E-xamit project maths 2013 Sample Papers Help

Leaving Cert Student

Hopefully this can be a thread for discussing the sample papers solutions because some of the questions don't have the answers anywhere which is rather frustrating...

I'll start with pg. 87, Q1 (B) de moivres theorem. I can't seem to make sense of it, and the answer is nowhere because it's a "show how" type of question.
Any help greatly appreciated.

StolzundStark

To Prove: cos3A = 4cos^3A -3cosA

Start with (cosA+isinA)^3
Using our friend De Moivre's method that gives us:
Cos3A + iSin3A

and we also know (a + b)3 = a3 + 3 a2 b + 3 a b2 + b3
so (cosA+isinA)^3 = cos^3A + 3cos^2AisinA + 3cosAi^2sin^2A +i^3sin^3A
Clean up the i^2 and make it -1 and equate the (cosA+isinA)^3 s giving us:

Cos3A + iSin3A = cos^3A + 3cos^2AisinA - 3cosAsin^2A -isin^3A
Now the part most people seem to forget. Equate like with like, ie real with real and imaginary with imaginary.

The Real: Cos3A = Cos^3A -3cosAsin^2A and we can ignore the imaginary as we are only looking for Cos3A. Look what we want to achieve in the proof and see what we have to change to get cos3A = 4cos^3A -3cosA. Simply change sin^2 to 1-cos^2. Now we have
Cos3A = Cos^3A - 3cosA(1-cos^2A), multiply out and you end up with
Cos3A = 4cos^3A-3cosA

Leaving Cert Student

StolzundStark wrote: »

To Prove: cos3A = 4cos^3A -3cosA

Start with (cosA+isinA)^3
Using our friend De Moivre's method that gives us:
Cos3A + iSin3A

and we also know (a + b)3 = a3 + 3 a2 b + 3 a b2 + b3
so (cosA+isinA)^3 = cos^3A + 3cos^2AisinA + 3cosAi^2sin^2A +i^3sin^3A
Clean up the i^2 and make it -1 and equate the (cosA+isinA)^3 s giving us:

Cos3A + iSin3A = cos^3A + 3cos^2AisinA - 3cosAsin^2A -isin^3A
Now the part most people seem to forget. Equate like with like, ie real with real and imaginary with imaginary.

The Real: Cos3A = Cos^3A -3cosAsin^2A and we can ignore the imaginary as we are only looking for Cos3A. Look what we want to achieve in the proof and see what we have to change to get cos3A = 4cos^3A -3cosA. Simply change sin^2 to 1-cos^2. Now we have
Cos3A = Cos^3A - 3cosA(1-cos^2A), multiply out and you end up with
Cos3A = 4cos^3A-3cosA

Nice very clean solution, I get it now just couldn't understand how to start the thing given that the question itself had no i terms in it, makes sense now cheers!

StolzundStark

No bother man, need help with any others?

Leaving Cert Student

Another one that is bugging me is q(b) on the next page 88, if anyone has any insight? I get c as an an answer and have no idea where c^2 seems to come into it?