Mathematics Challenge Marathon

[RuiAce]

Would the answer be 2pi/3 and 4pi/3?
( I couldn't upload a photo of my working, but basically divided both sides of the equation by sinx -cosx giving me sin^2x-cos^2x=1. Rearranged to get cos^2x-sin^2x=-1, therefore cos2x=-1, cos x=-1/2.

Firstly, you should've had sin²(x)+sin(x)cos(x)+cos²(x)-1 like HPL correctly achieved. This occurs due to the factorisation a³-b³=(a-b)(a²+ab+b²) This simplifies down to sin(x)cos(x), and he decided to take it one step further and rewrite it as sin(2x) using MX1 techniques.

Secondly, this was mentioned above. Who said it was permissible to cancel out (sin(x)-cos(x))?

[RuiAce]

I suppose as the question was never completed correctly (despite attempted), I will post up the solution.

NEXT QUESTION

Spoiler

Required knowledge: Preliminary Basic Arithmetic and Algebra, HSC Series

[Durnity]

D: Oh damn these are tough. After 3 attempts I finally got the 1st one lol...

[RuiAce]

I don't usually post questions when there's a previous unanswered but I suppose I will on this forum.

Spoiler

Required knowledge: Preliminary Basic Arithmetic and Algebra, Preliminary The Quadratic Polynomial and the Parabola

[Happy Physics Land]

I don't usually post questions when there's a previous unanswered but I suppose I will on this forum.

Spoiler

Required knowledge: Preliminary Basic Arithmetic and Algebra, Preliminary The Quadratic Polynomial and the Parabola

Im not gonna upload a picture because laptop is rigged
Let y = root x + root x + root x ...
square both sides
y^2 = x + (root x + root x + root x ....)
but y = root x + root x + root x ....
hence y^2 = x + y
So the whole thing becomes y^2 - y - x = 0

[RuiAce]

Im not gonna upload a picture because laptop is rigged
Let y = root x + root x + root x ...
square both sides
y^2 = x + (root x + root x + root x ....)
but y = root x + root x + root x ....
hence y^2 = x + y
So the whole thing becomes y^2 - y - x = 0

Basically the idea.

[lazydreamer]

they have something similar to this at BoS 

anyway, is Happy Physics Land just missing y>= 0 ?

Incorrect. The only thing that Jake justified was that this is true IF AND ONLY IF sin(x)-cos(x)≠0.

In this case, there is no exception to this.

so:
(sinx-cosx)(sin^2x+sinxcosx+cos^2x) = sinx - cosx
(sin^2x + sinxcosx + cos^2x) = 1 [where sinx-cosx does not = 0] <--so i would have to write this in exams yea?
because sin^2x + cos^2x = 1, therefore sinxcosx = 0
sinxsin(90-x) = 0, therefore:
sinx = 0, sin(90-x) = 0
x= 0, 180, 360

[RuiAce]

they have something similar to this at BoS 

anyway, is Happy Physics Land just missing y>= 0 ?


so:
(sinx-cosx)(sin^2x+sinxcosx+cos^2x) = sinx - cosx
(sin^2x + sinxcosx + cos^2x) = 1 [where sinx-cosx does not = 0] <--so i would have to write this in exams yea?
because sin^2x + cos^2x = 1, therefore sinxcosx = 0
sinxsin(90-x) = 0, therefore:
sinx = 0, sin(90-x) = 0
x= 0, 180, 360

Yes, but of course, you CAN'T write sinx-cosx does not = 0 because you were never told that!

[lazydreamer]

Yes, but of course, you CAN'T write sinx-cosx does not = 0 because you were never told that!

ooh no wonder, i thought both threads were just a coincidence  could you be the one who also started the BoS one?

aaah i'm sorry, idk what i'm doing sooooo then i would have to take everything to the left and then factor out sinx-cosx?

and btw did you ever post up the soln. for the 'prove a,b,c are arith. progression' question? so so hard i can't even lol

[pi]

You guys might enjoy some of the threads we have in the VCE board Challenging Question Threads

Feel free to nick questions from there or bump up those threads, many have been solved but perhaps you guys can find some more elegant proofs

[RuiAce]

ooh no wonder, i thought both threads were just a coincidence  could you be the one who also started the BoS one?

aaah i'm sorry, idk what i'm doing sooooo then i would have to take everything to the left and then factor out sinx-cosx?

and btw did you ever post up the soln. for the 'prove a,b,c are arith. progression' question? so so hard i can't even lol

No, I didn't 'start' any.

Yep. (sinx-cosx) is factored.

Hint for that question: There's a bit of simultaneous equations going on

[RuiAce]

You guys might enjoy some of the threads we have in the VCE board Challenging Question Threads

Feel free to nick questions from there or bump up those threads, many have been solved but perhaps you guys can find some more elegant proofs

Coolio. Will check when I run out of questions

[lazydreamer]

No, I didn't 'start' any.

Yep. (sinx-cosx) is factored.

Hint for that question: There's a bit of simultaneous equations going on

ooh are you lee? haha

(sinx-cosx)(sinxcosx+1) - (sinx-cosx) = 0
(sinx-cosx)(sinxcosx) = 0

sinx - cosx = 0, sinxcosx = 0

hence x = pi/4, 180+pi/4, or x = 0, pi/2, pi, 3pi/2, 2pi

?? hope that's it...

[RuiAce]

Spoiler

Required knowledge: Preliminary Basic Arithmetic and Algebra, HSC Series

[RuiAce]

I feel as though my tone was too toxic when I first signed up for these forums.

Anyway, here's another question. I originally wrote it for a treat but it was considered 'too hard'. So I decided to put it here. Warning: It is long.

Required knowledge

Preliminary Basic Arithmetic and Algebra, HSC Exponential and Logarithmic Functions, HSC Series, HSC Geometrical Applications of Differentiation

Hint for d)

It says to use part b). But avoiding using the sum of a G.P. MAY prove useful.