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??

- anonymous

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- anonymous

sinθ = 3/5
csc(π/2 - θ) = secθ = 1/cosθ = -4/5

- triciaal

|dw:1436736243879:dw|

- triciaal

given sin = 3/5 and in quad 2

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## More answers

- anonymous

@triciaal So what does that mean?

- triciaal

csc = 1/sin = hypotenuse / opposite
5/-4

- anonymous

The answer is -5/4 then?

- triciaal

|dw:1436737161908:dw|

- anonymous

What do I do with that?

- triciaal

your question "the exact form of csc (pi/2-theta) is?"

- triciaal

shouldn't it be between -1 and 1 only?

- freckles

"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

- triciaal

csc = hypotenuse/ opposite

- ybarrap

|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.

- triciaal

@zepdrix why/when is the value < -1 in general?

- zepdrix

Trish :)
Maybe you're thinking about sine and cosine, which have restricted output values between 1 and -1

- zepdrix

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.\[\Large\rm \csc \theta=\frac{hypotenuse}{opposite}=\frac{longer}{shorter}\gt1\]

- ybarrap

Does this make sense?
|dw:1436738761632:dw|

- triciaal

@zepdrix thanks just never thought about it
@ybarrap thanks for the CAST diagram but I was checking the limit ; not if positive or negative

- triciaal

@ybarrap no tan on your chart

- ybarrap

|dw:1436739672137:dw|

- anonymous

Sorry guys I'm back! So i think I got the answer this time around! @triciaal @zepdrix

- anonymous

-1/3/5 I think @zepdrix @triciaal @freckles

- freckles

what is that the answer to?

- freckles

are you looking at a new question?

- anonymous

No for the one I put on

- freckles

oh I guess you didn't understand what anyone posted :p

- anonymous

Yeah nothing at all XD

- anonymous

This is the only question I can't get!! :////

- freckles

ok let's start from the beginning
this shouldn't be long at all

- freckles

"sinθ = 3/5
csc(π/2 - θ) = secθ = 1/cosθ = -4/5"
do you remember this is what you first posted here?

- freckles

this was your first try here anyways

- anonymous

Yeah thats what I did, but I figured it was wrong

- freckles

it is a little wrong

- freckles

the first part is totally awesome

- freckles

the second part the part where you say 1/cos(theta)=-4/5 is a little off

- freckles

yes cos(theta)=-4/5
but what is 1/cos(theta)=?

- freckles

if I tell you a=2/5
then what is 1/a=?

- freckles

or
how about this if I tell you 1/a=5/2 then a=?

- freckles

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

- freckles

so if I say a=2/5 then 1/a=5/2
or if I say 1/a=5/2 then a=2/5

- freckles

so if cos(theta)=-4/5
then 1/cos(theta)=?

- freckles

if you want to replace cos(theta) with a you can
if a=-4/5
then 1/a=?

- anonymous

@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

- freckles

take your time

- anonymous

Wait so would it be -5/4 then??? *fingers crossed right now that its right** haha

- freckles

yes yes !!!

- anonymous

YYAYYY!!

- freckles

-5/4 is indeed the reciprocal of -4/5

- anonymous

So where do you go on from there? Is there more to be done??

- freckles

nope

- anonymous

Ohmygod that simple for a question like this? My mind is just blown right now .... XD

- freckles

\["\sinθ = 3/5 \\ \csc(π/2 - θ) = \secθ = 1/\cosθ = -4/5" \\ \text{ so the correcting your work would look like this: } \\ \csc(\frac{\pi}{2}-\theta)=\sec(\theta)=\frac{1}{\cos(\theta)}=\frac{1}{\frac{-4}{5}}=\frac{-5}{4}\]

- freckles

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!

- anonymous

@freckles Thanks again seriously!! Haha really helped! Literally am beating myself up not knowing lol

- freckles

i think we were all working on our algebra in trig and even in calculus

- freckles

algebra/arithemetic

- freckles

and/or

- freckles

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.

- anonymous

Haha true!! thanks again! I know with Calc I am going to need lots of help especially!!

- triciaal

math is fun just takes lots of practice

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