Find $ \int \cos^2(3x) \, dx $. A ferris wheel has a radius of 20 metres and is rotating at a rate of 1.5 radians per minute. The top of a carriage is $ h $ metres above the horizontal diameter of the ferris wheel. The angle of elevation of the top of the carriage from the centre of the ferris wheel is $ \theta $. (i) Show that $ \frac{dh}{d\theta} = 20 \cos \theta. $ (ii) At what speed is the top of the carriage rising when it is 15 metres higher than the horizontal diameter of the ferris wheel? Give your answer correct to one decimal place. Let $ f(x) = \sin^{-1} x + \cos^{-1} x. $ (i) Show that $ f'(x) = 0. $ (ii) Hence, or otherwise, prove $ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}. $ (iii) Hence, sketch $ f(x) = \sin^{-1} x + \cos^{-1} x. $

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Find $ \int \cos^2(3x) \, dx $ - HSC - SSCE Mathematics Extension 1 - Question 12 - 2018 - Paper 1

Question 12

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Find $ \int \cos^2(3x) \, dx $. A ferris wheel has a radius of 20 metres and is rotating at a rate of 1.5 radians per minute. The top of a carriage is $ h $ met... show full transcript

Worked Solution & Example Answer:Find $ \int \cos^2(3x) \, dx $ - HSC - SSCE Mathematics Extension 1 - Question 12 - 2018 - Paper 1

Step 1

Find $ \int \cos^2(3x) \, dx $

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Answer

To find this integral, we can use the identity for cosine squared:

$\cos^2(x) = \frac{1 + \cos(2x)}{2}$

Thus, we have:

$\int \cos^2(3x) \, dx = \int \frac{1 + \cos(6x)}{2} \, dx$

This splits our integral into two parts:

$= \frac{1}{2} \int 1 \, dx + \frac{1}{2} \int \cos(6x) \, dx$

Now, integrating:

( \int 1 , dx = x )
( \int \cos(6x) , dx = \frac{1}{6} \sin(6x) )

Combining these results gives:

$= \frac{1}{2} x + \frac{1}{12} \sin(6x) + C$

where ( C ) is the constant of integration.

Step 2

Show that $ \frac{dh}{d\theta} = 20 \cos \theta $

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Answer

Given the geometry of the situation, we can establish that the height ( h ) of the top of the carriage can be expressed as:

$h = 20 \sin(\theta).$

To find ( \frac{dh}{d\theta} ), we differentiate with respect to ( \theta ):

$\frac{dh}{d\theta} = 20 \cos(\theta).$

Thus, we have shown the required result.

Step 3

Let $ f(x) = \sin^{-1} x + \cos^{-1} x. $ Show that $ f'(x) = 0. $

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Answer

To find the derivative of ( f(x) ), we use the derivatives of the inverse trigonometric functions:

$f'(x) = \frac{d}{dx}(\sin^{-1} x) + \frac{d}{dx}(\cos^{-1} x)$

We know:

( \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}} )
( \frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1 - x^2}} )

Thus:

$f'(x) = \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - x^2}} = 0.$

Hence, we have shown that ( f'(x) = 0 ).

Step 4

Hence, prove $ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}. $

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Answer

Since we established that the derivative of ( f(x) = \sin^{-1} x + \cos^{-1} x ) is zero, it indicates that ( f(x) ) is constant across its domain.

To find this constant, evaluate for ( x = 0 ):

$f(0) = \sin^{-1}(0) + \cos^{-1}(0) = 0 + \frac{\pi}{2} = \frac{\pi}{2}.$

Consequently, since ( f(x) ) is constant and equal to ( \frac{\pi}{2} ), this confirms:

$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for all ( x ) in the valid range.

Step 5

Sketch $ f(x) = \sin^{-1} x + \cos^{-1} x. $

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Answer

The function ( f(x) = \sin^{-1} x + \cos^{-1} x ) is constant and equal to ( \frac{\pi}{2} ) on the interval ([-1, 1]).

As such, the graph of this function is a horizontal line at the level of ( \frac{\pi}{2} ).

The endpoints of the domain are (1, ( \frac{\pi}{2} )) and (-1, ( \frac{\pi}{2} )), indicating a constant value for all ( x \in [-1, 1] ).

Find \( \int \cos^2(3x) \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 12 - 2018 - Paper 1

Worked Solution & Example Answer:Find \( \int \cos^2(3x) \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 12 - 2018 - Paper 1

Find \( \int \cos^2(3x) \, dx \)

Show that \( \frac{dh}{d\theta} = 20 \cos \theta \)

Let \( f(x) = \sin^{-1} x + \cos^{-1} x. \) Show that \( f'(x) = 0. \)

Hence, prove \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}. \)

Sketch \( f(x) = \sin^{-1} x + \cos^{-1} x. \)

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