Author Topic: over9000 questions thread (Read 18118 times) Tweet Share

TrueTears · « **Reply #45 on:** March 08, 2009, 11:35:35 pm »

switch the x and y's around and solve for y

Over9000 · « **Reply #46 on:** March 08, 2009, 11:39:19 pm »

yeh, kinda knew that, I've already done that . I would just like to see how you did it so I can see if I got the answer right, cant check coz its a past Sac

TrueTears · « **Reply #47 on:** March 08, 2009, 11:42:41 pm »

Well after some workings i got $6(cos(\frac{\pi}{10}x)+1) = f^{-1}x$

Over9000 · « **Reply #48 on:** March 08, 2009, 11:46:35 pm »

Yeh, thats what I got, thanks. Just wanted to make sure I was right.

Over9000 · « **Reply #49 on:** March 09, 2009, 02:14:24 pm »

Did u get this for 1.d
$y= 12(1- cos(\frac{x\pi}{10}))^2$

TrueTears · « **Reply #50 on:** March 09, 2009, 02:20:19 pm »

1d) should be $6(1 - cos(\frac{\pi}{5}x))$

Over9000 · « **Reply #51 on:** March 09, 2009, 02:23:03 pm »

Oh, really, could you please show me how you got that

TrueTears · « **Reply #52 on:** March 09, 2009, 02:30:25 pm »

$(6cos(\frac{\pi}{10}x)+6) \times (2-2cos(\frac{\pi}{10}x)$ = $12(1 - cos^2(\frac{\pi}{10}x))$

but $cos(2(\frac{\pi}{10}x)) = 2cos^2(\frac{\pi}{10}x) -1$

solve for $cos^2(\frac{\pi}{10}x)$ yields $\frac{1}{2}(cos(\frac{\pi}{5}x)+1)$

sub this back in $12(1 - cos^2(\frac{\pi}{10}x))$ yields $12(1 - \frac{1}{2}cos(\frac{\pi}{5}x) - \frac{1}{2}) = 6(1-cos(\frac{\pi}{5}x))$

the questions states you have to put it into $6(1-cos(\frac{\pi}{5}x))$ so you can't just leave it as $12(1 - cos^2(\frac{\pi}{10}x))$

Over9000 · « **Reply #53 on:** March 10, 2009, 05:26:11 pm »

Theres this question, simplify $tan(sin^{-1}(\frac{-2}{3}))$
And for the written out asnwers to this question, it has the method for simplifying it as
$\frac{sin(sin^{-1}(\frac{-2}{3}))}{cos(sin^{-1}(\frac{-2}{3}))}$
Why would you multiply $sin^{-1}(\frac{-2}{3})$ by sin and cos, should you only multiply it by sin, isnt that like $\frac{1}{2}$ multiplied by 3 is $\frac{3}{6}$ which is $\frac{1}{2}$ ?

TrueTears · « **Reply #54 on:** March 10, 2009, 05:30:57 pm »

$tan(sin^{-1}(\frac{-2}{3}))$

so let $sin^{-1}(\frac{-2}{3}) = x$

$sinx = \frac{-2}{3}$

using the identity $\frac{1}{sin^2x} = 1 + \frac{1}{tan^2x}$

solving for tanx yields $tanx = \frac{2\sqrt{5}}{5}$ or $-\frac{2\sqrt{5}}{5}$

But here i'm not sure whether you choose negative or positive...

/0 · « **Reply #55 on:** March 10, 2009, 06:00:12 pm »

If you look at the $y=\sin^{-1}x$ graph, for $x\in \left[-1, 0\right]$ , you will get a negative value for y.

Therefore, the value $\sin^{-1}\left(\frac{-2}{3}\right)$ can take must be somewhere between $\left[-\frac{\pi}{2}, 0\right]$ , which is the fourth quadrant.

Hence, since the angle, $\sin^{-1}\left(\frac{-2}{3}\right)$ is in the fourth quadrant, $\tan{\left(\sin^{-1}{\left(\frac{-2}{3}\right)}\right)$ must be negative.

TrueTears · « **Reply #56 on:** March 10, 2009, 06:01:52 pm »

ah i get it, that clears up my problem too

thanks /0

Over9000 · « **Reply #57 on:** March 10, 2009, 06:02:04 pm »

/0 is right, the value is negative, but anyway, is the method I listed above wrong?

TrueTears · « **Reply #58 on:** March 10, 2009, 06:03:26 pm »

No, it's not wrong, i just find doing it non graphically is easier lol

Over9000 · « **Reply #59 on:** March 10, 2009, 06:05:00 pm »

Ok, thanks for the quick reply 8-)

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