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

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Nerdgirl
  • Nerdgirl
yeahhhh actually I don't
phi
  • 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
  • Nerdgirl
ok so far i'm following
phi
  • 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
  • phi
in other words, in your math career you should have seen \[ a^2 - b^2 = (a-b)(a+b) \]
Nerdgirl
  • 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
  • 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
  • phi
From a post up above, you know cos^2 theta = what ?
Nerdgirl
  • Nerdgirl
1 - sin^2 (theta)?
phi
  • 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
  • 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
  • Nerdgirl
a = 1 and b = sin? I think? Maybe? Possibly?
phi
  • 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
  • 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
  • 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
  • Nerdgirl
So I'm guessing it would look like 1^2 - sin(x)^2 = (1-sin(x))(1+sin(x))?
phi
  • phi
yes. though people write \(\sin^2(x) \) rather than \(\sin(x)^2 \) (but it means sin(x) * sin(x) either way)
phi
  • 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
  • phi
Last step. anything divided by itself is 1 you have a (1-sin x)/(1-sin x) which "cancel"
phi
  • phi
what is left over? that is the answer
Nerdgirl
  • Nerdgirl
I'm leaning toward either A or D
phi
  • phi
You don't have to lean. You should look at \[ \frac{ (1-\sin(x))(1+\sin(x))}{(1-\sin(x))} \]
Nerdgirl
  • Nerdgirl
OH!
Nerdgirl
  • Nerdgirl
I see now! Light bulb came on!
phi
  • phi
Do you understand that means 1-sin x divided by itself (leaving an extra factor of (1+sin x)
Nerdgirl
  • Nerdgirl
yep! ty ty ty ty ty you're awesome!!!
phi
  • phi
It is the same problem as \[ \frac{6}{3} = \frac{3 \cdot 2 }{3} = \frac{3}{3} \cdot 2 = 2\]
phi
  • 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
  • Nerdgirl
definitely. ty sooooo much again!!!!!

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