Preliminary mathematics marathon (1 Viewer)

ohexploitable · Jun 10, 2010

sin2x = 2t/(1+t²)

cos2x = (1-t²)(1+t²)

hscishard · Jun 10, 2010

hscishard said:

Let Angle EAD =x
Let angle EBD = y

$\frac{AD}{sin2y}=\frac{AB}{sin(180-(2y+x))}$
$\frac{BE}{sin2x}=\frac{AB}{sin(180-(2x+y))}$
$BE=\frac{AB(sin2x)}{sin(180-(2x+y))}$
$AD=\frac{AB(sin2y)}{sin(180-(2y+x))}$
AD = BE given...
$\frac{sin2y}{sin(180-(2y+x))}=\frac{sin2y}{sin(180-(2y+x))}$

ohexploitable said:
$\\\frac{\sin(2x-y)}{\sin2x}=\frac{\sin(2y+x)}{\sin2y} \\\\\text{suppose }y=x, \\\\\frac{\sin y}{\sin 2y}=\frac{\sin3y}{\sin2y} \\\\\sin y=\sin3y \\\\\text{sine is only equal in 1st/2nd and 3rd/4th,} \\\\y=180^{\circ}-3y \\y=45^{\circ} \\\\180^{\circ}+y=360^{\circ}-3y \\y=45^{\circ} \\\\\therefore y\neq x\text{ for all values, only for }x=45^{\circ}+180n^{\circ}.$

My attempt

Trebla said:
If x = y then
sin 2x.cos x - sin x.cos 2x = sin 2x.cos x + sin x.cos 2x
=> 2sin x cos 2x = 0
.: sin x = 0 or cos 2x = 0
If sin x = 0 then sin 2x = 0 which would provide an undefined expression (denominator is zero)
If cos 2x = 0 (i.e. x = kπ ± π/4 for integer k) then the equality can hold for x = y
Hence it is possible.

x=y is possible.
Therefore triangle ABC is possibly isosceles

hscishard · Jun 10, 2010

Omnipotence said:
$\120dpi \text{If tan} \theta = t, \text{express sin2}\theta \text{ and cos2}\theta \\\text{ in terms of t. Find the values of t for which} \\(k+1)sin2\theta +(k-1)cos2\theta =k+1$

If tanx = t...
You sure it isnt x/2?

Omnipotence · Jun 10, 2010

hscishard said:
If tanx = t...
You sure it isnt x/2?

Definite.

Omnipotence · Jun 10, 2010

ohexploitable said:
sin2x = 2t/(1+t²)

cos2x = (1-t²)(1+t²)

Lol, you answered the first part.

ohexploitable · Jun 10, 2010

hscishard said:
If tanx = t...
You sure it isnt x/2?

The 'conventional' one is tan(x/2) but it doesn't matter, you can modify it.

hscishard · Jun 10, 2010

Yea I didn't notice the 2theta

Well the obvious answer is when cos2theta=0

nikkifc · Jun 10, 2010

Trebla said:
Oh and btw, limits and continuity are in the course, just not treated formally. See section 8.2.

For simple stuff yes it is in the course (eg. showing discontinuities in graphs). But I don't think the question posted above would be examinable.

Omnipotence · Jun 11, 2010

Gnaw. No one answered me question.

edmundsung · Jun 11, 2010

Re: another question

hscishard said:
Is the answer a=4 and b=26?

nikkifc said:
This is not in the scope of the current HSC syllabus. I believe it was taken out in 1981 from the old course.

And yes the answer is (a, b) = (4, 26). Although simple, to do it properly, it requires the use of limits.

Oh... thanks.

but im still stuck on it ><

would u please tell me how you got it.

greatly appreciated

ohexploitable · Jun 11, 2010

Re: another question

edmundsung said:
Oh... thanks.

but im still stuck on it ><

would u please tell me how you got it.

greatly appreciated

$\\\lim_{x\to-1^{-}}g(x)=\lim_{x\to-1^{+}}g(x) \\\\\lim_{x\to-1^{-}}x^2=\lim_{x\to-1^{+}}ax+b \\\\1=-a+5 \\a=4$

Do the same for b.

nikkifc · Jun 11, 2010

Re: another question

ohexploitable said:
$\\\lim_{x\to-1^{-}}g(x)=\lim_{x\to-1^{+}}g(x) \\\\\lim_{x\to-1^{-}}x^2=\lim_{x\to-1^{+}}ax+b \\\\1=-a+5 \\a=4$

Do the same for b.

Precisely my point. It's not in the HSC course because left hand and right hand limits are not in the current 1981 syllabus.

Omnipotence · Jun 12, 2010

Re: another question

$\120dpi \text{Find the general solution to} \\ \\{Sin2x+sin4x=sin3x} \\ \\\text{Show working out.}$

fullonoob · Jun 12, 2010

Re: another question

check :
x = 3 + sin4t + root3 cos4t
express in x = x0 + asin(4t+A)

ans should be x = 3 + 2sin(4t+pi/6) am i correct

hscishard · Jun 12, 2010

Re: another question

Omnipotence said:
$\120dpi \text{Find the general solution to} \\ \\{Sin2x+sin4x=sin3x} \\ \\\text{Show working out.}$

Please show working out...It's pretty long.

The steps would be sin3x = sin(2x +x) then expand
sin4x= 2cos2xsin2x

ohexploitable · Jun 12, 2010

Re: another question

New Question:

Trebla · Jun 12, 2010

Re: another question

Omnipotence said:
$\120dpi \text{Find the general solution to} \\ \\{Sin2x+sin4x=sin3x} \\ \\\text{Show working out.}$

sin 2x + sin 4x = sin 3x
sin (3x - x) + sin (3x + x) = sin 3x
sin 3x cos x - sin x cos 3x + sin 3x cos x + sin x cos 3x = sin 3x
2sin 3x cos x = sin 3x
sin 3x (2cos x - 1) = 0

cos x = 1/2
=> x = 2nπ ± π/3 for integer n

sin 3x = 0
=> 3x = kπ
=> x = kπ/3 for integer k

Omnipotence · Jun 12, 2010

Re: another question

Trebla said:
sin 2x + sin 4x = sin 3x
sin (3x - x) + sin (3x + x) = sin 3x
sin 3x cos x - sin x cos 3x + sin 3x cos x + sin x cos 3x = sin 3x
2sin 3x cos x = sin 3x
sin 3x (2cos x - 1) = 0

cos x = 1/2
=> x = 2nπ ± π/3 for integer n

sin 3x = 0
=> 3x = kπ
=> x = kπ/3 for integer k

Your awesome.

random-1005 · Jun 12, 2010

Re: another question

ohexploitable said:
New Question:

lol from last yrs hsc, thats just remainder theorm and solving two simulataneous eqns its way too easy
p(1)=2
P(-2)=5

Preliminary mathematics marathon (1 Viewer)

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