Trig help please asap? Will fan and medal! :)
Simplify the trigonometric expression. cos^2 (theta)/1-sin(theta)
A. sin (theta)
B. 1 + sin (theta)
C. 1 - sin (theta)
D. 1 - sin (theta)/sin(theta)

- Nerdgirl

- jamiebookeater

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- Nerdgirl

@phi

- phi

do you know
\[ \cos^2 x + \sin^2 x = 1 \] ?

- Nerdgirl

uhhhh...... lol sorry I have my ditzy moments..... I think so?

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## More answers

- Nerdgirl

yeahhhh actually I don't

- phi

from which you can get (add -sin^2 to both sides)
\[ \cos^2 x = 1 - \sin^2 x \]
I would use that as the first step, to "get rid of" the cos^2

- Nerdgirl

ok so far i'm following

- phi

then notice that 1 is the same as 1*1
so you could write
\[ 1^2 - \sin^2 x\]
and that is a "difference of 2 squares"
which you should know how to factor

- phi

in other words, in your math career you should have seen
\[ a^2 - b^2 = (a-b)(a+b) \]

- Nerdgirl

yeah sorry I didn't reply quicker i'm multitasking.... sooooo..... it would beeeeeee..... hold on give me a min to think I've been doing math all day so my brain is kind of fried

- Nerdgirl

is there any way you could maybe just explain the whole thing to me then I can go back and look at it and do it on my own to make sure I understand the whole thing? @phi

- phi

From a post up above, you know
cos^2 theta = what ?

- Nerdgirl

1 - sin^2 (theta)?

- phi

that means you start with
\[ \frac{ \cos^2 (\theta)}{1-\sin(\theta)} \]
and replace the cos^2... (and let me use x instead of theta...easier to type)
\[ \frac{ 1-\sin^2 (x)}{1-\sin(x)} \]

- phi

now match 1^2 -sin^2 with a^2 - b^2
( remember 1 can be written as 1^2 )
what does "a" match up with ?
and what does "b" match up with?

- Nerdgirl

a = 1 and b = sin? I think? Maybe? Possibly?

- phi

definitely
now use
a^2 - b^2 = (a-b)(a+b)
replace the letters with what we have i.e. replace a with 1 and b with sin(x)
what do you get ?

- Nerdgirl

Um..... that's a good question.... a really good question.... (that's my way of saying I have no clue and sorry) :p

- phi

It does not take a clue
erase a and put in 1
erase b and write in sin(x)
and leave all the other stuff in the pattern alone.
a^2 - b^2 = (a-b)(a+b)

- Nerdgirl

So I'm guessing it would look like 1^2 - sin(x)^2 = (1-sin(x))(1+sin(x))?

- phi

yes. though people write \(\sin^2(x) \) rather than \(\sin(x)^2 \)
(but it means sin(x) * sin(x) either way)

- phi

so now you can write
\[ \frac{ 1-\sin^2 (x)}{1-\sin(x)} \\ \frac{ (1-\sin(x))(1+\sin(x))}{1-\sin(x)} \]

- phi

Last step. anything divided by itself is 1
you have a (1-sin x)/(1-sin x)
which "cancel"

- phi

what is left over? that is the answer

- Nerdgirl

I'm leaning toward either A or D

- phi

You don't have to lean. You should look at
\[ \frac{ (1-\sin(x))(1+\sin(x))}{(1-\sin(x))} \]

- Nerdgirl

OH!

- Nerdgirl

I see now! Light bulb came on!

- phi

Do you understand that means 1-sin x divided by itself (leaving an extra factor of (1+sin x)

- Nerdgirl

yep! ty ty ty ty ty you're awesome!!!

- phi

It is the same problem as
\[ \frac{6}{3} = \frac{3 \cdot 2 }{3} = \frac{3}{3} \cdot 2 = 2\]

- phi

math is filled with "light bulb" moments that are quite fun. Unfortunately it takes a bit of work to learn enough math to get to those moments.

- Nerdgirl

definitely. ty sooooo much again!!!!!

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