Author Topic: Holiday homeworkk (Read 10617 times) Tweet Share

TrueTears · « **Reply #75 on:** January 01, 2010, 05:24:38 pm »

Quote from: mandy on January 01, 2010, 05:22:23 pm

I'm having a lot of trouble understanding how to do these kinds of problems. For example, I look at this next question, and it looks like a rule, but it's like ... backwards.

Simplify the following:
a. $cot (\frac{\pi}{2} - y)$
For this one, the rule says: $tan(\frac{\pi}{2} - \theta) = cot x$
What am I meant to do?

b. $cosec(2\pi - x)$
And for this one, there isn't anything about $cosec$ in my rules at all :S

Don't just memorise formulas or rules, know your unit circle and what it actually $\text{means}$ . Circular functions is all about using the unit circle and deriving expressions for yourself.

Think about what you are doing in steps:

$\text{cosec}(x) \to \text{cosec}(-x) \to \text{cosec}(2\pi-x)$

brightsky · « **Reply #76 on:** January 01, 2010, 05:35:45 pm »

For questions like these, if you know the rules, it is just a matter of substitution:

a. $cot (\frac {\pi}{2} - y) = tan (\frac {\pi}{2} - (\frac {\pi}{2} - y))$

$= tan (y)$

brightsky · « **Reply #77 on:** January 01, 2010, 05:38:10 pm »

Quote from: TrueTears on January 01, 2010, 05:24:38 pm

Quote from: mandy on January 01, 2010, 05:22:23 pm
I'm having a lot of trouble understanding how to do these kinds of problems. For example, I look at this next question, and it looks like a rule, but it's like ... backwards.

Simplify the following:
a. $cot (\frac{\pi}{2} - y)$
For this one, the rule says: $tan(\frac{\pi}{2} - \theta) = cot x$
What am I meant to do?

b. $cosec(2\pi - x)$
And for this one, there isn't anything about $cosec$ in my rules at all :S
Don't just memorise formulas or rules, know your unit circle and what it actually $\text{means}$ . Circular functions is all about using the unit circle and deriving expressions for yourself.

Think about what you are doing in steps:

$\text{cosec}(x) \to \text{cosec}(-x) \to \text{cosec}(2\pi-x)$

+1

mandy · « **Reply #78 on:** January 01, 2010, 09:53:55 pm »

Quote from: brightsky on January 01, 2010, 05:35:45 pm

For questions like these, if you know the rules, it is just a matter of substitution:

a. $cot (\frac {\pi}{2} - y) = tan (\frac {\pi}{2} - (\frac {\pi}{2} - y))$

$= tan (y)$

How did you know what to substitute and into where?

GerrySly · « **Reply #79 on:** January 01, 2010, 10:35:30 pm »

To be honest I found that just by finding something I can reference, constantly applying it I could eventually figure out how it worked. Figuring it all out myself was just frustrating and wasn't fun at all.

http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Symmetry

Take a look at that mandy.

Quote from: mandy on January 01, 2010, 05:22:23 pm

Simplify the following:
a. $cot (\frac{\pi}{2} - y)$

For this one, you just use the rule you have

$\begin{align*}<br />\cot{(\frac{\pi}{2}-y)}&=\frac{1}{\tan{(\frac{\pi}{2}-y)}}\\<br />&=\frac{1}{\cot{(y)}}\\<br />&=\tan{(y)}<br />\end{align*}$

Quote from: mandy on January 01, 2010, 05:22:23 pm

b. $cosec(2\pi - x)$

For this one you should get it to a function you can work with ( $\sin{(x)}$ ) then work with that. So you have $\frac{1}{\sin{(2\pi-x)}}$ . Now if you take $2\pi$ and subtract an angle from it, it puts you in the 4th quadrant yeah? Just the same as if you took 0 and subtracted an angle from it.

