Author Topic: TrueTears question thread (Read 66136 times) Tweet Share

TrueTears · « **on:** December 26, 2008, 02:16:52 am »

I think i might be posting too much threads for just 1 question, so i'll make a separate thread XD (please excuse my noob questions haha). Here goes...

1. Simplify the following expression, in an exact form.

$sin( tan^{-1}(-2)) = \frac{2\sqrt{5}}{5}$

Let $tan^{-1}(-2) = x$

$\implies tan(x)=-2$

$\implies \frac{1}{sin^2(x)}=1 + \frac{1}{(-2)^2}$

$\implies sin^2(x) = 4/5$

$\implies sin(x) = \frac{2\sqrt{5}}{5}$

however the answers in my book say its $-\frac{2\sqrt{5}}{5}$ . How do i get the negative?

2. Let $sec \beta = b$ where $\beta \in [ \frac{\pi}{2},\pi]$ . Find, in terms of $\beta$ , 2 values of x in the range $[-\pi,\pi]$ which satisfies each of these equations:

a) $sec x = -b$

b) $cosec x = b$

Let $tan \gamma = c$ where $\gamma \in[ \pi,\frac{3\pi}{2}]$ . find in terms of $\gamma$ , 2 values of x in the range of $[0,2\pi]$ which satisfies

a) $cot x = c$

3. Given that $tan \alpha = p$ , where $\alpha$ is an acute angle, find each of the following in terms of $p$ .
a) $tan( \frac{3\pi}{2} + \alpha)$

vce08 · « **Reply #1 on:** December 26, 2008, 02:19:40 am »

You get the neg when u square root the sin^2(x) thingy.

Damo17 · « **Reply #2 on:** December 26, 2008, 08:01:08 am »

Quote from: TrueTears on December 26, 2008, 02:16:52 am

$\implies sin^2(x) = 4/5$

$\implies sin(x) = \frac{2\sqrt{5}}{5}$

however the answers in my book say its $-\frac{2\sqrt{5}}{5}$ . How do i get the negative?

$sin^2(x) = 4/5$
$\implies sin(x) =\pm \sqrt\frac{4}{5}$
$= \pm \frac{2\sqrt{5}}{5}$

TrueTears · « **Reply #3 on:** December 26, 2008, 01:27:05 pm »

yeah but why is the positive not one of the answers?

Matt The Rat · « **Reply #4 on:** December 26, 2008, 01:29:49 pm »

It'd be where the $tan^{-1}(-2)$ lies

TrueTears · « **Reply #5 on:** December 26, 2008, 01:43:24 pm »

but isn't $tan$ negative in the 2nd or 4th quadrant, so $tan(x)=-2$ could be either the 2nd or 4th quadrant. $sin$ is also positive in 2nd quadrant and negative in 4th quadrant, so shouldn't both positive and negative answers be included?

Matt The Rat · « **Reply #6 on:** December 26, 2008, 01:59:55 pm »

Remember where $tan^{-1}(x)$ is defined as a function ( $Range: (\frac{-\pi}{2},\frac{\pi}{2})$ ) and as that leaves only the fourth quadrant as possible then it must be a negative answer.

TrueTears · « **Reply #7 on:** December 26, 2008, 02:01:02 pm »

ahhh thanks Matt The Rat

Mao · « **Reply #8 on:** December 26, 2008, 02:03:52 pm »

quite right. but remember, the range of arctan is $\left( -\frac{\pi}{2},\frac{\pi}{2}\right)$ , so technically, you don't actually have the second quadrant to work with.

remember that arcsin and arctan only spits out values in 4th and 1st Q, and arccos only gives 1st and 2nd Q

Mao · « **Reply #9 on:** December 26, 2008, 02:09:02 pm »

the rest of the questions aren't hard to do, but very hard to explain without a drawing instrument, and i do not have my tablet nearby...
but the easiest way to do it is to draw graphs, then set up hypothetical points and see what works

TrueTears · « **Reply #10 on:** December 27, 2008, 03:49:14 pm »

Simplify $(cis(\frac{\pi}{2} - \alpha))^7$

$\implies cis( \frac{7\pi}{2} - 7\alpha)$

$\implies \frac{cis(\frac{7\pi}{2})}{cis(7\alpha)}$

$\implies \frac{cis(\frac{-\pi}{2})}{cis(7\alpha)}$

$\implies cis( \frac{-\pi}{2} - 7\alpha)$

however the answers in my book says its meant to be $cis( \frac{3\pi}{2} - 7\alpha)$

why is it $\frac{3\pi}{2}$ and not $\frac{-\pi}{2}$ ? $\frac{3\pi}{2}$ is not allowed in the interval of $(-\pi,\pi]$ for Argument, isn't it?

TrueTears · « **Reply #11 on:** December 27, 2008, 03:57:16 pm »

And also...

Show that $cos(\alpha) - sin(\alpha) i = cis(-\alpha)$

thanks

Damo17 · « **Reply #12 on:** December 27, 2008, 04:01:47 pm »

That interval would be true if the answer $\frac{-\pi}{2}$ was the actual Argument. However when you have the $-7\alpha$ in there as part of the Argument then most likely what they are doing is making it neater by making the $\frac{-\pi}{2}$ into a positive angle $\frac{3\pi}{2}$ . If you substituted a value for $\alpha$ then the answer you come up with would have to be within the interval of $(-\pi,\pi]$ .

TrueTears · « **Reply #13 on:** December 27, 2008, 04:07:03 pm »

ah i see, so how do i know when to leave the answers as they are or to change them to be within the interval of $(-\pi,\pi]$

eg, $cis(-2\pi+4\alpha)$ was simplified to $cis(4\alpha)$

however, $cis(-3\pi+6\alpha)$ was left as $cis(-3\pi+6\alpha)$ (according to book's answers)

so, how do i know when to keep simplifying or stop?

Damo17 · « **Reply #14 on:** December 27, 2008, 04:48:04 pm »

Quote from: TrueTears on December 27, 2008, 03:57:16 pm

And also...

Show that $cos(\alpha) - sin(\alpha) i = cis(-\alpha)$

thanks

Maybe use:

$\frac{1}{cis\theta}= cis(-\theta)$

$\frac{1}{cos\theta+sin\theta i}=cis(-\theta)$

$cos\theta-sin\theta i=cis(-\theta)$

Or

If $z=r cis(\theta)$ then $\tilde{z}= r cis(-\theta)$

in this case:
$z= cis(\theta)$
$= cos(\theta)+sin(\theta)i$

Therefore $cos(\theta)-sin(\theta)i=cis(-\theta)$

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