4 Unit Revising Marathon HSC '10 (1 Viewer)

adomad · Jan 18, 2010

jm01 said:
Assess the impacts of Leibniz' and Newton's contributions to calculus on society....

i pray to God that that will never pop up in the exam... it is in the outcomes isn't it? to appreciate the beauty and elegance of maths?

cutemouse · Jan 18, 2010

adomad said:
i pray to God that that will never pop up in the exam... it is in the outcomes isn't it? to appreciate the beauty and elegance of maths?

I don't think so. I was merely mocking HSC sciences

addikaye03 · Jan 18, 2010

vafa said:
Next Question,
--------------------
(reserved for my attempted solution)

Question: tan(A/2).tan(B/2)+tan(A/2).tan(C/2)+tan(B/2).tan(C/2)=1

addikaye03 · Jan 19, 2010

Question: tan(A/2).tan(B/2)+tan(A/2).tan(C/2)+tan(B/2).tan(C/2)=1

Get this thread moving again

jet · Jan 19, 2010

addikaye03 said:
(1-2sin^2a)+(1-2sin^2b)+(1-2sin^2c)=0

sin^2(a)+sin^2(b)+sin^2(c)=3/2 (1)

(1-cos^2a)+(1-cos^2b)+(1-cos^2c)=3/2

3-3/2=cos^2a+cos^2b+cos^2c (2)

Therefore (1)=(2) [sin^2a+sin^2b+sin^2c=cos^2a+cos^b+cos^c=3/2]

(cos^2a-sin^2a)+(cos^2b-sin^2b)+(cos^2c-sin^2c)=0

cos2a+cos2b+cos2c=0

Pretty noob solution, i'll think about the complex way now

Question: tan(A/2).tan(B/2)+tan(A/2).tan(C/2)+tan(B/2).tan(C/2)=1

There is a slight problem in that you assume the result in your first line.

I.e. you assume that cos(2a) + cos(2b) + cos(2c) = 0, when that is the result you want to actually prove in the first place.

adomad · Jan 19, 2010

jm01 said:
I don't think so. I was merely mocking HSC sciences

yeah, i knew that... that pops a lot.

addikaye03 · Jan 19, 2010

jetblack2007 said:
There is a slight problem in that you assume the result in your first line.

I.e. you assume that cos(2a) + cos(2b) + cos(2c) = 0, when that is the result you want to actually prove in the first place.

Thanks, that was quite idiotic lol

The Nomad · Jan 19, 2010

vafa said:
another alternative to a) is considering a polynomial, p(x) with roots such that

p(x)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc=0

Then because a, b and c are roots, they should satisfy p(x).

a^3-a^2(a+b+c)+a(ab+ac+bc)-abc=0
b^3-b^2(a+b+c)+b(ab+ac+bc)-abc=0
c^3-c^2(a+b+c)+c(ab+ac+bc)-abc=0

Then adding these three, would enable you to factorise it quite easily

There is also another approach for factorising this using determinants but because that confuses four unit students as it is not in the course, I will not mention it. Those interested in the determinant approach can email me.

Next Question,
--------------------

let a, b and c be real numbers such that

cos a+cos b+cos c=sin a+sin b+sin c=0

Prove that

cos 2a+cos 2b+cos 2c=sin 2a+sin 2b+sin 2c=0

(hint: You may use complex numbers to prove the result)

$\\ Cos a + Cos b + Cos c = Sin a + Sin b + Sin c = 0 \\ $Let $x = Cos a + iSina, y = Cos b + iSinb, z = Cos c + i Sin c\\ \therefore x=y=z=0\\ \therefore x^2 = y^2 + z^2 = 0$, and so $ x^2+y^2+z^2 = 0\\ $But $ x^2 = Cis 2a, y^2 = Cis 2b, z^2 = Cis 2c,$ $ \therefore Cis 2a + Cis 2b + Cis 2c = 0\\ $Equating real and imaginary parts: $ Cos 2a + Cos 2b + Cos 2c = Sin 2a + Sin2b + Sin2c = 0\\$

That is one sick question.

addikaye03 · Jan 19, 2010

The Nomad said:
$\\ Cos a + Cos b + Cos c = Sin a + Sin b + Sin c = 0 \\ $Let $x = Cos a + iSina, y = Cos b + iSinb, z = Cos c + i Sin c\\ \therefore x=y=z=0\\ \therefore x^2 = y^2 + z^2 = 0$, and so $ x^2+y^2+z^2 = 0\\ $But $ x^2 = Cis 2a, y^2 = Cis 2b, z^2 = Cis 2c,$ $ \therefore Cis 2a + Cis 2b + Cis 2c = 0\\ $Equating real and imaginary parts: $ Cos 2a + Cos 2b + Cos 2c = Sin 2a + Sin2b + Sin2c = 0\\$

That is one sick question.

Well done man, I couldn't get it. Try the Q i posted

study-freak · Jan 19, 2010

vafa said:
This is my solution.

