Mathematics Corner :)

Why was the 'a' not taken into account?

EDIT: Ah... I see my mistake. I ended up integrating parts. But I still don't understand about the 'a'
 
sorry..lol
The a is just a constant...
d/dt(y) simply = -2a(cos(t))(sin(t))

I hope I've cleared up the confusion
 
transition? said:
I believe they start calculus your third or fourth(the last) year in high school. So I'd say mainly 16/17/18 ish. On that note, I'm surprised you started at 14/15!

It really depends though which math track you're on. Some people are quite blessed, they go on a faster track. Others don't touch a real calculus until they've graduate their fourth year in high school and gone to a university.
Click to expand...

You're saying the final years of high school right? Years 10/11 are our final years, though I covered calculus in more detail when I studied for A-level Maths.
And promptly forgot everything. :D
 
sillygraph-1.jpg


I think I did a fantastic job with Paint. I'm rather proud of what I've done. Anyway... I have no clue where to start or what to do.

And I just noticed that the actual question seems to have an 'O' instead of 'y' in the y-axis. So, I suppose, it's an O-axis now.
 
Lol, wow, that IS a pretty amazing graph...

Okayy, you start by finding dy/dx; use the product rule...you get:

dy/dx = 2tan2x + (2x-1)(2sec^2(2x)) = 2tan2x + 4xsec^2(2x) - 2sec^2(2x)

At the stationary point P, x = k, and dy/dx = 0, so set the above expression ^^ to equal zero, and replace x with k:
2tan2k + 4xsec^2(2k) - 2sec^2(2k) = 0

Multiply throughout by cos^2(2k) - or divide by sec^2(2k) - to get:

2(cos^2(2k))(tan2k) + 4k - 2 = 0
2(cos2k)(sin2k) + 4k - 2 = 0 .......[because cosxtanx = sinx, so cos^2xtanx = cosxsinx]
sin4k + 4k - 2 = 0 ........[double angle formula for sinx]
Therefore 4k + sin4k - 2 = 0
 
Difficult stuff indeed. But maybe that's because I'm mesmerised by my art to focus on the differentiation. I think I'll take a break and then come back.
 
x satisfies this equation
cos(2x) + cos(3x) = 1
Show that it also satisfies this one:
2sin(2x) + 2sin(3x) = sin(4x) + 2sin(5x) + sin(6x)

hEr's one for ya, Raynn
 
The Ruler said:
Difficult stuff indeed. But maybe that's because I'm mesmerised by my art to focus on the differentiation. I think I'll take a break and then come back.
Click to expand...

Lol, yup, good idea...using the product rule gets wayy complicated in this example...

Whatsthepoint said:
x satisfies this equation
cos(2x) + cos(3x) = 1
Show that it also satisfies this one:
2sin(2x) + 2sin(3x) = sin(4x) + 2sin(5x) + sin(6x)

hEr's one for ya, Raynn
Click to expand...

Lol, it looked like this was gonna take foreverr, but it wasn't too bad...so, we're trying to show:
2sin(2x) + 2sin(3x) = sin(4x) + 2sin(5x) + sin(6x)

Starting with the right side, and used the double angle formula for sinx,
sin4x + 2sin5x + sin6x
= 2sin2xcos2x + 2(sin2xcos3x + sin3xcos2x) + 2sin3xcos3x ......[the only odd angle is the sin5x; divide that into 2x and 3x]
= (2sin2xcos2x + 2sin2xcos3x) + (2sin3xcos2x + 2sin3xcos3x).....[regrouping]
= 2sin2x(cos2x + cos3x) + 2sin3x(cos2x + cos3x)......[factorizing]

We know from the previous identity thing that (cos2x + cos3x) = 1
Therefore,
sin4x + 2sin5x + sin6x = 2sin2x(1) + 2sin3x{1)
sin4x + 2sin5x + sin6x = 2sin2x + 2sin3x

There we are!
 
Last edited:
~Raynn~ said:
Lol, yup, good idea...using the product rule gets wayy complicated in this example...



Lol, it looked like this was gonna take foreverr, but it wasn't too bad...so, we're trying to show:
2sin(2x) + 2sin(3x) = sin(4x) + 2sin(5x) + sin(6x)

Starting with the right side, and used the double angle formula for sinx,
sin4x + 2sin5x + sin6x
= 2sin2xcos2x + 2(sin2xcos3x + sin3xcos2x) + 2sin3xcos3x ......[the only odd angle is the sin5x; divide that into 2x and 3x]
= (2sin2xcos2x + 2sin2xcos3x) + (2sin3xcos2x + 2sin3xcos3x).....[regrouping]
= 2sin2x(cos2x + cos3x) + 2sin3x(cos2x + cos3x)......[factorizing]

We know from the previous identity thing that (cos2x + cos3x) = 1
Therefore,
sin4x + 2sin5x + sin6x = 2sin2x(1) + 2sin3x{1)
sin4x + 2sin5x + sin6x = 2sin2x + 2sin3x

There we are!
Click to expand...
Yep.
when I was supposed to know it I didn't know all these formulas, like sin (alpha + beta).
You're officially an aspie, Raynn.:skeleton:
 
~Raynn~ said:
2(cos2k)(sin2k) + 4k - 2 = 0 .......[because cosxtanx = sinx, so cos^2xtanx = cosxsinx]
sin4k + 4k - 2 = 0 ........[double angle formula for sinx]
Therefore 4k + sin4k - 2 = 0
Click to expand...

Those are the bits I'm having difficulty understanding. Are those just formulae I'm supposed to know?
 
a) sin^2 x + cos^2 x = 1, show that cosec^2 x - cot^2 x = 1

I could do that.

b) Hence, or otherwise, prove that

cosec^4 x - cot^4 x = cosec^2 x + cot^2 x

... And I'm stuck. It's only for 2 marks so I think I'm missing something obvious here.

It's a proof, by the way.
 
Last edited:
cosec^2 x - cot^4 x = cosec^2 x + cot^2 x

check this again
it aint right cause cosec^s from both sides will cancel leaving you with an impossible identity
-cot^4 = cot^2
 

Similar Threads

Back
Top