Maths Game! (2 Viewers)

[bored.of.u]

Alright, let's start things off simple then

Find in terms of t, where t = tan (x/2)

cos (x) + sin² (x/2)

haha ok my solution:

(1-t^2)/(1+t^2) + [t/√(1+t^2)]^2
= (1-t^2)/(1+t^2) + (t^2/1+t^2)
= 1/1+t^2

is this ok gurmies?? =D

ok i shall post up another question (easy):

Use a suitable substitution to change x^3 + 12x^2 + 30x + 4 = 0 into a cubic equation of the form u^3 + cu + d = 0

 

[gurmies]

Correct bored.of.u =) Did you use the triangle to solve it?

 

[bored.of.u]

Correct bored.of.u =) Did you use the triangle to solve it?

Yep i used the triangle <--- makes my life so much easier with the t-substitution btw i challenge you to be the first one to answer my question =D

 

[gurmies]

Yikes, that was tough:

x^3 + 12x^2 + 30x + 4 = 0

let x = u - 4

(u - 4)^3 + 12(u - 4)^2 + 30(u - 4) + 4 = 0

u^3 - 12u^2 + 48u - 64 + 12(u^2 - 8u + 16) + 30u - 120 + 4 = 0

u^3 - 12u^2 + 48u - 64 + 12u^2 - 96u + 192 + 30u - 120 + 4 = 0

u^3 - 18u + 12 = 0

Is that correct?

 

[bored.of.u]

Yikes, that was tough:

x^3 + 12x^2 + 30x + 4 = 0

let x = u - 4

(u - 4)^3 + 12(u - 4)^2 + 30(u - 4) + 4 = 0

u^3 - 12u^2 + 48u - 64 + 12(u^2 - 8u + 16) + 30u - 120 + 4 = 0

u^3 - 12u^2 + 48u - 64 + 12u^2 - 96u + 192 + 30u - 120 + 4 = 0

u^3 - 18u + 12 = 0

Is that correct?

Yep! Good Work =D although you could have done it much simpler:

Alternatively:

x^3 + 12x^2 + 30x + 4 = 0
(x+4)^3 + 30x + 4 = 48x + 64
(x+4)^3 - 18x -60 = 0
(x+4)^3 - 18(x+4) + 12 = 0
This is in the form u^3 + cu + d = 0
Where, u = x + 4
.'. u^3 - 18u + 12 =0

Dont forget to post up a question gurmies =D

 

[Trebla]

Here's an interesting and somewhat challenging one:

(a) Show that (without induction)

sin x = 2ⁿ.sin (x/2ⁿ).cos (x/2ⁿ)............cos (x/8).cos (x/4).cos (x/2)

for some integer n


(b) Hence show that as n --> ∞

(sin x) / x = cos (x/2).cos (x/4).cos (x/8)................


(c) Deduce that:

[maths]\frac{2}{\pi} = \frac{\sqrt{2}}{2} . \frac{\sqrt{2 + \sqrt{2}}}{2} . \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} .......[/maths]

Have fun!

 

[jet]

With the locus:
|z + (1/z)| = 2
Hence, |z^2 + 1| = 2|z|

Let z = x + iy
Therefore |z| = sqrt(x^2 + y^2)

Now, z^2 = (x^2 - y^2) + 2xyi [This is in real and imaginary parts)
Hence, z^2 + 1 = (x^2 + 1 - y^2) + 2xyi
Therefore |z^2 + 1| = sqrt((x^2 + 1 - y^2)^2 + 4x^2 y^2)

I kept on going (I have no time now) and I got the same solution.

If you were confused at this point, (x^2 + y^2 - 1 + 2y)(x^2 + y^2 - 1 -2y) = 0

Say therefore either
x^2 + y^2 + 2y = 1
x^2 + y^2 -2y = 1

Then complete the square and you get the 2 circles.
The only stipulation is that x,y cannot equal zero

 

[study-freak]

Here's an interesting and somewhat challenging one:

(a) Show that (without induction)

sin x = 2ⁿ.sin (x/2ⁿ).cos (x/2ⁿ)............cos (x/8).cos (x/4).cos (x/2)

for some integer n


(b) Hence show that as n --> ∞

(sin x) / x = cos (x/2).cos (x/4).cos (x/8)................


(c) Deduce that:

[maths]\frac{2}{\pi} = \frac{\sqrt{2}}{2} . \frac{\sqrt{2 + \sqrt{2}}}{2} . \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} .......[/maths]

Have fun!

