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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: droodles on January 31, 2008, 06:02:32 pm

Title: i need halp plez
Post by: droodles on January 31, 2008, 06:02:32 pm: $y=2sin(x-\frac{\pi}{4}) + 1$ between $[-2{\pi},2{\pi}]$

x - ints

what
Title: Re: i need halp plez
Post by: dcc on January 31, 2008, 06:13:15 pm: x-intercepts are when y=0

$0 = 2sin(x - \dfrac{\pi}{4}) + 1$

Subtracting 1 from both sides:

$-1 = 2sin(x - \dfrac{\pi}{4})$

Dividing by 2:

$sin(x - \dfrac{\pi}{4}) = -\dfrac{1}{2}$

$\mbox{Let u }= x - \dfrac{\pi}{4}$

$sin(u) = -\dfrac{1}{2}$

Now to figure out these values, you can use ArcSin() to figure out a value, but you should just know the unit circle and that $sin(\dfrac{\pi}{6}) = \dfrac{1}{2}$ and how to relate this to negative 0.5

$u = -\dfrac{13\pi}{6} , -\dfrac{5\pi}{6} , -\dfrac{\pi}{6} , \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$

$x - \dfrac{\pi}{4} = -\dfrac{13\pi}{6} , -\dfrac{5\pi}{6} , -\dfrac{\pi}{6} , \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$

Note we remove the solution $x = \dfrac{11\pi}{6} + \dfrac{\pi}{4}$ as it is out of the domain $[-2\pi, 2\pi]$

$x = -\dfrac{23\pi}{12}, -\dfrac{7\pi}{12}, \dfrac{\pi}{12}, \dfrac{17\pi}{12}$
Title: Re: i need halp plez
Post by: droodles on January 31, 2008, 06:18:11 pm: PUT IT IN RADIANS OR I WILL BASH U
Title: Re: i need halp plez
Post by: droodles on January 31, 2008, 06:37:04 pm: i like that in a man
Title: Re: i need halp plez
Post by: droodles on January 31, 2008, 07:45:20 pm: what is the y-intercept of

$2cos3(x-\frac{\pi}{4})$
Title: Re: i need halp plez
Post by: Collin Li on January 31, 2008, 07:56:09 pm: The y-intercept is when $x=0$ :

$\implies 2\cos \frac{3\pi}{4} = -2\cdot\frac{\sqrt{2}}{2} = -\sqrt{2}$
Title: Re: i need halp plez
Post by: bucket on January 31, 2008, 07:58:18 pm: i love you math fanatics.
Title: Re: i need halp plez
Post by: ?HI? on January 31, 2008, 08:15:14 pm: just in case ppl don't know, I'd like to explain a step in Coblin's working (hope you don't mind ;))...note that this is one way of understanding it, among several other ways.

when x=0, you get: 2cos(-3pi/4)

however, due to symmetry, cos(-x)=cos(x) --> this can be seen by drawing the unit circle and triangles in it

thus you get what coblin has --> 2cos(3pi/4)

hehe, just in case ppl don't see that link
Title: Re: i need halp plez
Post by: shaun1990 on February 01, 2008, 08:04:34 pm: hey jeff, u work at subway in cranny and used to work at doveton makkaz didnt ya?
Title: Re: i need halp plez
Post by: kido_1 on February 02, 2008, 10:44:11 am: Try to memorise exact values, symmetry, and do a lot of those questions and you'll easily get the hang of trig.
Title: Re: i need halp plez
Post by: nak on February 02, 2008, 03:29:50 pm: remember the 2 triangles and that circle thingy. (forgot the name.. its been a while :P)
Title: Re: i need halp plez
Post by: Collin Li on February 02, 2008, 03:56:01 pm: The unit circle. What "2 triangles" are you talking about?
Title: Re: i need halp plez
Post by: Mao on February 02, 2008, 04:04:28 pm: probably the way he visualises the symmetry, draw two triangles in the appropriate quadrants...?
Title: Re: i need halp plez
Post by: nak on February 02, 2008, 04:14:10 pm: ummm ill draw them on paint..
Title: Re: i need halp plez
Post by: nak on February 02, 2008, 04:20:16 pm: [img]http://www.uploadgeek.com/uploads456/0/tri.JPG[/img]
here's the triangles...
Title: Re: i need halp plez
Post by: midas_touch on February 02, 2008, 05:26:23 pm: nice artwork there :P.
Title: Re: i need halp plez
Post by: Collin Li on February 02, 2008, 06:26:23 pm: Oh! I don't bother with that, I prefer to remember the exact values by rote. It's really easy because you just remember it like this:

$\sin 0 = \frac{\sqrt{0}}{2} = 0$

$\sin \frac{\pi}{6} = \frac{\sqrt{1}}{2} = \frac{1}{2}$

$\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$

$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$

$\sin \frac{\pi}{2} = \frac{\sqrt{4}}{2} = 1$

Notice how it progresses like $\frac{\sqrt{n}}{2}$ for $n = 0,1,2,3,4$ ?

