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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: dooodyo on April 17, 2011, 11:17:11 pm

Title: Dooodyo's noob questions
Post by: dooodyo on April 17, 2011, 11:17:11 pm: Hey guys,

Just have one question how would you find all intercepts for sin 3x =1 for ( -2pie, 2pie) ?

THanks in advance
Title: Re: Dooodyo's noob questions
Post by: luken93 on April 17, 2011, 11:21:50 pm: 3x = sin^-1(1)
3x = pi/2
x = pi/6

Then add/subtract period (2pi/3)

x = pi/6 - 4pi/6, pi/6, pi/6 + 4pi/6
= -3pi/6, pi/6, 5pi/6
Title: Re: Dooodyo's noob questions
Post by: Greatness on April 17, 2011, 11:22:03 pm: sin(3x)=1 has a reference angle of pi/2
therefore: 3x=pi/2
x=pi/6 -now to get it within the domain. the period is 2pi/3, so add & minus 4pi/6 from pi/6
thus: x= -11/pi, -7pi/6, -pi/2, pi/6, 5pi/6, 3pi/2
Title: Re: Dooodyo's noob questions
Post by: dooodyo on April 17, 2011, 11:25:22 pm: Oh haha thanks heaps Swarley and Luken

And do you always add or subtract period?

Or sometimes I thought you just add or subtract 2pie from the 1st quad angle? soooo confused :-\
Title: Re: Dooodyo's noob questions
Post by: Greatness on April 17, 2011, 11:29:01 pm: It depends what you want to do. If you were to find solutions in a restricted domain then find the first 2 solutions, then you can add, subtract the period from/to the solutions. It just depends on what thte domain is.
If you have 23pi/4 but the questoin wants it in [0,2pi] then 13pi/6 - 2pi = pi/6.
Title: Re: Dooodyo's noob questions
Post by: dooodyo on April 17, 2011, 11:31:30 pm: Wait so does that mean you add 2pie or add the period each time?
Title: Re: Dooodyo's noob questions
Post by: Greatness on April 17, 2011, 11:36:09 pm: oh sry for the confusion. When you're finding solutions you add the period. However if you want to find an equvialent angle then you add/subtract multiples of 2pi.
I thinks that right
Title: Re: Dooodyo's noob questions
Post by: dooodyo on April 25, 2011, 04:51:23 pm: Hey guys,

I have another noob question any help would be appreciated.

How would you find the general solution to sqrt(3)tan(pi/6-3x)-1=0 ?

why is it (pi x n)/3 and not (-pi x n)/3 ?
Title: Re: Dooodyo's noob questions
Post by: brightsky on April 25, 2011, 05:09:39 pm: They are the same, since n is any integer.
Title: Re: Dooodyo's noob questions
Post by: Aurelian on April 25, 2011, 05:59:31 pm: Quote from: dooodyo on April 25, 2011, 04:51:23 pm
Hey guys,

I have another noob question any help would be appreciated.

How would you find the general solution to sqrt(3)tan(pi/6-3x)-1=0 ?

why is it (pi x n)/3 and not (-pi x n)/3 ?
Quote from: brightsky on April 25, 2011, 05:09:39 pm
They are the same, since n is any integer.

Provided your solution is followed by "n E Z"

You could, however, say +/- both those options, but you'd need to say "n E N U O"
Title: Re: Dooodyo's noob questions
Post by: brightsky on April 25, 2011, 06:04:27 pm: Quote from: Aurelian on April 25, 2011, 05:59:31 pm
You could, however, say +/- both those options, but you'd need to say "n E N U O"

There are infinite many ways to write the 'general solution'. I'd say just write one, with n E Z at the end as Aurelian said. I'm pretty sure all the 'variations' would be accepted, as all of them are technically correct.
Title: Re: Dooodyo's noob questions
Post by: Aurelian on April 25, 2011, 06:11:52 pm: Yeah, I'd stick to the n E Z variation like brightsky said - less room for silly mistakes.

Also, this might be of help;

http://vce.atarnotes.com/forum/index.php/topic,35962.msg382191.html#msg382191
Title: Re: Dooodyo's noob questions
Post by: dooodyo on April 25, 2011, 10:14:42 pm: wait does " n E N U O " mean element of natural no.s union 0 ?
Title: Re: Dooodyo's noob questions
Post by: Aurelian on April 25, 2011, 11:16:56 pm: Quote from: dooodyo on April 25, 2011, 10:14:42 pm
wait does " n E N U O " mean element of natural no.s union 0 ?

Yep :) I'd have used latex but I don't know how haha sorry :P
Title: Re: Dooodyo's noob questions
Post by: dooodyo on April 26, 2011, 10:40:23 am: Thanks heaps :D
Title: Re: Dooodyo's noob questions
Post by: dooodyo on May 14, 2011, 07:15:26 pm: Hey guys stuck on this question :tickedoff:

how would you prove (1 - sin2x)/(sinx - cosx) = sinx - cosx ?

Any help would be much appreciated.
Title: Re: Dooodyo's noob questions
Post by: luken93 on May 14, 2011, 07:38:27 pm: $\frac{1 - sin(2x)}{sin(x) - cos(x)}$

Using $sin^2(x) + cos^2(x) = 1$ and $sin(2x) = 2sin(x)cos(x)$

$\frac{sin^2(x) + cos^2(x) - 2sin(x)cos(x)}{sin(x) - cos(x)}$

Recognising that it's DOPS

$\frac{(sin(x) - cos(x))^2}{sin(x) - cos(x)}$

$sin(x) - cos(x)$ , As required.
Title: Re: Dooodyo's noob questions
Post by: dooodyo on May 14, 2011, 07:39:52 pm: Oh haha cool thanks heaaaaaaapss :D