(a) RHS=2ⁿ.sin (x/2ⁿ).cos (x/2ⁿ)............cos (x/8).cos (x/4).cos (x/2)
=2^n-1.sin (x/2^n-1).cos (x/2^n-1)............cos (x/8).cos (x/4).cos (x/2) [using sin2x=2sinxcosx]
repeating this,
RHS=sinx=LHS

(b) (sin x) / x = cos (x/2).cos (x/4).cos (x/8)...cos (x/2ⁿ).sin (x/2ⁿ).2ⁿ / x
But as n --> ∞, sin (x/2^∞) / (x / 2^∞) --> 1
and sin (x/2^∞) / (x / 2^∞).cos (x/2ⁿ)=sin (x/2^∞-1) / (x / 2^∞-1) --> 1 [using sin2x=2sinxcosx]
and so on.
Hence, as n --> ∞, (sin x) / x = cos (x/2).cos (x/4).cos (x/8)...cos (x/2ⁿ).sin (x/2ⁿ).2ⁿ / x --> cos (x/2).cos (x/4).cos (x/8)................

(c)Let x= [maths]\frac{pi}{\22}[/maths].
Then from (b), as n --> ∞,
(sin x) / x = cos (x/2).cos (x/4).cos (x/8)................
=[maths]\frac{\sqrt{2}}{2} . \frac{\sqrt{2 + \sqrt{2}}}{2} . \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} .......[/maths]
using exact values of trigonometric expressions.


Wow... a good question... Took me awhile to type this up.(coz I didn't know how to use those [maths] or ⁿ stuff. Just learnt it by looking at yours, Trebla, lol)
and it seems I can't fix sign in (c) for pi to appear properly as a mathematical symbol... anyway, anyone tell me if I got something wrong.



My Question: Integrate from pi/2 to 0, cosx / (cosx + sinx) in respect to x.

 

[Dragonmaster262]

Hi a new maths game ^^

Rules: Answer a question and then ask a question, anywhere from 2unit prelim to 3unit. Have fun guys.

I'll post the first question,

If, x + y=1 and x^3 + y^3=19 find the value of x^2 + y^2

Are you an accelerant?

 

[lolokay]

My Question: Integrate from pi/2 to 0, cosx / (cosx + sinx) in respect to x.

Spoiler

let x = pi/2 - u
dx = -du
when x = pi/2, u=0
when x = 0, u = pi/2
cosx = sinu
sinx = cosu

pi/2,0 Int( cosx / (cosx + sinx) .dx ) = 0,pi/2 Int(sinu / (sinu+cosu) .-du)
= pi/2,0 Int( sinu / (sinu + cosu) .du )
= pi/2,0 Int( sinx / (cosx + sinx) .dx )

so
pi/2,0 Int( cosx / (cosx + sinx) .dx )
= 1/2 pi/2,0 Int( sinx + cosx / (cosx + sinx) .dx )
= 1/2 pi/2,0 Int(1)
= pi/4

my notation is a bit messy

 

[Pwnage101]

With the locus:
|z + (1/z)| = 2
Hence, |z^2 + 1| = 2|z|

Let z = x + iy
Therefore |z| = sqrt(x^2 + y^2)

Now, z^2 = (x^2 - y^2) + 2xyi [This is in real and imaginary parts)
Hence, z^2 + 1 = (x^2 + 1 - y^2) + 2xyi
Therefore |z^2 + 1| = sqrt((x^2 + 1 - y^2)^2 + 4x^2 y^2)

I kept on going (I have no time now) and I got the same solution.

If you were confused at this point, (x^2 + y^2 - 1 + 2y)(x^2 + y^2 - 1 -2y) = 0

Say therefore either
x^2 + y^2 + 2y = 1
x^2 + y^2 -2y = 1

Then complete the square and you get the 2 circles.
The only stipulation is that x,y cannot equal zero

Just curious - Why are you doing 17 units for your HSC?

 

[bored.of.u]

Are you an accelerant?

huh?? i dont quite get u...

 

[shaon0]

huh?? i dont quite get u...

He means: Have you accelerated in Maths (or any other subject) this year?

 

[bored.of.u]

He means: Have you accelerated in Maths (or any other subject) this year?

no...wat makes him think im an accelerant??

 

[gurmies]

The fact that you haven't even started preliminary mathematics yet and know so much

 

[rellenayyy]

i don't think i'd ever find this game fun.
unless they're easy ass 1+1 questions and involve some sort of prize at the end.

 

[shaon0]

I believe my solutions are correct, until somebody takes a logical and coherent stab at them. What you have said has made no sense...

You're solutions correct.

Question:
S [limits:a to -a] sqrt(a^2-x^2) dx
I'll give a hint. Use x=cos(u) as a substituion.

 

[tacogym27101990]

You're solutions correct.

Question:
S [limits:a to -a] sqrt(a^2-x^2) dx
I'll give a hint. Use x=cos(u) as a substituion.

or you could just realise its a semi circle radius a
therefore S [limits:a to -a] sqrt(a^2-x^2) dx
= pi.a^2/2

 

[random-1005]

Are you an accelerant?

isn't it obvious that he is, hes only going into yr11 this year, and he has allready finished the entire 2/3 unit course for the hsc. thats friggen crazy being so far ahead

 

[Dragonmaster262]

no...wat makes him think im an accelerant??

Did you teach yourself this stuff? Or dd you have a Private Tutor?