$\cos x$ just goes in the opposite direction, and $\tan x$ is defined by both $\sin x$ and $\cos x$ .
Title: Re: i need halp plez
Post by: nak on February 02, 2008, 06:30:06 pm: Quote from: midas_touch on February 02, 2008, 05:26:23 pm
nice artwork there :P.

hehhe i love paint :D
Title: Re: i need halp plez
Post by: kido_1 on February 02, 2008, 09:30:16 pm: Oh, I thought you meant the symettry of the two triangles on the Unit Circle.
Title: Re: i need halp plez
Post by: dcc on February 02, 2008, 10:50:36 pm: Another way of remembering exact values:

Imagine you have your left arm outstretched in front of you, with your palm facing your face (so you can see the insides of your hand). Now, Imagine you wanted to find $sin(\dfrac{\pi}{6})$ . Put down the first finger (your index finger) on your hand and then, count the number of fingers to the left of that (in this case, there is only 1 finger to the left of your index finger. Now, what you next do, is take the number of fingers on the left, square root it, and divide by two.

I.e. For $sin(\dfrac{\pi}{6})$ :

1 finger to the left, so

$sin(\dfrac{\pi}{6}) = \dfrac{\sqrt{1}}{2} = \dfrac{1}{2}$

Now lets say you wanted to find $sin(\dfrac{\pi}{4})$ :

There are two fingers to the left (as your middle finger is $\dfrac{\pi}{4}$ ), so:

$sin(\dfrac{\pi}{4}) = \dfrac{\sqrt{2}}{2}$

Now, Index = pi/6, Middle = pi/4, ring = pi/3:

If you wanted to find a Cosine instead of a sine, its simple, you just count the number of fingers to the RIGHT of the finger which you wish to find:

i.e. to find $cos(\dfrac{\pi}{3})$ :

There is 1 finger to the right of your ring finger, so we have:

$cos(\dfrac{\pi}{3}) = \dfrac{\sqrt{1}}{2} = \dfrac{1}{2}$

Now, whats even better is finding the Tangent of an angle, since we know that:
$tan(x) = \dfrac{sin(x)}{cos(x)}$

Then we can figure out $tan(x)$ by counting the number of fingers to the left of the specified finger, square rooting it, then dividing it by the square root of the number of fingers to the right of the specified finger.

i.e. to find $tan(\dfrac{\pi}{4})$ :

There are two fingers to the left of your middle finger, and two to the right so:

$tan(\dfrac{\pi}{4}) = \dfrac{\sqrt{2}}{\sqrt{2}} = 1$
Title: Re: i need halp plez
Post by: unknown id on February 02, 2008, 11:04:49 pm: wow, thats cool! how'd u figure that out? :S
Title: Re: i need halp plez
Post by: kido_1 on February 02, 2008, 11:05:45 pm: Yeh, that is pretty cool.
I just memorised exact values by doing a load of questions.
Title: Re: i need halp plez
Post by: Collin Li on February 02, 2008, 11:21:15 pm: Quote from: unknown id on February 02, 2008, 11:04:49 pm
wow, thats cool! how'd u figure that out? :S

It's just a logical consequence of the rules I stated above, except it cleverly converts that counting onto your hand.

Quote from: kido_1 on February 02, 2008, 11:05:45 pm
Yeh, that is pretty cool.
I just memorised exact values by doing a load of questions.

Yeah, me too. I recommend this way. I think in both degrees and radians, they are second-nature to me (including their exact values).
Title: Re: i need halp plez
Post by: dcc on February 02, 2008, 11:24:47 pm: Quote from: coblin on February 02, 2008, 11:21:15 pm
Quote from: unknown id on February 02, 2008, 11:04:49 pm
wow, thats cool! how'd u figure that out? :S

It's just a logical consequence of the rules I stated above, except it cleverly converts that counting onto your hand.

No, it's just that i'm amazing, and coblin isn't.
Title: Re: i need halp plez
Post by: AppleXY on February 02, 2008, 11:29:41 pm: Quote from: coblin on February 02, 2008, 11:21:15 pm
Quote from: unknown id on February 02, 2008, 11:04:49 pm
wow, thats cool! how'd u figure that out? :S

It's just a logical consequence of the rules I stated above, except it cleverly converts that counting onto your hand.

Quote from: kido_1 on February 02, 2008, 11:05:45 pm
Yeh, that is pretty cool.
I just memorised exact values by doing a load of questions.

Yeah, me too. I recommend this way. I think in both degrees and radians, they are second-nature to me (including their exact values).

Yeah. I just memorised it. I knew if $\sin(\frac{\pi}{6})$ = $\frac{1}{2}$ then $\cos(\frac{\pi}{6})$ would be $\dfrac{\sqrt{3}}{2}$ .
Title: Re: i need halp plez
Post by: Collin Li on February 02, 2008, 11:46:01 pm: Quote from: AppleXY on February 02, 2008, 11:29:41 pm
Yeah. I just memorised it. I knew if $\sin(\frac{\pi}{6})$ = $\frac{1}{2}$ then $\cos(\frac{\pi}{6})$ would be $\dfrac{\sqrt{3}}{2}$ .

Just a note on LaTeX, instead of using: $\sin(\frac{\pi}{6})$ , try:

Code: [Select]
[tex]\sin \left( \frac{\pi}{6} \right)[/tex] $\sin \left( \frac{\pi}{6} \right)$

The brackets re-size with the largest thing, so it looks a lot nicer.