1. ## Re: International Baccalaureate Marathon 2016

Originally Posted by davidgoes4wce

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

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

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

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

2. ## 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.

3. ## Re: International Baccalaureate Marathon 2016

Originally Posted by davidgoes4wce
$sin (5x) = sin (4x+x)$

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

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

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

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

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

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

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

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

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

That's my final simplified step.
Sign errors all over the place.

5. ## Re: International Baccalaureate Marathon 2016

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

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

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

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

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

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

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

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

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

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

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

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

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

That's my final simplified step.

6. ## 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

7. ## 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

8. ## 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.

9. ## Re: International Baccalaureate Marathon 2016

OK Thanks I see it try to fix it again

10. ## 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?

11. ## Re: International Baccalaureate Marathon 2016

Originally Posted by davidgoes4wce
$sin (5x) = sin (4x+x)$

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

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

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

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

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

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

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

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

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

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

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

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

Is that right?
If we sub. in x = pi/2 to the final answer, since sin(pi/2) = 1, we get -32 + 72 -24 + 5 = 21, which is outside the range of the LHS, which is sin(5x), which is between -1 and 1 always.

This is the check that parad0xica mentioned in an earlier comment.

12. ## Re: International Baccalaureate Marathon 2016

Originally Posted by davidgoes4wce
$sin (5x) = sin (4x+x)$

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

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

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

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

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

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

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

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

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

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

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

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

Is that right?
Not possible, sin5x is a polynomial in sinx of degree 5.

Some power in there is broken.

14. ## Re: International Baccalaureate Marathon 2016

Originally Posted by davidgoes4wce
$sin (5x) = sin (4x+x)$

$\textbf{L1. } sin(4x)cos(x)+sin(x)cos(4x)$

$\textbf{L2. } [2 sin (2x) cos (2x)] cos x+sin x (1-2sin^{2}2x)$

$\textbf{L3. } [ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times sin 2x \times sin 2x)$

$\textbf{L4. } [ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times 2\ sin x \ cos x \ \times 2sinx \ cos x)$

$\textbf{L5. } 4 \ cos^{2} x \ sin x \ cos(2x)+sin x(1-8 sin^{2}x \ cos^{2} x)$

$\textbf{L6. } 4 (1-sin^{2}x) sin x (1-2sin^{2}2x)+sin \ x[1-8 sin^{2}x (1-sin^{2}x)]$
Looking at the first term only when going from line 5 to line 6, you did this:

$\cos 2x = 1 - 2 \sin ^2 2x$

I think it was an issue of focus, not of lack of knowledge

15. ## Re: International Baccalaureate Marathon 2016

I'll try it again after watching a bit of the Euro 2016 action

16. ## Re: International Baccalaureate Marathon 2016

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

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

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

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

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

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

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

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

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

$16 sin^{5} x-20sin^{3}x+5 sin x$

17. ## Re: International Baccalaureate Marathon 2016

Maths HL Introductory Problem:

My thinking was to integrate the circumference of a circle in order to get the area of a circle.

18. ## Re: International Baccalaureate Marathon 2016

Originally Posted by davidgoes4wce
Maths HL Introductory Problem:

My thinking was to integrate the circumference of a circle in order to get the area of a circle.
Yeah that's essentially one way to do it. The typical HSC method would be to do something like integrate sqrt(r^2 - x^2) between 0 and r (this getting the area of the quarter-disk), and then multiplying by 4. The integration would be carried out using a trig. substitution (or you could use integration by parts but for that definite integral, trig. sub. is nicer).

19. ## Re: International Baccalaureate Marathon 2016

Something like

$\int 2 \pi r dr = \pi {r^2} + C$

20. ## Re: International Baccalaureate Marathon 2016

Originally Posted by davidgoes4wce
Something like

$\int 2 \pi r dr = \pi {r^2} + C$
Pretty much. And the area of a circle of radius 0 is 0, which makes C = 0. Not sure how rigorous you are required to be though with regards to explaining the integral.

Here's a paper discussing when the perimeter of a shape will be the derivative of its area, or area the derivative of volume: https://www.math.byu.edu/~mdorff/docs/DorffPaper07.pdf .

21. ## Re: International Baccalaureate Marathon 2016

I was a bit confused on this Introductory Question in Maths HL on Differentiation page 527 of Cambridge.

Im thinking the function is

$C=\frac{12+v^2}{100}$

We have to use differentiation and set $\frac{dC}{dv}$ to some value to solve for v.

Anyone with any ideas to get a kick start to this question?

22. ## Re: International Baccalaureate Marathon 2016

Originally Posted by davidgoes4wce

I was a bit confused on this Introductory Question in Maths HL on Differentiation page 527 of Cambridge.

Im thinking the function is

$C=\frac{12+v^2}{100}$

We have to use differentiation and set $\frac{dC}{dv}$ to some value to solve for v.

Anyone with any ideas to get a kick start to this question?
$\noindent If the speed travelled at is v, then the time taken is distance divided by speed, i.e. \frac{50}{v}. So the total cost (in terms of litres of fuel used) of the journey is the time multiplied by the rate of fuel usage, so it is C = \frac{50}{v}\times \frac{12+v^2}{100}. So this is the cost function to minimise.$

23. ## Re: International Baccalaureate Marathon 2016

I totally didn't see it like that, after you explained that I got it.

24. ## Re: International Baccalaureate Marathon 2016

Originally Posted by InteGrand
$\noindent If the speed travelled at is v, then the time taken is distance divided by speed, i.e. \frac{50}{v}. So the total cost (in terms of litres of fuel used) of the journey is the time multiplied by the rate of fuel usage, so it is C = \frac{50}{v}\times \frac{12+v^2}{100}. So this is the cost function to minimise.$
I had a go at finishing off this question

$By simplifying terms we get, C=\frac{12+v^2}{2v}$

$\frac{dC}{dv}=\frac{(2v)(2v)-(12+v^2)(2)}{4v^2}$

$\frac{dC}{dv}=0$

$0=4v^2-24-2v^2$

$0=2v^2-24$

$0=2(v^2-12)$

$0=v^2-12$

$v=2 \ \sqrt{3} \ miles \ per \ hour$

25. ## Re: International Baccalaureate Marathon 2016

Intro Problem Chapter 16

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