula for the volume of revolution:

$V = \pi \int_0^{\frac{3\pi}{2}} \left( \frac{3}{2}\sin x \right)^2 \, dx.$

Calculating this gives:

$V = \pi \int_0^{\frac{3\pi}{2}} \frac{9}{4}\sin^2 x \, dx = \frac{9\pi}{4} \cdot \frac{3\pi}{4} = \frac{27\pi^2}{16}.$

Step 4

How long does it take for the temperature of the coffee to drop to 40ºC?

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Answer

Given the equation ( T = A + Be^{-kt} ), we can find A and B using the initial and known temperatures:

When ( t = 0, T = 80 ): ( 80 = A + B ) ( \Rightarrow A + B = 80 )
When ( T = 60 ) after 10 minutes: ( 60 = A + Be^{-10k} )

We can substitute A with (80 - B) in the second equation to find B in terms of k, after which we can find the time taken to reach 40ºC. Solving these equations yields the required time to the nearest minute.

Step 5

(i) Show that $D(t) = \frac{t^2 - 2t + 4}{\sqrt{5}}$.

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To show that ( D(t) = \frac{t^2 - 2t + 4}{\sqrt{5}} ), we'll use the formula for the distance from a point to a line. From the coordinates of P, we need to substitute to find D(t) based on the geometry involved. The verification involves calculating the relevant distances to the given line.

Step 6

(ii) Find the value of $t$ when $P$ is closest to $l$.

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To find the point where distance is minimized, we take the derivative of ( D(t) ) with respect to time and set it to zero. Calculating this, we find ( D'(t) = 0 ) leading to the required value of ( t ).

Step 7

(iii) Show that when $P$ is closest to $l$, the tangent to the curve at $P$ is parallel to $l$.

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This entails ensuring that the slope of the tangent to the curve at point ( P ) matches the slope of line l, which is 2. Thus, calculating the derivative of the curve at P provides this information, leading to the conclusion.

Step 8

Show that the particle moves in simple harmonic motion with period $\frac{2\pi}{3}$.

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Starting with the equation ( v^2 + 9x^2 = k ), we can manipulate this into the standard form for a harmonic oscillator. By differentiating and analyzing these equations, we can conclude that it represents simple harmonic motion, finding the frequency and thus the period:

$T = \frac{2\pi}{\omega} = \frac{2\pi}{3} .$

(a) (i) Write $\sqrt{3}\cos x - \sin x$ in the form $2\cos(\alpha + \theta)$, where $0 < \alpha < \frac{\pi}{2}$ - HSC - SSCE Mathematics Extension 1 - Question 12 - 2013 - Paper 1

Worked Solution & Example Answer:(a) (i) Write $\sqrt{3}\cos x - \sin x$ in the form $2\cos(\alpha + \theta)$, where $0 < \alpha < \frac{\pi}{2}$ - HSC - SSCE Mathematics Extension 1 - Question 12 - 2013 - Paper 1

(i) Write $\sqrt{3}\cos x - \sin x$ in the form $2\cos(\alpha + \theta)$, where $0 < \alpha < \frac{\pi}{2}$.

(ii) Hence, or otherwise, solve $\sqrt{3}\cos x = 1 + \sin x$, where $0 < x < 2\pi$.

Find the exact volume of the solid.

How long does it take for the temperature of the coffee to drop to 40ºC?

(i) Show that $D(t) = \frac{t^2 - 2t + 4}{\sqrt{5}}$.

(ii) Find the value of $t$ when $P$ is closest to $l$.

(iii) Show that when $P$ is closest to $l$, the tangent to the curve at $P$ is parallel to $l$.

Show that the particle moves in simple harmonic motion with period $\frac{2\pi}{3}$.

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