Using that logic you know that you are working with $\sin{}$ in the 4th quadrant, what is $\sin{}$ in the 4th quadrant? It is negative

$\therefore \csc{(2\pi-x)}&=\frac{1}{\sin{(2\pi-x)}}&=\frac{1}{-\sin{(x)}}&=-\csc{(x)}$

mandy · « **Reply #80 on:** January 01, 2010, 11:07:09 pm »

That was a really good explanation GerrySly, thanks so much :]
I think I'm starting to understand it now after looking through that link you gave me.
Thank youuuuu

brightsky · « **Reply #81 on:** January 01, 2010, 11:26:45 pm »

Quote from: mandy on January 01, 2010, 09:53:55 pm

Quote from: brightsky on January 01, 2010, 05:35:45 pm
For questions like these, if you know the rules, it is just a matter of substitution:

a. $cot (\frac {\pi}{2} - y) = tan (\frac {\pi}{2} - (\frac {\pi}{2} - y))$

$= tan (y)$

How did you know what to substitute and into where?

The rule states that:

$cot (x) = tan (\frac {\pi}{2} - x)$

The 'x' in this case is $\frac {\pi}{2} - y$ because the question asks you to evaluate for $cot (\frac {\pi}{2} - y)$ .

Hence you substitute $x = \frac {\pi}{2} - y$ into your formula and your done.

mandy · « **Reply #82 on:** January 03, 2010, 01:51:41 pm »

Find the Argument of each of the following, correct to two decimal places.
a. 5 + 12 $i$

How come I'm getting this wrong?
Don't I use $tan\theta = \frac{12}{5}$ ?
And the use inverse tan to find the theta?

The correct answer is 1.18.

Edit: Sorry guys, my calculator was in degrees mode. I got the answer now. Thanks anyway

mandy · « **Reply #83 on:** January 03, 2010, 03:43:28 pm »

Find the Argument of the following, correct to two decimal places.
a. -8 + 15 $i$

How come I can't do this now

jimmy999 · « **Reply #84 on:** January 03, 2010, 04:10:20 pm »

Quote from: mandy on January 03, 2010, 03:43:28 pm

Find the Argument of the following, correct to two decimal places.
a. -8 + 15 $i$

How come I can't do this now

Chances are you're not looking in the correct quadrant. If you see, the real part is negative whilst the imaginary part is positive, therefore the angle is in the 2nd quadrant. So just calculate $\theta = \pi-\arctan(\frac{15}{8})$

mandy · « **Reply #85 on:** January 03, 2010, 04:16:35 pm »

Quote from: jimmy999 on January 03, 2010, 04:10:20 pm

Quote from: mandy on January 03, 2010, 03:43:28 pm
Find the Argument of the following, correct to two decimal places.
a. -8 + 15 $i$

How come I can't do this now

Chances are you're not looking in the correct quadrant. If you see, the real part is negative whilst the imaginary part is positive, therefore the angle is in the 2nd quadrant. So just calculate $\theta = \pi-\arctan(\frac{15}{8})$

Yeah, I knew the angle is located in the 2nd quadrant, but then what does $arctan$ mean?

jimmy999 · « **Reply #86 on:** January 03, 2010, 04:25:55 pm »

arctan is the same thing as tan inverse. It's just a different name. Examiners tend to use either one. I prefer using arctan

mandy · « **Reply #87 on:** January 03, 2010, 04:44:14 pm »

Oh okay, thank you jimmy999, I don't see how I couldn't do that before ==

mandy · « **Reply #88 on:** January 03, 2010, 09:07:03 pm »

Convert the following complex numbers from the cartesian form $a + bi$ into the form $r cis \theta$ , where $\theta = Arg (a + bi)$ .
a. -1 - $i$

This is what I did:
- Found the r value (modulus) = $\sqrt2$
- Found the $\theta$ = tan ^-1(1)= $\frac{\pi}{4}$

The Argument is meant to be $-\frac{3\pi}{4}$ .
Why am I getting this wrong?

kamil9876 · « **Reply #89 on:** January 03, 2010, 09:09:58 pm »

-1-i is in the 3rd quadrant, so it is clearly not $\frac{\pi}{4}$ . You're better off doing this by sketching and using trigonometry. The arctan formula is not true in all quadrants.

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