Author Topic: Circular functions help (Read 3049 times) Tweet Share

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 =]

pHysiX · « **Reply #1 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 $\frac{5\pi}{6}$

d0minicz · « **Reply #2 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.

/0 · « **Reply #3 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, $b \in \left(0, \frac{\pi}{2}\right)$

$\therefore k =0$

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

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

Glockmeister · « **Reply #4 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?

Mao · « **Reply #5 on:** April 04, 2009, 10:58:25 pm »

along with every other topic?

Glockmeister · « **Reply #6 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.

Mao · « **Reply #7 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

/0 · « **Reply #8 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)

d0minicz · « **Reply #9 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 =]

Flaming_Arrow · « **Reply #10 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

Raymond · « **Reply #11 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

d0minicz · « **Reply #12 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!

TrueTears · « **Reply #13 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.

Flaming_Arrow · « **Reply #14 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

Search the forums now!

We have moved!

Author Topic: Circular functions help (Read 3049 times) Tweet Share

d0minicz

Circular functions help

pHysiX

Re: Circular functions help

d0minicz

Re: Circular functions help

/0

Re: Circular functions help

Glockmeister

Re: Circular functions help

Mao

Re: Circular functions help

Glockmeister

Re: Circular functions help

Mao

Re: Circular functions help

/0

Re: Circular functions help

d0minicz

Re: Circular functions help

Flaming_Arrow

Re: Circular functions help

Raymond

Re: Circular functions help

d0minicz

Re: Circular functions help

TrueTears

Re: Circular functions help

Flaming_Arrow

Re: Circular functions help

Recent Posts

User:
Password: