Author Topic: dcc help me thread (Read 22061 times) Tweet Share

droodles · « **on:** February 18, 2008, 07:10:40 pm »

for $\pi < a < \frac {3\pi}{2} with \ cos (a) = -sin (b) \ where \ 0<b< \frac {\pi}{2}$ find a in terms of b

how is it $\pi+b$

ALERT ALERT IF YOU ARE GOING TO LOOK AT MY THREAD PLEASE SOLVE DONT LURK OR I WILL SHOOT YOU

Collin Li · « **Reply #1 on:** February 18, 2008, 09:31:04 pm »

I'm not sure whether my approach is what you're looking for, but I don't end up with your conclusion:

$\cos(a) = \sin(a + \frac{\pi}{2}) = -\sin(b)$

$\implies b = 2\pi - \left(a+\frac{\pi}{2}\right)$

$a = \frac{3\pi}{2} - b$

dcc · « **Reply #2 on:** February 18, 2008, 09:38:50 pm »

For all the crap in those quadrants, I THINK:

First of all, we know that, by the relationship between sin & cos:

$cos(a) = sin(a + \dfrac{\pi}{2})$

Also:

$-sin(b) = sin(2\pi - b)$

Therefore, subbing into our above relationship:

$sin(a + \dfrac{\pi}{2}) = sin(2\pi - b)$

$a + \dfrac{\pi}{2} = 2\pi - b$

$a = \dfrac{3\pi}{2} - b$

Collin Li · « **Reply #3 on:** February 19, 2008, 01:56:55 pm »

Quote from: droodles on February 18, 2008, 07:10:40 pm

for $\pi < a < \frac {3\pi}{2} with \ cos (a) = -sin (b) \ where \ 0<b< \frac {\pi}{2}$ find a in terms of b

how is it $\pi+b$

If $a = \pi + b$ , $\implies \frac{3\pi}{2} - b = \pi + b \implies b = \frac{\pi}{4}$ from both our solutions. That is not the only answer, so $\pi+b$ is not the correct answer (given that dcc and I are correct).

Mao · « **Reply #4 on:** February 20, 2008, 08:40:13 pm »

Quote from: droodles on February 18, 2008, 07:10:40 pm

for $\pi < a < \frac {3\pi}{2} with \ cos (a) = -sin (b) \ where \ 0<b< \frac {\pi}{2}$ find a in terms of b

how is it $\pi+b$

here's my crack at it

$cos(a) = sin(a+\frac{\pi}{2})$

$-sin(b) = sin(-b)$

$-b + 2n \cdot \pi = a+\frac{\pi}{2}$ where n is any integer

$-b - \frac{\pi}{2} + 2n \cdot \pi = a$

for a to be within its constraing (long story short), n=1

$a=\frac{3 \pi}{2} -b$

riiight, the book definitely stuffed up somewhere =\

Actually, for the answer the book gave you (supposedly)
this implies:

$cos(\pi + b) = -sin(b)$

$-cos(b) = -sin(b)$

$sin(b)=cos(b)$ ?? this is clearly not true.

however, this is based on the assumption that $cos(a)=-sin(b)$ for the entire interval of a and b, if the question is asking for a simple solution, then we only need to match the constraints (without worring about the equation at all), in that case, the answer will be $a=\pi + b$

*note, the above paragraph is not written in perfect maths lingo, please correct me if i used the wrong terminologies

droodles · « **Reply #5 on:** February 24, 2008, 03:56:24 pm »

hi help what what

Toothpaste · « **Reply #6 on:** February 24, 2008, 04:05:13 pm »

$sin(\pi + \theta)$
= $-sin \theta$
= $-b$

$cos(125\pi - \theta)$
= $- cos \theta$
= $-a$

droodles · « **Reply #7 on:** February 24, 2008, 04:07:09 pm »

toothpick i got my english handbook cool'd today and it cost $4.10 like you said but the girl who did it was really slack and the pages look all up and down

droodles · « **Reply #8 on:** February 29, 2008, 10:10:05 pm »

find $cos(x)\ when \ \tan(x) = 2, from \ 0 \to \pi$

i know it involves pythagorean identity and SOHCAHTOA but how

Collin Li · « **Reply #9 on:** February 29, 2008, 10:18:05 pm »

From $[0,\, \pi]$ , $\tan x = 2$ for $x$ in quadrant 1. We know this because in quadrant 2, from $\left(\frac{\pi}{2},\, \pi\right]$ , $\tan x < 0 \neq 2$ . This means that $\cos x > 0$ . This is important, because the value we get from the triangle method used below needs to account for the sign.

$\tan x = \frac{O}{A} = 2$

Draw a triangle with the angle $x$ , and the 'opposite' as length 2, and the 'adjacent' as length 1, so that the triangle corresponds to $\tan x = 2 = \frac{2}{1}$ .

Now, we can find the hypotenuse: $H = \sqrt{2^2 + 1^2} = \sqrt{5}$

Therefore a solution is: $\cos x = \frac{A}{H} = \frac{1}{\sqrt{5}}$

Possible solutions for $\cos x$ are $\pm \frac{1}{\sqrt{5}}$ but as confirmed above, $x$ lies in quadrant 1, and hence $\cos x > 0\; \therefore \cos x = \frac{1}{\sqrt{5}}$

droodles · « **Reply #10 on:** March 01, 2008, 08:44:44 pm »

Find $x:\ 2sin^2 +3sinx=2,\ 0<x<360$

Is this even a do-able question? How?

brendan · « **Reply #11 on:** March 01, 2008, 08:48:11 pm »

yes it is doable. hint: let a = sinx

droodles · « **Reply #12 on:** March 01, 2008, 08:48:39 pm »

Mao · « **Reply #13 on:** March 01, 2008, 08:49:45 pm »

now use the general quadratic formula

unknown id · « **Reply #14 on:** March 01, 2008, 08:52:53 pm »

$2sin^2 + 3sinx - 2 = 0$

You have to recognise that this equation is in the form of a quadratic polynomial

$(2sinx - 1)(sinx + 2) = 0$

$sinx = \frac{1}{2} or sinx = -2$

Since $-1 <= sinx <= 1$ , $sinx = \frac{1}{2}$

$x = 30^0, 150^0$ since $sinx$ is positive in quadrants 1 and 2

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