Author Topic: Holiday homeworkk (Read 10660 times) Tweet Share

mandy · « **Reply #30 on:** December 10, 2009, 11:10:13 pm »

Quote from: mandy on December 06, 2009, 09:53:05 pm

DOUBLE ANGLE FORMULAE

Find x such that:

4sinx^o = secx^o, 0^o<x<360^o

- Note: These signs "< " are meant to be less than or equal to signs, but I don't know how to make them :[

I'm trying to do this question again, and I have no idea how to do it. I don't understand crappy's explanation anymore. I got up to the sin2x = 1/2 bit.

The answers are: 15, 75, 195, 255 :S

Cataclysmic · « **Reply #31 on:** December 10, 2009, 11:17:06 pm »

$sec(x) = \frac{1}{cosx}$

so $\frac{1}{4sinx} = cos{x}$

$1 = 4sinxcosx$

Formula: $sin(2x) = 2sinxcosx$

so $1 = 2sin(2x)$

$1/2 = sin(2x)$

$sin^{-1}(1/2) = 2x$ over the domain of [0,2pi]

$2x=\frac{\pi}{6},\frac{5\pi}{6},\frac{13\pi}{6},\frac{17\pi}{6}$

$2x = 30, 150, 390, 510$

$x = 15,75,75,195,255$

(to know where you stop finding solutions, you can see that it's 2x, so all the solutions will be divided by 1/2 so you can expand the domain to [0*2,2pi*2] = [0,4pi] and find solutions within that domain ( thats why I've got it going to 510 degrees)

btw, don't you use radians?

mandy · « **Reply #32 on:** December 10, 2009, 11:23:12 pm »

Thanks so much Cataclysmic, it makes sense now, hahaa.
Um, I have no idea why this qn isnt in radians, but it usually is, lol :]

mandy · « **Reply #33 on:** December 11, 2009, 12:15:40 am »

I've got a really evil qn.

Find x such that:
4tanx = 6tan2x, 0 $\leq$ x $\leq$ 2 $\pi$

This is what I did, have no idea really:
4tanx = 6 $(\frac{2tanx}{1-tan^2x})$

4tanx = $\frac{12tanx}{1-tan^2x}$

$(4tanx)(1-tan^2x)$ = $12tanx$

$4tanx - 4tan^3x - 12tanx = 0$

$-4tan^3x - 8tanx = 0$

$-4tanx(tan^2x + 2) = 0$

Let $tanx = a,$
$-4a(a^2 + 2) = 0$

$-4a^3 - 8a = 0$

And then solve for a ?
I don't even know if this way is real, I just had a try :[

crappy · « **Reply #34 on:** December 11, 2009, 12:27:37 am »

Quote from: mandy on December 11, 2009, 12:15:40 am

I've got a really evil qn.

Find x such that:
4tanx = 6tan2x, 0 $\leq$ x $\leq$ 2 $\pi$

This is what I did, have no idea really:
4tanx = 6 $(\frac{2tanx}{1-tan^2x})$

4tanx = $\frac{12tanx}{1-tan^2x}$

$(4tanx)(1-tan^2x)$ = $12tanx$

$4tanx - 4tan^3x - 12tanx = 0$

$-4tan^3x - 8tanx = 0$

$-4tanx(tan^2x - 2) = 0$

Let $tanx = a,$
$-4a(a^2 - 2) = 0$

$-4a^3 + 8a = 0$

And then solve for a ?
I don't even know if this way is real, I just had a try :[

yeah, what you've done is perfectly fine. once you find the solutions for that equation, which is 0, root(2), and -root(2), you then make:
tan x=0
tan x=root(2)
tan x=-root(2)

and solve for x, and then take the solutions that fit in the domain

mandy · « **Reply #35 on:** December 11, 2009, 12:32:36 am »

Does that give the answers 0, $\pi$ , $2\pi$ ?
Because I don't know how to get to those.

crappy · « **Reply #36 on:** December 11, 2009, 12:36:23 am »

Quote from: mandy on December 11, 2009, 12:32:36 am

Does that give the answers 0, $\pi$ , $2\pi$ ?
Because I don't know how to get to those.

