IB Maths Marathon (4 Viewers)

davidgoes4wce · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

Maths SL Cambridge Ex 6A Q6

This was the solution:

Now where I was stuck on was part (b) (ii)

I did it a slightly different way, when looking at the RHS

Like the book I assumed:

$u_n=2^n , u_{n-1}=2^{n-1}, u_{n+1}=2^{n+1}$

The only difference where I didn't understand was this step, I included the 2 in front of the bracket as mentioned in the question

$2(3 u_n -2 u_{n-1})$

$2[ 3 \times 2^n-2^1 \times 2^{n-1} ]$

$6 \times 2^n-2 \times 2^n$

$$ By factorising $ 2^n(6-2)$

$$ I got $ 4 \times 2^n$

davidgoes4wce · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

This is a larger size of the question if anybody couldn't see the smaller version:

InteGrand · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

davidgoes4wce said:
Maths SL Cambridge Ex 6A Q6

This was the solution:

Now where I was stuck on was part (b) (ii)

I did it a slightly different way, when looking at the RHS

Like the book I assumed:

$u_n=2^n , u_{n-1}=2^{n-1}, u_{n+1}=2^{n+1}$

The only difference where I didn't understand was this step, I included the 2 in front of the bracket as mentioned in the question

$2(3 u_n -2 u_{n-1})$

$2[ 3 \times 2^n-2^1 \times 2^{n-1} ]$

$6 \times 2^n-2 \times 2^n$

$$ By factorising $ 2^n(6-2)$

$$ I got $ 4 \times 2^n$

$\noindent The presence of the $2$ outside the brackets seems to be a typo there. It should just be without that $2$ I think (in other words the given recurrence equation at the start of the question).$

davidgoes4wce · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

I thought it was a typo as well.

davidgoes4wce · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

Slight difference between IB is they call the first term $u_1$

HSC call is $a$

But its the same thing

davidgoes4wce · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

In IB they call the last term $u_n$

In HSC they call the last term $l$

eg

IB $S_n=\frac{n}{2}(u_1+u_n)$

HSC $S_n=\frac{n}{2}(a+l)$

davidgoes4wce · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

Maths HL Complex Numbers

How do you do the Introductory Problem?

InteGrand · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

davidgoes4wce said:
Maths HL Complex Numbers

How do you do the Introductory Problem?

$\noindent Note that $\sin 5\theta = \Im \left(z^5\right)$, where $z=\cos \theta + i\sin \theta$ (due to De Moivre's theorem). But $z^5 = \left(\cos \theta +i\sin \theta\right)^5$ can be expanded using the Binomial Theorem, and its imaginary part found, which will then lead to the answer.$

InteGrand · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

$\noindent Actually realised they probably expected you to do that one without use of complex numbers. One way to do this is to expand $\sin 5x = \sin \left(4x+x\right) = \sin 4x \cos x + \cos 4x \sin x$. Now, $\sin 4x = 2\sin 2x \cos 2x$ and $\cos 4x = 1-2\sin^2 2x$. Each of these can be reduced to being in terms of $\sin x$ and $\cos x$ now using the double angle formulas, and then subbed back into $\sin 5x = \sin 4x \cos x + \cos 4x \sin x$ to help get the answer. Remember to use $\cos^2 x = 1-\sin^2 x$ if needed (to get the final answer just in terms of $\sin x$).$

davidgoes4wce · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

InteGrand said:
$\noindent Actually realised they probably expected you to do that one without use of complex numbers. One way to do this is to expand $\sin 5x = \sin \left(4x+x\right) = \sin 4x \cos x + \cos 4x \sin x$. Now, $\sin 4x = 2\sin 2x \cos 2x$ and $\cos 4x = 1-2\sin^2 2x$. Each of these can be reduced to being in terms of $\sin x$ and $\cos x$ now using the double angle formulas, and then subbed back into $\sin 5x = \sin 4x \cos x + \cos 4x \sin x$ to help get the answer. Remember to use $\cos^2 x = 1-\sin^2 x$ if needed (to get the final answer just in terms of $\sin x$).$

this is what I did based on your advice:

$sin (5x) = sin (4x+x)$

$sin(4x)cos(x)+sin(x)cos(4x)$

$[2 sin (2x) cos (2x)] cos x+sin x (1-2sin^{2}2x)$

Im guessing that we have to break that double again down again ?

