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Assuming 54 actually was an identity, you would do it like any other one, just multiply them. It would look like this: tan(x+45)tan(x-45)=(tanx + tan45)/(1-tanxtan45) TIMES (tanx - tan45)/(1+tanxtan45) =(tanx+1)/(1-tanx) TIMES (tanx-1)/(1+tanx) [tan45=1] =(tan^2(x)-1)/(1-tan^2(x))

All right, give me a minute and I'll type it up... Will update as I go :) The formula I will be using is cos(2x)=2cos^2(x)-1: <a href="http://www.codecogs.com/eqnedit.php?latex=cos4x=cos(2x@plus;2x)\\ =2cos^2(2x)-1 \\ =2(2cos^2(x)-1)^2-1 \\ =2(4cos^4(x)-4cos^2(x)@plus;1)-1 \\...

The problem with that is that the first isn't an identity so by definition you can't prove it. Which book did you get that question from?

Sorry, do you want us to solve these or to prove them as an identity? The reason is that the first one isn't an identity (sub in x=30 for example). 43 though is. The way to do 43 (assuming you don't do 4U and can't use De Moivre's theorem) is to turn cos4x into cos(2x+2x), expand that to get...

Re: 2012 HSC MX2 Marathon The answer to my question, a few posts back

Re: 2012 HSC MX2 Marathon Is this a past HSC question? I remember doing this exact question about two days ago. I'll let someone else post a solution though. With your previous post, the answer is a nice integer. You made a mistake in the second line where you changed (1-w)(1-w)(1-w^2)^2 into...

Re: 2012 HSC MX2 Marathon n=6 (assuming n=0 isn't allowed). Given that w is the complex root of z^3=1 with smallest positive argument, evaluate: (1-w)(1-w^2)(1-w^4)(1-w^8)

You can't just divide both sides by sinx because you remove the solution sinx=0. As a general rule, you shouldn't divide by a function of x unless you know that x=0 isn't a solution. A neater way to do the first one is to do: sin3x-3in2x=0 2cos(5x/2)sin(x/2)=0 (factorising) cos(5x/2)=0 or...

Is the domain restricted or do you want a general answer? The reason is I don't want to spend half an hour typing something in latex (I'm very unfluent) if that's not the answer you're looking for :)

I came first at a selective school and I'm white. The main thing is that (trying to avoid stereotyping) most asians have a totally different perspective to people of European background. Say on the weekend when I might go to the beach, some of my asian friends will ask how I can do that without...

I wish our school (Syd Tech) was like that. We got Terry Lee, but I had to buy my own copy of Cambridge from a previous Yr 12 because our school just had enough for one class...

Honestly though even if you can, I wouldn't as I doubt the markers would look favourably on it as they would assume that you couldn't be bothered to analyse something yourself/lacked originality.

Sydney Tech - Our class uses Terry Lee. Others use a mixture of Patel and Coroneos.

This. I know determine wrote over 2000 words for one of his HSC essays and over 1500 for the other. That, and the fact that they were actually good (read:quality) meant that he got 20/20 for both essays. Form my experience, in my prelim exam I probably wrote about 1500 words and got 20/20. The...

Mate, you sound like a boss. I haven't finished my belonging essay yet but when I do, I'll be sure to swing it past you. Cheers:)

Is this the question you are referring to: Discuss the influence of international organisations and contemporary trading blocs and agreements in promoting globalisation. In your answer, use the economic information provided. (2008 HSC). Regardless, what I would talk about is pretty much that...

Mate those marks are incredible (partcularly english in my opinion). Out of curiosity, roughly how many pages did you write for each eco essay and how did you feel about your essays when you finished?

Re: 2012 HSC MX2 Marathon Thanks guys. The way I did it was Carrotsticks way but I don't get how that can only be a one mark question...

Its like any other graph. Say if you've graphed y=cosx. If you want to graph y=2cosx, then at every y value you multiply it by two. This means that the amplitude will double, but the period will remain the same because obviously when cosx=0, 2cosx will also equal zero. The same principle applies...

Re: 2012 HSC MX2 Marathon The answer is pi/4. The way I did it was find the gradient of the line passing through the origin that was a tangent to the parabola since that would have the minimum argument of z and then used the fact that the gradient of a line equals the tangent of the argument to...