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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)

Mathematics
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  • phi
do you know \[ \cos^2 x + \sin^2 x = 1 \] ?
uhhhh...... lol sorry I have my ditzy moments..... I think so?

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Other answers:

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
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 \[ a^2 - b^2 = (a-b)(a+b) \]
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
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 ?
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?
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 ?
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)
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
I'm leaning toward either A or D
  • phi
You don't have to lean. You should look at \[ \frac{ (1-\sin(x))(1+\sin(x))}{(1-\sin(x))} \]
OH!
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)
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.
definitely. ty sooooo much again!!!!!

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