4 Unit Revising Marathon HSC '10 (2 Viewers)

[addikaye03]

Next Question,
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(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]

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]

(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]

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

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

 

[addikaye03]

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]

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,
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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)

That is one sick question.

 

[addikaye03]

That is one sick question.

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

 

[study-freak]

This is my solution.

View attachment 20187

nice solution!

 

[cutemouse]

New question:

 

[study-freak]

lols, beaten although vafa, you made one stupid mistake of subbing in an incorrect value for y!

 

[addikaye03]

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]

lols, beaten although vafa, you made one stupid mistake of subbing in an incorrect value for y!

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]

Find the general solutions to:


[6 m]

 

[cutemouse]

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]

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

They're angles in a triangle.

 

[The Nomad]

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]

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]

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

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

 

[The Nomad]

Can you provide a mathematical proof for this statement? We all wish, it could be that easy, but it is not. your hypothesis is cosa+cosb+cos c=sin a+sin b+sin c=0, but your point is just one of the cases, it could be the case that cosa, cosb and cos c and sin a, sin b and sin c, all add up to zero. Right? and in that case, neither of the terms could be zero but their addition would be zero.

Oh deary me...I completely misread the question. My mistake, I know what you mean now.

 

[shaon0]

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

I got it out like this, but i'm wondering whether a polynomial proof is possible. The polynomial is degree 3 with roots tan(A/2), tan(B/2), tan(C/2) and sum of roots=product of roots.