Current Year 12 Thread Mark I!

Oh right, that book is apparently for "beginners".
I spams slice turns
It's method is also so unorganised that I don't understand it.

You made me get excited and all

Sorry

Do the calculation seperately

Hey

Well, if it helps i feel horrible and sort of wish that i'd just got my predicted grades instead. I think that the whole thing was just a case of beginners luck.

You're the super smart one. You can speak over 200 different languages (or thereabouts), worked out the nationality of those people that dabby was talking about just from the spelling of their names, taught me what a Pagan was, and what linguistics is and countless other things.

Oh, and are you the one who said they had an idioglossia? Because i finally got round to googling what it actually means, and if it was you then you are officially one of the coolest people i've known, ever!!

Ooh, I made it in time for the new thread.

What do you mean?

Yeah, it's time to push for A-Levels

It is for AQA B, it may not be in other specifications.

Yeah thats right, I messed up getting the -11sin term explain plz

*Maths stuff written with Latex*

Hi! Have you had a good day?

OMG We are year 12s



This surreal for anyone else?

OMG We are year 12s



This surreal for anyone else?

OMG We are year 12s



This surreal for anyone else?

*does the haka*

I think we're doing that spec, but it's likely to be similar for other boards anyway

I'm a year 13.

Sure thing.



But

So:

$\dfrac{1}{\sin \theta} (3\cos 2\theta +7) + 11 = 0$

Multiplying brackets out gives:

$\dfrac{3\cos 2\theta + 7}{\sin \theta} + 11 = 0$

But $\cos 2A = 1 - 2\sin^2 A$ (one of the three forms of the cos double angle formulae).

So:

$\dfrac{3(1 - 2\sin^2 \theta) + 7}{\sin \theta} + 11 = 0$

$\dfrac{3 -6\sin^2 \theta +7}{\sin \theta} +11 = 0$

$\dfrac{10 -6\sin^2 \theta}{\sin \theta} +11 = 0$

Multiplying all terms by $\sin \theta$ gives:

$10 -6\sin^2 \theta + 11\sin \theta = 0$

$-6\sin^2 \theta + 11\sin \theta +10 = 0$

Additional step of multiplying every term by -1:

$6\sin^2 \theta - 11\sin \theta -10 = 0$