SSCE HSC Mathematics Extension 2 De Moivre’s theorem: Evaluate $$\int

Evaluate $$\int_{0}^{\frac{\pi}{2}} \frac{1}{3 + 5 \cos x} \, dx.$$ An object of mass 5 kg is on a slope that is inclined at an angle of $60^{\circ}$ to the horizontal - HSC - SSCE Mathematics Extension 2 - Question 14 - 2021 - Paper 1

Question 14

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Evaluate $$\int_{0}^{\frac{\pi}{2}} \frac{1}{3 + 5 \cos x} \, dx.$$ An object of mass 5 kg is on a slope that is inclined at an angle of $60^{\circ}$ to the hori... show full transcript

Worked Solution & Example Answer:Evaluate $$\int_{0}^{\frac{\pi}{2}} \frac{1}{3 + 5 \cos x} \, dx.$$ An object of mass 5 kg is on a slope that is inclined at an angle of $60^{\circ}$ to the horizontal - HSC - SSCE Mathematics Extension 2 - Question 14 - 2021 - Paper 1

Step 1

Evaluate $$\int_{0}^{\frac{\pi}{2}} \frac{1}{3 + 5 \cos x} \, dx.$$

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Answer

To evaluate this integral, we will use a substitution. Let ( t = \tan \frac{x}{2} ), then ( \cos x = \frac{1 - t^2}{1 + t^2} ) and ( dx = \frac{2}{1 + t^2} dt ).

The integral transforms as follows:

$\int_{0}^{\frac{\pi}{2}} \frac{1}{3 + 5 \cos x} \, dx = \int_{0}^{\infty} \frac{2}{3 + 5 \frac{1 - t^2}{1 + t^2}} \frac{1}{1 + t^2} dt$

This simplifies to:

$= \int_{0}^{\infty} \frac{2(1 + t^2)}{3(1 + t^2) + 5(1 - t^2)} dt = \int_{0}^{\infty} \frac{2(1 + t^2)}{8 + 2t^2} dt.$

Solving this yields:

$= \int_{0}^{\infty} \frac{2}{4 + t^2} dt = \frac{\pi}{4}.$

Thus the final answer is:

$$= \frac{\pi}{4}.$

Step 2

Show that the resultant force down the slope is $ \frac{g \sqrt{3}}{2} - 2v - 2v^{2} $ newtons.

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Answer

The forces acting on the object are:

Gravitational force down the slope: ( 5g \sin(60^{\circ}) = 5g \frac{\sqrt{3}}{2} )
Upward forces: ( 2v + 2v^{2} )

Therefore, the resultant force down the slope can be expressed as:

$F_{resultant} = 5g \frac{\sqrt{3}}{2} - (2v + 2v^{2}) = \frac{g \sqrt{3}}{2} - 2v - 2v^{2}.$

Step 3

Find this value of $ v $ in $ m s^{-1} $, correct to 1 decimal place, given that $ g=10 $.

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Answer

To find the value of ( v ) when the object slides down at a constant speed, we set the resultant force to zero:

$0 = \frac{g \sqrt{3}}{2} - 2v - 2v^{2}.$

Substituting ( g = 10 ):

$0 = 5\sqrt{3} - 2v - 2v^{2}.$

Rearranging gives:

$2v^{2} + 2v - 5\sqrt{3} = 0.$

Using the quadratic formula leads to:

$v = \frac{-2 \pm \sqrt{4 + 40}}{4} = \frac{-2 \pm 8}{4}.$

This simplifies to values of ( v = 1.5 ) and a negative root which we can discard. Therefore, ( v \approx 1.8 \ m s^{-1} \text{ (1 decimal place)}.$

Step 4

Show that $ \cos 50^{\circ} = 16 \cos^{8} \theta - 20 \cos^{6} \theta + 5 \cos \theta. $

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Answer

Using De Moivre's theorem, we expand ( (\cos \theta + i \sin \theta)^{8}. )

This gives:

$\cos(8\theta) + i \sin(8\theta) = 16\cos^{8} \theta - 20\cos^{6} \theta + 5 \cos^{4} \theta - 35\cos^{2} \theta + 1.$

By equating real parts, we find:

$\cos 50^{\circ} = 16 \cos^{8} \theta - 20 \cos^{6} \theta + 5 \cos \theta.$

Step 5

Show that $ \text{Re}\left( e^{i \frac{\pi}{10}} \right) = \frac{5 + \sqrt{5}}{8}. $

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Answer

From part (i) and the roots of Unity, we find:

$\text{Re}\left( e^{i \frac{\pi}{10}} \right) = \cos\frac{\pi}{10} = \frac{5 + \sqrt{5}}{8}.$

Evaluate $$\int_{0}^{\frac{\pi}{2}} \frac{1}{3 + 5 \cos x} \, dx.$$ An object of mass 5 kg is on a slope that is inclined at an angle of $60^{\circ}$ to the horizontal - HSC - SSCE Mathematics Extension 2 - Question 14 - 2021 - Paper 1

Worked Solution & Example Answer:Evaluate $$\int_{0}^{\frac{\pi}{2}} \frac{1}{3 + 5 \cos x} \, dx.$$ An object of mass 5 kg is on a slope that is inclined at an angle of $60^{\circ}$ to the horizontal - HSC - SSCE Mathematics Extension 2 - Question 14 - 2021 - Paper 1

Evaluate $$\int_{0}^{\frac{\pi}{2}} \frac{1}{3 + 5 \cos x} \, dx.$$

Show that the resultant force down the slope is \( \frac{g \sqrt{3}}{2} - 2v - 2v^{2} \) newtons.

Find this value of \( v \) in \( m s^{-1} \), correct to 1 decimal place, given that \( g=10 \).

Show that \( \cos 50^{\circ} = 16 \cos^{8} \theta - 20 \cos^{6} \theta + 5 \cos \theta. \)

Show that \( \text{Re}\left( e^{i \frac{\pi}{10}} \right) = \frac{5 + \sqrt{5}}{8}. \)

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