View attachment 20187

nice solution!

cutemouse · Jan 19, 2010

New question:

$If \ y=(1+x^3)^{1+x^3} \ find \ \frac{dy}{dx}$

study-freak · Jan 20, 2010

$\\y=(1+x^3)^{1+x^3} \\log_{1+x^3}y=1+x^3 \\\frac{lny}{ln(1+x^3)}=1+x^3 \\lny=(1+x^3)ln(1+x^3) \\\frac{\frac{dy}{dx}}{y}=3x^2ln(1+x^3)+3x^2 \\\frac{dy}{dx}=3x^2(1+ln(1+x^3))y \\\frac{dy}{dx}=3x^2(1+ln(1+x^3))(1+x^3)^{1+x^3}$
lols, beaten although vafa, you made one stupid mistake of subbing in an incorrect value for y!

addikaye03 · Jan 20, 2010

vafa said:
What is exactly your question? Just proving that? What are A, B, and C?

A, B and C are the angles of a triangle.

Trebla · Jan 20, 2010

study-freak said:
$\\y=(1+x^3)^{1+x^3} \\log_{1+x^3}y=1+x^3 \\\frac{lny}{ln(1+x^3)}=1+x^3 \\lny=(1+x^3)ln(1+x^3) \\\frac{\frac{dy}{dx}}{y}=3x^2ln(1+x^3)+3x^2 \\\frac{dy}{dx}=3x^2(1+ln(1+x^3))y \\\frac{dy}{dx}=3x^2(1+ln(1+x^3))(1+x^3)^{1+x^3}$
lols, beaten although vafa, you made one stupid mistake of subbing in an incorrect value for y!

vafa said:
View attachment 20189

Just to let people here know, the technique employed in finding the derivative as shown above is known as logarithmic differentiation which involves taking the natural logarithm of both sides and implicitly differentiating. This technique can actually make some normally complicated functions much simpler to differentiate because the log laws can be used to separate out composite parts of a function...may be potentially useful in exam questions (though quite rare lol)?

cutemouse · Jan 20, 2010

Find the general solutions to:

$cos4\theta+cos2\theta=\sqrt2 cos^2\theta+\frac{1}{\sqrt2}sin2\theta$
[6 m]

cutemouse · Jan 20, 2010

Trebla said:
Just to let people here know, the technique employed in finding the derivative as shown above is known as logarithmic differentiation which involves taking the natural logarithm of both sides and implicitly differentiating. This technique can actually make some normally complicated functions much simpler to differentiate...may be potentially useful in exam questions (though quite rare lol)?

Yeah I thought of that as well. Esp, in curve sketching which require multiple applications of the product rule (eg. to find y'' when y is a 'complicated' function)

But I figured that it would be too much "mucking around" at 2U/3U level. But in 4U you never know

shaon0 · Jan 20, 2010

addikaye03 said:
A, B and C are the sides of a triangle.

They're angles in a triangle.

The Nomad · Jan 20, 2010

vafa said:
My brain can not understand how x=y=z=0, could you please clarify this for me?

Cos a = 0 and Sin a = 0, so x = Cos a + i Sin a = 0 + 0i = 0. Same goes for y and z.

And your solution is so long

jet · Jan 20, 2010

The Nomad said:
Cos a = 0 and Sin a = 0, so x = Cos a + i Sin a = 0 + 0i = 0. Same goes for y and z.

And your solution is so long

I'm not quite sure whether that is correct. I think there may be a few different cases.

The Nomad · Jan 20, 2010

addikaye03 said:
Next Question,
--------------------
(reserved for my attempted solution)

Question: tan(A/2).tan(B/2)+tan(A/2).tan(C/2)+tan(B/2).tan(C/2)=1

$\\$Sides of a triangle add to 180 degrees, so $ A + B + C = 180.\\ A = 180 - (B + C), Tan(\frac{A}{2}) = Tan(90 - \frac{B + C}{2}) = Cot(\frac{B + C}{2})\\ Tan\frac{A}{2}Tan\frac{B}{2}\cdot Tan\frac{A}{2}Tan\frac{C}{2}\cdot Tan\frac{B}{2}Tan\frac{C}{2} = \\(Tan\frac{B}{2}+Tan\frac{C}{2})Tan\frac{A}{2} + Tan\frac{B}{2}Tan\frac{C}{2}=\\ (Tan\frac{B}{2} + Tan\frac{C}{2})Cot\frac{B + C}{2} + Tan\frac{B}{2}Tan\frac{C}{2} = \\(Tan\frac{B}{2} + Tan\frac{C}{2})\frac{1-Tan\frac{B}{2}Tan\frac{C}{2}}{Tan\frac{B}{2}+Tan\frac{C}{2}} + Tan\frac{B}{2}Tan\frac{C}{2} = \\(1 - Tan\frac{B}{2}Tan\frac{C}{2}) + Tan\frac{B}{2}Tan\frac{C}{2} = 1.$

That was so annoying to type. And what do you mean Jetblack?

4 Unit Revising Marathon HSC '10 (1 Viewer)

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