ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Studyinghard on March 26, 2011, 12:18:51 pm

Title: Forgotten basic methods!
Post by: Studyinghard on March 26, 2011, 12:18:51 pm: Solve |sin(x/3)| = 1/2

Completely blanked on the general formula for solving this :S
Title: Re: Forgotten basic methods!
Post by: pi on March 26, 2011, 12:42:00 pm: $|\sin(\frac{x}{3})| = \frac{1}{2}$
$\sqrt{(\sin(\frac{x}{3}))^2} = \frac{1}{2}$
$(\sin(\frac{x}{3}))^2 = \frac{1}{4}$
$\sin(\frac{x}{3}) = \frac{1}{2} \ or \ \sin(\frac{x}{3}) = -\frac{1}{2}$
$\frac{x}{3} = \frac{\pi}{6}, \frac{5\pi}{6}, ... \ or \ \frac{x}{3} = \frac{7\pi}{6}, \frac{11\pi}{6}, ...$
$x = \frac{\pi}{2}, \frac{5\pi}{2}, ... \ or \ x = \frac{7\pi}{2}, \frac{11\pi}{2}, ...$
$x = \frac{\pi}{2} + 3n\pi \ or \ x = \frac{5\pi}{2} + 3n\pi, \ n \in Z \$

I hope thats right (there are infinite solutions as you didn't specify a domain, but just add or subtract $3n\pi$ , as shown), I too haven't used the general formula for ages
Title: Re: Forgotten basic methods!
Post by: Studyinghard on March 26, 2011, 12:58:47 pm: dont you have to add or subtract the period in the end, and the period is 6pi
Title: Re: Forgotten basic methods!
Post by: pi on March 26, 2011, 01:03:52 pm: Quote from: Studyinghard on March 26, 2011, 12:58:47 pm
dont you have to add or subtract the period in the end, and the period is 6pi

oops, knew I was forgetting something. Fixed
Title: Re: Forgotten basic methods!
Post by: taiga on March 26, 2011, 01:06:47 pm: ENG 1090? :P

If you think about it; it pretty much means that whatever is inside the brackets has to equal 1/2 or -1/2 in order for the modulus of it to equal 1/2. Hence Sin(x/3) = 1/2 and sin(x/3) = -1/2 are our two solutions
Title: Re: Forgotten basic methods!
Post by: Studyinghard on March 26, 2011, 01:10:54 pm: wait what where did you get 3npi from in your edited post :S
Title: Re: Forgotten basic methods!
Post by: taiga on March 26, 2011, 01:13:26 pm: I think he means 2npi

[edit] - my bad didnt read his working out, skipped to final line xD
Title: Re: Forgotten basic methods!
Post by: pi on March 26, 2011, 01:13:33 pm: Quote from: Studyinghard on March 26, 2011, 01:10:54 pm
wait what where did you get 3npi from in your edited post :S

Instead of $x = \frac{\pi}{2} + 3n\pi, \frac{5\pi}{2} + 3n\pi \ or \ x = \frac{7\pi}{2} + 3n\pi, \frac{11\pi}{2} + 3n\pi$ , I simplified it down to just two solutions. btw, as it a modulus, the period is actually 3pi (ie x/3 = ...+npi, x = ...+n3pi), not 6pi
Title: Re: Forgotten basic methods!
Post by: Studyinghard on March 26, 2011, 03:51:34 pm: why is the period different for a mod function? When was this rule created :S
Title: Re: Forgotten basic methods!
Post by: pi on March 26, 2011, 03:58:05 pm: Quote from: Studyinghard on March 26, 2011, 03:51:34 pm
why is the period different for a mod function? When was this rule created :S

Because there are no negative ordinates (all negative ordinates turned to positive due to modulus).

The graph looks like a series of 'bumps' (ie no troughs) with a repetition (ie period) of 3pi.

Temporary image:
(http://www4b.wolframalpha.com/Calculate/MSP/MSP382419f0564g3bh0260300003c16fae088c75ci5?MSPStoreType=image/gif&s=36&w=370&h=204)
Title: Re: Forgotten basic methods!
Post by: Menang on March 26, 2011, 04:35:30 pm: There's a reason why I'm so screwed for methods. -_-
Title: Re: Forgotten basic methods!
Post by: Studyinghard on April 09, 2011, 10:37:21 pm: BUMP. just saying that the period was wrong. it was actually 6pi
Title: Re: Forgotten basic methods!
Post by: luken93 on April 10, 2011, 12:16:16 am: Quote from: Studyinghard on April 09, 2011, 10:37:21 pm
BUMP. just saying that the period was wrong. it was actually 6pi
I don't think it's 6pi, think about the graph, it is half of a normal sine graph...

If it were a normal sin(x/3) graph, then yes it would be 6pi, but because it is |sin(x/3)| the period is 6pi/2 = 3pi
Title: Re: Forgotten basic methods!
Post by: Studyinghard on April 10, 2011, 08:20:29 am: Well this was for an assignment and i got it marked wrong. Can we get more confirmations for this so I can argue this ?
Title: Re: Forgotten basic methods!
Post by: TrueTears on April 10, 2011, 05:16:57 pm: $\frac{\frac{2\pi}{\frac{1}{3}}}{2} = 3\pi$ is the period.