InteGrand · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

davidgoes4wce said:
this is what I did based on your advice:

$sin (5x) = sin (4x+x)$

$sin(4x)cos(x)+sin(x)cos(4x)$

$[2 sin (2x) cos (2x)] cos x+sin x (1-2sin^{2}2x)$

Im guessing that we have to break that double again down again ?

Yep.

davidgoes4wce · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

Curious little fact about negative numbers mentioned in the book

"In Europe, it took until the 17th century for negative numbers to be accepted."

An Indian mathematician Brahmagupta , who's name comes up fairly frequently was the first to introduce the idea.

Paradoxica · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

davidgoes4wce said:
$sin (5x) = sin (4x+x)$

$sin(4x)cos(x)+sin(x)cos(4x)$

$[2 sin (2x) cos (2x)] cos x+sin x (1-2sin^{2}2x)$

$[ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times sin 2x \times sin 2x)$

$[ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times 2\ sin x \ cos x \ \times 2sinx \ cos x)$

$4 \ cos^{2} x \ sin x \ cos(2x)+sin x(1-8 sin^{2}x \ cos^{2} x)$

$4 (1-sin^{2}x) sin x (2 sin^{2}x-1)+sin \ x[1-8 sin^{2}x (1-sin^{2}x)]$

$(4-4sin^{2}x_)(2sin^{3}x-sinx)+sin x[1-8sin^{2}x+8sin^{4}x]$

$8 sin^{3}x-4sinx-8sin^{5}x+4sin^{3}x+sinx-8sin^{3}x+8sin^{5}x$

$4sin^{3}x-3sin x$

That's my final simplified step.

Sign errors all over the place.

davidgoes4wce · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

Ahhh I made a mistake

davidgoes4wce · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

I had a go again:

$sin (5x) = sin (4x+x)$

$sin(4x)cos(x)+sin(x)cos(4x)$

$[2 sin (2x) cos (2x)] cos x+sin x (1-2sin^{2}2x)$

$[ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times sin 2x \times sin 2x)$

$[ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times 2\ sin x \ cos x \ \times 2sinx \ cos x)$

$4 \ cos^{2} x \ sin x \ cos(2x)+sin x(1-8 sin^{2}x \ cos^{2} x)$

$4 (1-sin^{2}x) sin x (1-2sin^{2}2x)+sin \ x[1-8 sin^{2}x (1-sin^{2}x)]$

$(4-4sin^{2}x_)(sinx-2 \times 2 \ sinx cos x \ 2 sin x cos x)+sin x[1-8sin^{2}x+8sin^{4}x]$

$(4-4sin^{2}x_)(sinx-8 sin^{2}x \ cos^{2} x)+sin x[1-8sin^{2}x+8sin^{4}x]$

$(4-4sin^{2}x_)[sinx-8 sin^{2}x \ (1-sin^{2}x)]+sin x[1-8sin^{2}x+8sin^{4}x]$

$(4-4sin^{2}x_)[sinx-8 sin^{2}x +8 sin^{4} x]+sin x[1-8sin^{2}x+8sin^{4}x]$

$4 sin x-32 sin^{2}x+32sin^{4}x-4sin^{3}x+32 sin^{4}x-32sin^{6}x+4sin^{3}x+sinx-8sin^{3}x+8sin^{5}x$

$-32sin^{6}x+8sin^{5}x+64sin^{4}x+8sin^{3}x-32sin^{2}x+4sin{x}$

That's my final simplified step.

parad0xica · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

To the above working out, you can verify the solution by plugging in x = pi/2. RHS yields 20 which exceeds the range of the sine function. So you probably made a small error somewhere

pi/2 because sin(pi/2) is easily calculated

davidgoes4wce · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

$sin (5x) = sin (4x+x)$

$sin(4x)cos(x)+sin(x)cos(4x)$

$[2 sin (2x) cos (2x)] cos x+sin x (1-2sin^{2}2x)$

$[ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times sin 2x \times sin 2x)$

