If sin theta=3/5 and theta is in quad 2 the exact form of csc (pi/2-theta) is...? Just want to make sure if this works??

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If sin theta=3/5 and theta is in quad 2 the exact form of csc (pi/2-theta) is...? Just want to make sure if this works??

Mathematics
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sinθ = 3/5 csc(π/2 - θ) = secθ = 1/cosθ = -4/5
|dw:1436736243879:dw|
given sin = 3/5 and in quad 2

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@triciaal So what does that mean?
csc = 1/sin = hypotenuse / opposite 5/-4
The answer is -5/4 then?
|dw:1436737161908:dw|
What do I do with that?
your question "the exact form of csc (pi/2-theta) is?"
shouldn't it be between -1 and 1 only?
"sinθ = 3/5 csc(π/2 - θ) = secθ = 1/cosθ = -4/5" in your work here the only mistake you made was not flipping -4/5 that cos(theta)=-4/5 so 1/cos(theta) is the reciprocal of that
csc = hypotenuse/ opposite
|dw:1436737547332:dw| However, since we are in the 2nd quadrant $$ -\csc\left (\frac{\pi}{2}-\theta\right )=\frac{{\bf\color{red}{-}}5}{~~4} $$ Note the negative corresponding to the value of \(x\) in the 2nd quadrant.
@zepdrix why/when is the value < -1 in general?
Trish :) Maybe you're thinking about sine and cosine, which have restricted output values between 1 and -1
That's the only way they will satisfy a triangle `hypotenuse longer than other sides`\[\large\rm \sin \theta=\frac{opposite}{hypotenuse}=\frac{shorter}{longer}\lt1\]But for Cosecant, I guess we have the reverse going on, ya? We always need to be 1 or larger.\[\Large\rm \csc \theta=\frac{hypotenuse}{opposite}=\frac{longer}{shorter}\gt1\]
Does this make sense? |dw:1436738761632:dw|
@zepdrix thanks just never thought about it @ybarrap thanks for the CAST diagram but I was checking the limit ; not if positive or negative
@ybarrap no tan on your chart
|dw:1436739672137:dw|
Sorry guys I'm back! So i think I got the answer this time around! @triciaal @zepdrix
what is that the answer to?
are you looking at a new question?
No for the one I put on
oh I guess you didn't understand what anyone posted :p
Yeah nothing at all XD
This is the only question I can't get!! :////
ok let's start from the beginning this shouldn't be long at all
"sinθ = 3/5 csc(π/2 - θ) = secθ = 1/cosθ = -4/5" do you remember this is what you first posted here?
this was your first try here anyways
Yeah thats what I did, but I figured it was wrong
it is a little wrong
the first part is totally awesome
the second part the part where you say 1/cos(theta)=-4/5 is a little off
yes cos(theta)=-4/5 but what is 1/cos(theta)=?
if I tell you a=2/5 then what is 1/a=?
or how about this if I tell you 1/a=5/2 then a=?
a and 1/a are reciprocals of one another that means they are the "flippings of each other" like 2 and 1/2 are reciprocals and 2/5 and 5/2 are reciprocals and -2/5 and -5/2 are reciprocals
so if I say a=2/5 then 1/a=5/2 or if I say 1/a=5/2 then a=2/5
so if cos(theta)=-4/5 then 1/cos(theta)=?
if you want to replace cos(theta) with a you can if a=-4/5 then 1/a=?
@freckles Alright I think I need a little bit to take this all in cuz you just shot me a whole bunch of stuff at me XD I think I get what your saying let me see what I can do
take your time
Wait so would it be -5/4 then??? *fingers crossed right now that its right** haha
yes yes !!!
YYAYYY!!
-5/4 is indeed the reciprocal of -4/5
So where do you go on from there? Is there more to be done??
nope
Ohmygod that simple for a question like this? My mind is just blown right now .... XD
\["\sinθ = 3/5 \\ \csc(π/2 - θ) = \secθ = 1/\cosθ = -4/5" \\ \text{ so the correcting your work would look like this: } \\ \csc(\frac{\pi}{2}-\theta)=\sec(\theta)=\frac{1}{\cos(\theta)}=\frac{1}{\frac{-4}{5}}=\frac{-5}{4}\]
by the way you know that first thing you used it called a cofunction identity most people forget that one so you are doing awesome!
@freckles Thanks again seriously!! Haha really helped! Literally am beating myself up not knowing lol
i think we were all working on our algebra in trig and even in calculus
algebra/arithemetic
and/or
guess what you are going to have to know so much algebra for calculus you are going to think you never learned algebra :p well that has been my experience with some anyways I'm kind a kidding. I know you are going do okay especially since you can always get help if you need it.
Haha true!! thanks again! I know with Calc I am going to need lots of help especially!!
math is fun just takes lots of practice

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