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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: d0minicz on February 22, 2009, 10:52:50 am

Title: Circular functions help
Post by: d0minicz on February 22, 2009, 10:52:50 am: For $\frac {\pi}{2} \ {<} \ {x} \ {<} {\pi}$ with $\sin {x} = \sin \frac {\pi}{6}$ find the value of x (do not evaluate $\sin \frac {\pi}{6}$ )

thanks =]
Title: Re: Circular functions help
Post by: pHysiX on February 22, 2009, 11:38:25 am: ok.

so from the expression, we should note that $x=\frac{\pi}{6}$

now this is the first quadrant solution. the domain is for a quadrant 2 angle.

therefore, angle = $\pi-x$
which is $\frac{5\pi}{6}$
Title: Re: Circular functions help
Post by: d0minicz on April 04, 2009, 03:11:47 pm: For $\frac {\pi}{2} < a < \pi$ with $sin(a) = cos(b)$ where $0<b< \frac {\pi}{2}$ , find $a$ in terms of $b$ .
thanks.
Title: Re: Circular functions help
Post by: /0 on April 04, 2009, 08:21:36 pm: $\sin{a}=\cos{b} \Rightarrow\cos{\left(-\frac{\pi}{2}+a\right)}=\cos{b}$

$\therefore -\frac{\pi}{2} + a = b + 2k\pi \ , \ k \in Z$

But $-\frac{\pi}{2}+a \in \left(0, \frac{\pi}{2}\right)$

Also, $b \in \left(0, \frac{\pi}{2}\right)$

$\therefore k =0$

$\Rightarrow -\frac{\pi}{2}+a = b$

$\therefore a = \frac{\pi}{2}+b$
Title: Re: Circular functions help
Post by: Glockmeister on April 04, 2009, 10:40:50 pm: Quote from: /0 on April 04, 2009, 08:21:36 pm
$\sin{a}=\cos{b} \Rightarrow\cos{\left(-\frac{\pi}{2}+a\right)}=\cos{b}$

$\therefore -\frac{\pi}{2} + a = b + 2k\pi \ , \ k \in Z$

But $-\frac{\pi}{2}+a \in \left(0, \frac{\pi}{2}\right)$

Also, $b \in \left(0, \frac{\pi}{2}\right)$

$\therefore k =0$

$\Rightarrow -\frac{\pi}{2}+a = b$

$\therefore a = \frac{\pi}{2}+b$

yeah we seem to get this sort of question every year. Perhaps someone should sticky this up?
Title: Re: Circular functions help
Post by: Mao on April 04, 2009, 10:58:25 pm: along with every other topic? :P
Title: Re: Circular functions help
Post by: Glockmeister on April 05, 2009, 01:50:21 am: no no, im just saying, this particular question

sin a = cos b , find a in terms of b question (or its variations) seems to always pop up.
Title: Re: Circular functions help
Post by: Mao on April 05, 2009, 02:05:37 am: but everything else pop up too! like transformations, differentiations, etc etc

maybe we should write a FAQ for all of these but then we'd just be creating a new textbook/study-guide, lol
Title: Re: Circular functions help
Post by: /0 on April 05, 2009, 02:08:04 am: This particular variation took a bit of thinking.

Normally I can just say $\sin a = \cos b \Rightarrow \cos{\left(\frac{\pi}{2}-a\right)} = \cos b$ but that doesn't work here. (As far as I can tell)
Title: Re: Circular functions help
Post by: d0minicz on May 07, 2009, 08:11:06 pm: The satisfaction factor of a customer can be modelled by :
S= $s= sin\frac {\pi x}{n}$ , where s is the satisfaction factor, n is the number of tomatoes purchased and x is the cost in dollars.
1. For an outlay of $10, determine the numbers of tomatoes that a customer should buy for a satisfaction factor of AT LEAST 0.75.
2. How much should 30 tomatoes cost a customer if they wish to have a satisfaction factor of at least 0.90.

i can get half of the answers, jjust the other halves i cant get =="
thanks =]
Title: Re: Circular functions help
Post by: Flaming_Arrow on May 07, 2009, 08:16:13 pm: for 1 you should do

$0.75 \le \sin\frac{10\pi}{n}$

$\sin^{-1}(0.75) \le \frac{10\pi}{n}$ and $(\pi-\sin^{-1}(0.75)) \le \frac{10\pi}{n}$ because sin is positive in quadrants 1 and 2 i think you can do the rest
Title: Re: Circular functions help
Post by: Raymond on May 13, 2009, 04:04:46 pm: y(t)=12+e^(kt)sin (pi(t)/4) find the largest possible value of k in exact values
Title: Re: Circular functions help
Post by: d0minicz on August 21, 2009, 09:48:07 pm: How do i work out the range and domain of a function like $y = -3|sin2x| + 3$ ?

thanks!
Title: Re: Circular functions help
Post by: TrueTears on August 21, 2009, 09:49:24 pm: Do a quick sketch yeah?

just do sin(2x), flip the negative part over the positive side, then apply transformations.
Title: Re: Circular functions help
Post by: Flaming_Arrow on August 21, 2009, 09:54:25 pm: isn't domain just R?
Range is just [0,3]

to work out the range i would look at the max/min values of sin which is 1/0

sub them in for sin 2x to get ur range
Title: Re: Circular functions help
Post by: kamil9876 on August 21, 2009, 10:09:28 pm: Because TT despises graphs and adores algebraic alternatives:
$<br />-1\leq sin2x \leq 1$ (can take on any values in this range)

$0 \leq |sin2x| \leq 1$
$-3 \leq -3|sin2x| \leq 0$

adding 3 to both sides:

$0 \leq -3|sin2x| + 3 \leq 3$
Title: Re: Circular functions help
Post by: TrueTears on August 21, 2009, 10:10:02 pm: Quote from: kamil9876 on August 21, 2009, 10:09:28 pm
Because TT despises graphs and adores algebraic alternatives:
$<br />-1\leq sin2x \leq 1$ (can take on any values in this range)

$0 \leq |sin2x| \leq 1$
$-3 \leq -3|sin2x| \leq 0$

adding 3 to both sides:

$0 \leq -3|sin2x| + 3 \leq 3$
LOL I just showed him that on msn :P