$[ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times 2\ sin x \ cos x \ \times 2sinx \ cos x)$

$4 \ cos^{2} x \ sin x \ cos(2x)+sin x(1-8 sin^{2}x \ cos^{2} x)$

$4 (1-sin^{2}x) sin x (1-2sin^{2}2x)+sin \ x[1-8 sin^{2}x (1-sin^{2}x)]$

$(4-4sin^{2}x_)(sinx-2 \times 2 \ sinx cos x \ 2 sin x cos x)+sin x[1-8sin^{2}x+8sin^{4}x]$

$(4-4sin^{2}x_)(sinx-8 sin^{2}x \ cos^{2} x)+sin x[1-8sin^{2}x+8sin^{4}x]$

$(4-4sin^{2}x_)[sinx-8 sin^{2}x \ (1-sin^{2}x)]+sin x[1-8sin^{2}x+8sin^{4}x]$

$(4-4sin^{2}x_)[sinx-8 sin^{2}x +8 sin^{4} x]+sin x[1-8sin^{2}x+8sin^{4}x]$

$4 sin x-32 sin^{2}x+32sin^{4}x-4sin^{3}x+32 sin^{4}x-32sin^{6}x+4sin^{3}x+sinx-8sin^{3}x+8sin^{5}x$

$-32sin^{6}x+8sin^{5}x+64sin^{4}x+8sin^{3}x-32sin^{2}x+5sin{x}$

Think I found it

parad0xica · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

This:

$4 (1-sin^{2}x) sin x (1-2sin^{2}2x)+sin \ x[1-8 sin^{2}x (1-sin^{2}x)]$

To this:

$(4-4sin^{2}x)(sinx-2 \times 2 \ sinx cos x \ 2 sin x cos x)+sin x[1-8sin^{2}x+8sin^{4}x]$

But it should be:

$(4-4sin^{2}x)(sinx-2 \times \underline{sin x} \times 2 \ sinx cos x \ 2 sin x cos x)+sin x[1-8sin^{2}x+8sin^{4}x]$

Also, your final result is still wrong because substituting x = pi/2 yields 20.

davidgoes4wce · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

OK Thanks I see it try to fix it again

davidgoes4wce · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

$sin (5x) = sin (4x+x)$

$sin(4x)cos(x)+sin(x)cos(4x)$

$[2 sin (2x) cos (2x)] cos x+sin x (1-2sin^{2}2x)$

$[ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times sin 2x \times sin 2x)$

$[ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times 2\ sin x \ cos x \ \times 2sinx \ cos x)$

$4 \ cos^{2} x \ sin x \ cos(2x)+sin x(1-8 sin^{2}x \ cos^{2} x)$

$4 (1-sin^{2}x) sin x (1-2sin^{2}2x)+sin \ x[1-8 sin^{2}x (1-sin^{2}x)]$

$(4-4sin^{2}x_)(sinx-2 \ sin x \ \times 2 \ sinx cos x \ 2 sin x cos x)+sin x[1-8sin^{2}x+8sin^{4}x]$

$(4-4sin^{2}x_)(sinx-8 sin^{3}x \ cos^{2} x)+sin x[1-8sin^{2}x+8sin^{4}x]$

$(4-4sin^{2}x_)[sinx-8 sin^{3}x \ (1-sin^{2}x)]+sin x[1-8sin^{2}x+8sin^{4}x]$

$(4-4sin^{2}x_)[sinx-8 sin^{3}x +8 sin^{5} x]+sin x[1-8sin^{2}x+8sin^{4}x]$

$4 sin x-32 sin^{3}x+32sin^{5}x-4sin^{3}x+32 sin^{5}x-32sin^{7}x+4sin^{3}x+sinx-8sin^{3}x+8sin^{5}x$

$-32sin^{7}x+72sin^{5}x-24sin^{3}x+5sin{x}$

Is that right?

Make a Difference – Donate to Bored of Studies!

IB Maths Marathon (4 Viewers)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 4)