HSC 2013 Maths Marathon (archive) (1 Viewer)

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Re: HSC 2013 2U Marathon

Arc length= 5pi/3 = circumference of cone
Hence radius of cone= 5/6
Radius of sector= 5 = slant height of cone
Therefore height of cone, H^2= 5^2 - 25/36
h= root(875/36)
Volume cone= 1/3pi R^2 h
= 1/3 pi 25/36 x root(875/36) ?
The answer is

125 root(35) pi / 648 cm^3
 

Menomaths

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Re: HSC 2013 2U Marathon

Prove that:

 
Last edited:

andybandy

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Re: HSC 2013 2U Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{sin\theta&space;cos^{2}\theta&space;-sin^{3}\theta&space;}{2cos^{3}\theta&space;-cos\theta&space;}&space;=&space;\frac{sin\theta&space;(cos^{2}\theta&space;-sin^{2}\theta)&space;}{cos\theta&space;(2cos^{2}\theta&space;-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{sin\theta&space;cos^{2}\theta&space;-sin^{3}\theta&space;}{2cos^{3}\theta&space;-cos\theta&space;}&space;=&space;\frac{sin\theta&space;(cos^{2}\theta&space;-sin^{2}\theta)&space;}{cos\theta&space;(2cos^{2}\theta&space;-1)}" title="\frac{sin\theta cos^{2}\theta -sin^{3}\theta }{2cos^{3}\theta -cos\theta } = \frac{sin\theta (cos^{2}\theta -sin^{2}\theta) }{cos\theta (2cos^{2}\theta -1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta&space;}{cos\theta&space;}.\frac{cos2\theta&space;}{cos2\theta&space;}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta&space;}{cos\theta&space;}.\frac{cos2\theta&space;}{cos2\theta&space;}" title="= \frac{sin\theta }{cos\theta }.\frac{cos2\theta }{cos2\theta }" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;tan\theta&space;.1&space;=&space;tan\theta" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;tan\theta&space;.1&space;=&space;tan\theta" title="= tan\theta .1 = tan\theta" /></a>

Edit: Realised this is the 3 unit method, for the 2 unit method, you would change


<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta(cos^{2}\theta&space;-&space;(1&space;-&space;cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta(cos^{2}\theta&space;-&space;(1&space;-&space;cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" title="= \frac{sin\theta(cos^{2}\theta - (1 - cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta(2cos^{2}\theta&space;-&space;1)}{cos\theta(2cos^{2}\theta-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta(2cos^{2}\theta&space;-&space;1)}{cos\theta(2cos^{2}\theta-1)}" title="= \frac{sin\theta(2cos^{2}\theta - 1)}{cos\theta(2cos^{2}\theta-1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta}{cos\theta}&space;=&space;tan\theta" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta}{cos\theta}&space;=&space;tan\theta" title="= \frac{sin\theta}{cos\theta} = tan\theta" /></a>
 
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HeroicPandas

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Re: HSC 2013 2U Marathon

^U used 3U double angles

A hint is to never write a '1' as a '1'
 

Menomaths

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Re: HSC 2013 2U Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{sin\theta&space;cos^{2}\theta&space;-sin^{3}\theta&space;}{2cos^{3}\theta&space;-cos\theta&space;}&space;=&space;\frac{sin\theta&space;(cos^{2}\theta&space;-sin^{2}\theta)&space;}{cos\theta&space;(2cos^{2}\theta&space;-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{sin\theta&space;cos^{2}\theta&space;-sin^{3}\theta&space;}{2cos^{3}\theta&space;-cos\theta&space;}&space;=&space;\frac{sin\theta&space;(cos^{2}\theta&space;-sin^{2}\theta)&space;}{cos\theta&space;(2cos^{2}\theta&space;-1)}" title="\frac{sin\theta cos^{2}\theta -sin^{3}\theta }{2cos^{3}\theta -cos\theta } = \frac{sin\theta (cos^{2}\theta -sin^{2}\theta) }{cos\theta (2cos^{2}\theta -1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta&space;}{cos\theta&space;}.\frac{cos2\theta&space;}{cos2\theta&space;}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta&space;}{cos\theta&space;}.\frac{cos2\theta&space;}{cos2\theta&space;}" title="= \frac{sin\theta }{cos\theta }.\frac{cos2\theta }{cos2\theta }" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;tan\theta&space;.1&space;=&space;tan\theta" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;tan\theta&space;.1&space;=&space;tan\theta" title="= tan\theta .1 = tan\theta" /></a>

Edit: Realised this is the 3 unit method, for the 2 unit method, you would change


<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta(cos^{2}\theta&space;-&space;(1&space;-&space;cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta(cos^{2}\theta&space;-&space;(1&space;-&space;cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" title="= \frac{sin\theta(cos^{2}\theta - (1 - cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta(2cos^{2}\theta&space;-&space;1)}{cos\theta(2cos^{2}\theta-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta(2cos^{2}\theta&space;-&space;1)}{cos\theta(2cos^{2}\theta-1)}" title="= \frac{sin\theta(2cos^{2}\theta - 1)}{cos\theta(2cos^{2}\theta-1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta}{cos\theta}&space;=&space;tan\theta" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta}{cos\theta}&space;=&space;tan\theta" title="= \frac{sin\theta}{cos\theta} = tan\theta" /></a>
It's so much easier with the 3U method, wish we could use it



For this question why can't we use
 
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