Question

(1 - sin theta) ÷ (sin theta) = (cot^2 theta) ÷ (1 + csc theta)

Answer 1

Hmm that's a strange one :b thinking....

Answer 2

wait, the first side, the bottom denominator is just sine theta

Answer 3

\[\Large \frac{1-\sin \theta}{\sin \theta}\quad=\quad \frac{\cot^2\theta}{1+\csc \theta}\]Oh hah :3 my bad

Answer 4

Ok ok ok this one isn't too bad.
Just a few weird steps to take :o

We're going to manipulate the left side, and try to match it to the right.
Let's start by multiplying the top and bottom by (1+sin theta)\[\Large \frac{1-\sin \theta}{\sin \theta}\left(\frac{1+\sin \theta}{1+\sin \theta}\right)\quad=\quad \frac{?}{\sin \theta(1+\sin \theta)}\]

Answer 5

Do you understand what happens in the numerator?

Answer 6

Yes! That is the first step that I took also, but could not finish.. :)

Answer 7

So that puts us here, yes?\[\Large \frac{1-\sin^2\theta}{\sin \theta(1+\sin \theta)}\]Does the top simplify any further?

Answer 8

yes

Answer 9

to cos^2 theta

Answer 10

\[\Large \frac{\cos^2\theta}{\sin \theta(1+\sin \theta)}\]Hmm ok good.

Answer 11

Recall that:\[\Large \cot \theta\quad=\quad \frac{\cos \theta}{\sin \theta}\qquad\implies\qquad \cot^2\theta\quad=\quad \frac{\cos^2\theta}{\sin^2\theta}\]

Answer 12

Hmm see how we have a cos^2 on top, and a sin on bottom?
If we only had one more sine in the bottom, we could write our numerator as cot^2.

Hmm we're so close! Where can we get another sine from?

Answer 13

distributing the sine on the bottom?

Answer 14

Hmm we actually want to do the opposite!
We want to steal a sine from inside of the brackets.
So we'll `factor out` sine from each term in the brackets.

Answer 15

\[\Large \frac{\cos^2\theta}{\sin \theta(1+\sin \theta)}\quad=\quad \frac{\cos^2\theta}{\sin \theta\cdot \sin \theta\left(\frac{1}{\sin \theta}+1\right)}\]Understand what I did there? :o

Answer 16

not really :O

Answer 17

where did the 1 over sine come from?

Answer 18

\[\LARGE\color{royalblue}{1+\sin \theta\qquad=\qquad \sin \theta\left(\frac{1}{\sin \theta}+\frac{\sin \theta}{\sin \theta}\right)}\]I factored a sine out of each term inside the brackets.

Answer 19

Multiplying the sines together in the denominator gives us\[\Large \frac{\cos^2\theta}{\sin \theta(1+\sin \theta)}\quad=\quad \frac{\cos^2\theta}{\sin ^2\theta\left(\frac{1}{\sin \theta}+1\right)}\]

Answer 20

Using our identity from earlier:\[\Large\frac{\color{orangered}{\cos^2\theta}}{\color{orangered}{\sin ^2\theta}\left(\frac{1}{\sin \theta}+1\right)}\quad=\quad\frac{\color{red}{\cot^2\theta}}{\left(\frac{1}{\sin \theta}+1\right)}\]Yes?

Answer 21

Hey @zepdrix just wanted to let you know I'm still on this question and will be back to finish it, just got really busy after the other day. Sorry i didn't respond :o

Answer 22

care to help me on it a bit later? :)

Answer 23

sure :3

Answer 24

thanks!

Answer 25

k, I'm back!

Answer 26

could you explain though, why when dividing a sine in the parentheses of: sine(1+sine)
that sine becomes sine^2 . I just don't understand that part... is it just something I'm just supposed to know?

Answer 27

From this step?\[\Large \frac{\cos^2\theta}{\sin \theta(1+\sin \theta)}\]Factoring a sine out of each term in the brackets gives us,\[\Large \frac{\cos^2\theta}{\sin \theta\cdot \sin \theta\left(\frac{1}{\sin \theta}+1\right)}\]

Answer 28

That's where the extra sine is coming from, still confused on that? :o

Answer 29

yes from that step :) what does factoring out something mean again??

Answer 30

Here's a quick recap on factoring.
Factoring is 2 steps:
~We divide a value out of each term,
~We multiply by that same value, except we save the multiplication, we don't multiply it out.

Example:\[\Large 3x+9\]Step 1 is:
~Divide 3 out of each term,\[\Large \frac{3x}{3}+\frac{9}{3}\]But in order to keep the equation balanced, we have to also multiply by 3.\[\Large 3x+9\quad = \quad 3\left(\frac{3x}{3}+\frac{9}{3}\right)\quad=\quad 3(x+3)\]

Answer 31

ookkayy that makes sense! thanks for the example really helped.

Answer 32

\[\Large 1+\sin \theta\]So in our brackets we're dividing each term by sin theta:\[\Large \frac{1}{\sin \theta}+\frac{\sin \theta}{\sin \theta}\]But we must also multiply by sin theta to keep it balanced, yes?\[\Large \sin \theta\left(\frac{1}{\sin \theta}+\frac{\sin \theta}{\sin \theta}\right)\]

Answer 33

Ya it's a weird process :) factoring is really important though! make sure you understand it.

Answer 34

Yes for sure... I will be sure to remember it now! :)

Answer 35

k so i guess that it is it? we will just change the cos^2/sine^2 into cot^2 over (1 + sine) ?

Answer 36

So you created a cot^2, good.
But we also need to find a secant somewhere, don't we?

Answer 37

I mean csc*

Answer 38

yes, nvm that isn't it :0

Answer 39

\[\Large\frac{\cot^2\theta}{\left(\frac{1}{\sin \theta}+1\right)}\]

Answer 40

It almost perfectly matches up, if it wasn't for one piece. Hmmm
Can you think of any identities that might help? +_+

Answer 41

1/sine equals to csc ... : )

Answer 42

so we have it! oh that wasn't so bad after all : )

Answer 43

yayyy \c:/ gj strawberry shortcake!

Answer 44

you are too fuuny hah : )

Answer 45

No I'll admit, that was a pretty nasty problem.
Required quite a bit of work to get through :3

Gotta really be comfortable with your identities to get through a problem like that.

Answer 46

I am mastering them slowly but hey thanks so much for thee help, greatly appreciated! : )

Answer 47

np c: