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HSC 2015 MX2 Marathon (archive) (5 Viewers)

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Kaido

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Re: HSC 2015 4U Marathon

hence show that cos 9 and sin 9 satisfy the equation
16x^5-20x^3+5x-1/sqrt(2) = 0
angles are in degrees please convert to radians
no brackets, hard to tell which polynomial is where
 

porcupinetree

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Re: HSC 2015 4U Marathon

good try but i'm afraid it is invalid.
When you wrote this line:
|z|+|z-2i|=|z-4+2i|+|z-2i|
Did you deduce that |z| = |z-4+2i| because the distance from (4, -2) to the line y = 2x - 5 is equal to the distance from (0, 0) to the line? Thanks.
 

FrankXie

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Re: HSC 2015 4U Marathon

When you wrote this line:
|z|+|z-2i|=|z-4+2i|+|z-2i|
Did you deduce that |z| = |z-4+2i| because the distance from (4, -2) to the line y = 2x - 5 is equal to the distance from (0, 0) to the line? Thanks.
because that is given
 

Kaido

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Re: HSC 2015 4U Marathon

9 = pi/20
we want to express in terms of 5x -> so becomes pi/4

16x^5-20x^3+5x-1/sqrt(2)
16x^5-20x^3+5x=1/sqrt(2)
LHS=cos5x/sin5x (same thing)=pi/4
=1/sqrt(2)
=RHS

(since sin(pi/4)=cos(pi/4) at pi/4)
 
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t_davis9

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Re: HSC 2015 4U Marathon

NEXT QUESTION
Sorry, can't tell if the old one was finished.
http://imgur.com/k3Xgd2N
This was taken from the final page of my complex numbers test
 
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Re: HSC 2015 4U Marathon

NEXT QUESTION

this has been answered already, but i have a quick question
why can't you just find the perp distances of (0,0) and (0,2) to the locus determined using the given info, and then add them together??
thanks :D
 

InteGrand

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Re: HSC 2015 4U Marathon

NEXT QUESTION
Sorry, can't tell if the old one was finished.
http://imgur.com/k3Xgd2N
This was taken from the final page of my complex numbers test


It suffices to show all the results assuming , due to the periodicity and powers of the functions in question. So assume .

(i)





(since sin2 t + cos2 t = 1 ∀t ∈ ℝ)

.

If cos θ ≠ -1, ℝe(1+z) = 1+cos θ > 1+(-1) = 0, so that





, using the identity for cos θ ≠ -1 (this can be proved using t-formulae or half-angle identities)

, since , because we assumed that , and .

So for cos θ ≠ -1,

If cos θ = -1 (that is, if ), then 1+z = (1-1)+i.sin(±π) = 0, and hence arg(1+z) is undefined. However, the above formula still holds when 1+z = 0 ⇔ cos θ = -1.

So we have .

(ii) (1+z)4 = 1 + 4z+6z2 + 4z3 + z4, by the binomial theorem.

But by de Moivre's theorem and part (i),





, since (from double angle identity for cosine)

.

Equating both expressions for (1+z)4,



The identity required to be shown follows from taking real parts of the above equation and recalling that by de Moivre's theorem, and that 'real part' is a linear operator, so that the real part of a sum is the sum of real parts.

The second expression is given by , which follows in a similar way, this time by taking imaginary parts.

(iii)

The first identity follows by dividing Equation (*) by the identity required to prove in part (ii) (and recalling that ), for (so that the factor of can be cancelled from the numerator and denominator).

For the second identity, we rearrange the identities proven in part (ii) to get:



and

.

The identity follows upon dividing (1) by (2), provided that .
 
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FrankXie

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Re: HSC 2015 4U Marathon

this has been answered already, but i have a quick question
why can't you just find the perp distances of (0,0) and (0,2) to the locus determined using the given info, and then add them together??
thanks :D
No the perpendicular distance is totally different from the distance between two points. The question actually asked us to find one point P from the line y=2x-5 so that the sum of AP and BP (where A(0,0) and B(0,2), AP and BP mean distances between points NOT perp distances) is minimized. And this is a classic geometry problem (maybe referred to as bend path probelm?), where in the solution the optimal point P is chosen such that the angle of incidence equals the angle of reflection.
 

FrankXie

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Re: HSC 2015 4U Marathon

Here is my try





and


Dividing equations (1) and (2),



On both sides of equation (3) multiplying by the denominator,



Therefore,
 
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Kaido

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Re: HSC 2015 4U Marathon

Interesting solve by Frank, esp the first part.
And looking at the amount of latex Integrand used, you can tell this guy's a beast ;)
 

Kaido

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Re: HSC 2015 4U Marathon

Better just to use a geometrical representation (so much easier and is probably the preferred method anyways)
 

braintic

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Re: HSC 2015 4U Marathon

Why not the negative y-axis?
There is no requirement that arguments must be principal values.

Just check z=-2i:

arg(z-2)+arg(z+2) = arg (-2-2i) + arg (2-2i) = 5pi/4 - pi/4 = pi

If the equation is satisfied by a complex number using any of its infinite number of arguments, then that point lies on the locus.
 

RealiseNothing

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Re: HSC 2015 4U Marathon

NEXT QUESTION
Sorry, can't tell if the old one was finished.
http://imgur.com/k3Xgd2N
This was taken from the final page of my complex numbers test
Eh a geometrical way to prove the first part (and honestly anytime you see , doing it geometrically is probs the way to go).

By sine rule

But and

So

And obviously so we get
 
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