Author Topic: I broke maths. (Read 15675 times) Tweet Share

Hutchoo · « **Reply #60 on:** April 21, 2011, 05:03:59 am »

Thanks for that guys.

I have another question 8-)

See the attachment below.

Thanks

luffy · « **Reply #61 on:** April 21, 2011, 09:07:33 am »

Below is how I would do the question - Sorry if I made any mistakes.

$\textup{let }z = 1 + i\tan{(\theta)}$

$\therefore Arg{(z)} = \tan^{-1}{(\frac{\tan{(\theta)}}{1})}$

$\therefore Arg{(z)} = \theta$

$|z| = \sqrt{1^2 + \tan^2{(\theta})}$

Recall that $\tan^2{(\theta)} + 1 = \sec^2{(\theta)}$

$\therefore |z| = \sqrt{\sec^2{(\theta)}$

$\therefore |z| = \sec{(\theta)}$

$\therefore z = |z|cis(\theta)$

$\therefore z = \sec{(\theta)}cis(\theta)$

Hope I helped.

nacho · « **Reply #62 on:** April 21, 2011, 11:22:48 am »

is that spesh?

xZero · « **Reply #63 on:** April 21, 2011, 11:57:49 am »

yup, complex numbers

Hutchoo · « **Reply #64 on:** April 21, 2011, 01:58:05 pm »

Quote from: luffy on April 21, 2011, 09:07:33 am

Below is how I would do the question - Sorry if I made any mistakes.

$\textup{let }z = 1 + i\tan{(\theta)}$

$\therefore Arg{(z)} = \tan^{-1}{(\frac{\tan{(\theta)}}{1})}$

$\therefore Arg{(z)} = \theta$

$|z| = \sqrt{1^2 + \tan^2{(\theta})}$

Recall that $\tan^2{(\theta)} + 1 = \sec^2{(\theta)}$

$\therefore |z| = \sqrt{\sec^2{(\theta)}$

$\therefore |z| = \sec{(\theta)}$

$\therefore z = |z|cis(\theta)$

$\therefore z = \sec{(\theta)}cis(\theta)$

Hope I helped.

Thanks man, you did help

.
I'm just confused with sec/csc/cot identities*. Is it all assumed knowledge? How do you know if something is one of those identities*?
Does anyone have something I can read on them?

Thanks.

* = May not be the right term to describe sec/csc/cot etc.

luffy · « **Reply #65 on:** April 21, 2011, 02:07:38 pm »

In methods, they aren't really assumed knowledge. But, its definitely good to know. If you haven't figured it out by now, I'll just go through them with you.

$\sec{(\theta)} = \frac{1}{\cos{(\theta)}}$

$\csc{(\theta)} = \frac{1}{\sin{(\theta)}}$

$\cot{(\theta)} = \frac{1}{\tan{(\theta)}}$

Recall that:

$\sin^2{(\theta)} + \cos^2{(\theta)} = 1$

If you divide both sides by $\cos^2{(\theta)}$ , you will get:

$\frac{\sin^2{(\theta)}}{\cos^2{(\theta)}} + \frac{\cos^2{(\theta)}}{\cos^2{(\theta)}} = \frac{1}{\cos^2{(\theta)}}$

$\therefore \tan^2{(\theta)} + 1 = \sec^2{(\theta)}$ .

Thats all you really need to know about them. Hope I helped.

Andiio · « **Reply #66 on:** April 21, 2011, 04:07:47 pm »

Quote from: luffy on April 21, 2011, 02:07:38 pm

In methods, they aren't really assumed knowledge. But, its definitely good to know. If you haven't figured it out by now, I'll just go through them with you.

$\sec{(\theta)} = \frac{1}{\cos{(\theta)}}$

$\csc{(\theta)} = \frac{1}{\sin{(\theta)}}$

$\cot{(\theta)} = \frac{1}{\tan{(\theta)}}$

Recall that:

$\sin^2{(\theta)} + \cos^2{(\theta)} = 1$

If you divide both sides by $\cos^2{(\theta)}$ , you will get:

$\frac{\sin^2{(\theta)}}{\cos^2{(\theta)}} + \frac{\cos^2{(\theta)}}{\cos^2{(\theta)}} = \frac{1}{\cos^2{(\theta)}}$

$\therefore \tan^2{(\theta)} + 1 = \sec^2{(\theta)}$ .

Thats all you really need to know about them. Hope I helped.

Yup, it's moreso Specialist knowledge.

Hutchoo · « **Reply #67 on:** April 27, 2011, 06:43:22 pm »

Question is attached.

Thanks guys

TrueTears · « **Reply #68 on:** April 27, 2011, 06:54:57 pm »

$2b^3+a = (2^{\frac{1}{3}}b)^3 + (a^{\frac{1}{3}})^3$

now $(x^3+y^3) = (x+y)(x^2-xy+y^2)$

sub in and ur done

Hutchoo · « **Reply #69 on:** May 05, 2011, 01:46:58 am »

"The hour hand of a clock is 7 cm long. Calculate the area it measures out in 5 hours."

TT told me that the angle would be at 2pi/5 (assuming that it moves from 12 to 5 on the clock).
..
WHAT NOW?

brightsky · « **Reply #70 on:** May 05, 2011, 08:13:07 pm »

recall the area of the circle to be pi r^2. since 12 hrs = 360 degrees, then 5 hours = 360 * 5/12.
the hour hand is 7 cm, so in this case r = 7. so the area when the hand measures out in 12 hrs would be 49 pi. However we only won't 5/12 th of that, so 49pi * 5/12 is the answer. simplify as applicable.

Hutchoo · « **Reply #71 on:** May 19, 2011, 08:39:56 pm »

Just a straight-forward probability question:

If a card is drawn at random from a normal deck of cards, ﬁnd the probability that it is:
a) red or a king
b) an ace or a heart
c) a picture card or red
d) a club or a number card or ace

Thanks.

brightsky · « **Reply #72 on:** May 19, 2011, 08:47:00 pm »

count the number of cards that are "desired" (for instance, in a), count the number of cards that are red or king) --> watch out for overlap here. then divide that by 52 (normal deck of cards).

Hutchoo · « **Reply #73 on:** May 19, 2011, 09:01:56 pm »

Quote from: brightsky on May 19, 2011, 08:47:00 pm

count the number of cards that are "desired" (for instance, in a), count the number of cards that are red or king) --> watch out for overlap here. then divide that by 52 (normal deck of cards).

Thanks man.

T_T Just realised it was ridiculously easy.

Hutchoo · « **Reply #74 on:** June 06, 2011, 04:14:04 pm »

YO. I'm back.

Can someone show me their workings for this following question?:

Factorise: $125(w-3)^3 + 27(4+w)^3.$

Thanks a lot.

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Author Topic: I broke maths. (Read 15675 times) Tweet Share

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