Question

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)

Answer 1

@phi

Answer 2

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

Answer 3

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

Answer 4

yeahhhh actually I don't

Answer 5

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

Answer 6

ok so far i'm following

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

1 - sin^2 (theta)?

Answer 13

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)
\[ \frac{ 1-\sin^2 (x)}{1-\sin(x)} \]

Answer 14

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?

Answer 15

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

Answer 16

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 ?

Answer 17

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

Answer 18

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)

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

what is left over? that is the answer

Answer 24

I'm leaning toward either A or D

Answer 25

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

Answer 26

OH!

Answer 27

I see now! Light bulb came on!

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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.

Answer 32

definitely. ty sooooo much again!!!!!