Yeah that should be right, although the calc didnt give me 2pi, but by subbing it in, it is right.
Umm, which part are you unsure of? solving for tanx or solving that quadratic eqn?

mandy · « **Reply #37 on:** December 11, 2009, 12:38:48 am »

Quote from: crappy on December 11, 2009, 12:36:23 am

Quote from: mandy on December 11, 2009, 12:32:36 am
Does that give the answers 0, $\pi$ , $2\pi$ ?
Because I don't know how to get to those.

Yeah that should be right, although the calc didnt give me 2pi, but by subbing it in, it is right.
Umm, which part are you unsure of? solving for tanx or solving that quadratic eqn?

Nah, I know how to solve for tanx, I got what you did.
But how does that turn into 0, $\pi$ , $2\pi$ ?

When I put it into the calc, I got all these weird decimals.

crappy · « **Reply #38 on:** December 11, 2009, 12:44:45 am »

oh lol, I think you made a mistake in your original working, check the part where you took out 4tanx. It should be +2, not -2.

either this, or my brain is fried lol

crappy · « **Reply #39 on:** December 11, 2009, 12:46:07 am »

which would be mean that 0 is the only solution hence, x=0, pi, 2pi

mandy · « **Reply #40 on:** December 11, 2009, 12:49:12 am »

Quote from: crappy on December 11, 2009, 12:44:45 am

oh lol, I think you made a mistake in your original working, check the part where you took out 4tanx. It should be +2, not -2.

either this, or my brain is fried lol

lmao, yeah its meant to be +2. gosh, its been a long day of math and im dead, haha.
okaaay, thanks alot for your help! makes sense now :]

trinon · « **Reply #41 on:** December 11, 2009, 03:35:21 am »

lol mandy you confused me so much with your working. I thought that what you did was right after someone else approved it, and I couldn't figure out why I was constantly getting +2 and not -2.

crappy · « **Reply #42 on:** December 11, 2009, 04:52:20 am »

my bad, I thought her working looked right so I didn't bother checking everything lol

mandy · « **Reply #43 on:** December 11, 2009, 09:13:37 pm »

Quote from: trinon on December 11, 2009, 03:35:21 am

lol mandy you confused me so much with your working. I thought that what you did was right after someone else approved it, and I couldn't figure out why I was constantly getting +2 and not -2.

Quote from: crappy on December 11, 2009, 04:52:20 am

my bad, I thought her working looked right so I didn't bother checking everything lol

Sorry guys, thanks for picking it up :]

mandy · « **Reply #44 on:** December 11, 2009, 09:38:40 pm »

Using half angle formulae, find the values of $cos\frac{\pi}{8}$ , $sin\frac{\pi}{8}$ and tan $\frac{\pi}{8}$ in simplest form.

With this question, I have no idea what they're even asking me to do or find. And I've never even heard of the 'half angle formulae', aren't I doing the Double Angle Formulae exercise now ...
Help me please :[

Search the forums now!

We have moved!

Author Topic: Holiday homeworkk (Read 10660 times) Tweet Share

mandy

Re: Holiday homeworkk

Cataclysmic

Re: Holiday homeworkk

mandy

Re: Holiday homeworkk

mandy

Re: Holiday homeworkk

crappy

Re: Holiday homeworkk

mandy

Re: Holiday homeworkk

crappy

Re: Holiday homeworkk

mandy

Re: Holiday homeworkk

crappy

Re: Holiday homeworkk

crappy

Re: Holiday homeworkk

mandy

Re: Holiday homeworkk

trinon

Re: Holiday homeworkk

crappy

Re: Holiday homeworkk

mandy

Re: Holiday homeworkk

mandy

Re: Holiday homeworkk

Recent Posts

User:
Password: