Given tan theta = -2/3, where pi/2 < 0 < pi
and
csc B=2 where 0

Mathematics
**
**- lilsis76

Given tan theta = -2/3, where pi/2 < 0 < pi
and
csc B=2 where 0**
**

Mathematics
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- lilsis76

Given tan theta = -2/3, where pi/2 < 0 < pi
and
csc B=2 where 0**
**

Mathematics
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- lilsis76

|dw:1350692195063:dw|

- anonymous

|dw:1350692395385:dw|

- lilsis76

based on the information i know it will be first quad.

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- anonymous

csc B = 1/sin B

- lilsis76

oh duh haha its negative, so its going backwards right?

- anonymous

yes

- lilsis76

|dw:1350692578128:dw| right?

- anonymous

yes, next

- lilsis76

dang it, im going to need identities right? cuz it kinda looks like it :/

- lilsis76

im trying to find the identities

- anonymous

1st find theta and beta
sin (0 + B) = sin 0 cos B + cos 0 sin B
cos (0 â€“ B) = cos 0 cos B + sin 0 sin B

- lilsis76

UGH!!! okay im back lol
okay, are these the identities I should know for finding theta and beta then? just so i save it?

- anonymous

\[\tan \theta=\frac{ -2 }{ 3 }\]
\[\tan^{-1} (\tan \theta)=\tan^{-1} (\frac{ -2 }{ 3 })\]

- lilsis76

okay so far i know:
sin = 2/x cos =2/x and tan is the same -2/3

- anonymous

what does x = ?

- anonymous

|dw:1350693950387:dw|

- anonymous

by the way what grade is this

- lilsis76

okay hold on, this is precal. im in a community college, but im horrible at math haha

- lilsis76

x = sqrt 13

- anonymous

been a while since i been in the class
anyways x is correct

- lilsis76

oh really? did u graduate?
and YAY okay i got the x, now will that be the theta?

- lilsis76

okay so the cos=3sqrt13 /26

- anonymous

right now im working on my|dw:1350694287644:dw|

- anonymous

can you calculate theta now

- lilsis76

uh.... well the one i had or placed on here was wrong? the
okay so the cos=3sqrt13 /26

- lilsis76

is the 2 a regular 2, or a negative 2? since its going downward of the unit circle?

- lilsis76

wait im lost:
i want to go back a lil to the soh cah toa
cah is adj/hyp
that would be 3 / sqrt13 |dw:1350694753458:dw| right?

- anonymous

thats the cos of thata, dont you want theta?

- lilsis76

yes the theta of tangent

- lilsis76

wait. cos theta = the 3sqrt13 / 13

- lilsis76

im so lost

- anonymous

can i do it on a piece of paper then send it to you, im tired and want to go to sleep; yes i know its only 9:20 lol

- lilsis76

lol yes you may. and thanks, im getting a headache from this too

- anonymous

alright so first we need to figure out where tan is negative. Using this circle
|dw:1350696636481:dw|

- anonymous

these are when the functions are positive. for instance, in quadrant 1 , all function are positive

- lilsis76

okay so tan would be negative in...................the cosine area

- anonymous

and also the sine , because tangent is
\[\frac{sin\theta}{cos\theta}\]... so anytime either one is negative tan will be negative ... when sin and cos are both negative ,tan is positive which is this quadrant|dw:1350696881960:dw|

- lilsis76

okay so the 2nd and 3rd piece on the left sides

- anonymous

next we use the information it gives about theta
\[\pi/2<\theta<\pi\]
can you mark these angles on my picture

- lilsis76

|dw:1350697001280:dw|

- anonymous

so theta is less than \[\pi/2\] but greater than \[\pi\]|dw:1350697010218:dw|

- anonymous

this means \[\theta is in your second quadrant

- lilsis76

okay i see that now

- anonymous

alright so now lets solve the equation
\[tan(\theta)=\frac{-2}{3}\]

- anonymous

how would you get theta by itself

- anonymous

do you know about inverse trig functions?

- lilsis76

im back server was werid. let me look

- lilsis76

no i dont know about the inverse trig functions unless your
talkin about it being 1/tan or like the cos/sin

- anonymous

inverse function work as "cancelers" of trig functions they're usually denoted as
\[arcsin,arctan,arccos\]
or
\[sin^{-1},cos^{-1}\]
you know them because you know 1/tanx. I think you just didn't know what they were called

- anonymous

these cancel out the same trig function so
\[arcsin(sin(x))=x\]

- anonymous

so what would you do in your case
\[tan(\theta)=\frac{-2}{3}\]

- lilsis76

no we have not gone over this :/
and so the arctan one would be
arctan(tan(x))= -2/3 ?

- anonymous

if you do arctan on one side you have to do it on the other =]

- lilsis76

-2/3arctan(tan(x))=-2/3

- anonymous

ok say
\[tan(x)=\frac{-2}{3}\]
you know to get rid of tan you have to use arctan, however if you multiply the left side by arctan, you have to multiply the right side also
\[arctan(tan(x)=arctan(\frac{-2}{3})\]

- anonymous

since arctan cancels tan you get
\[x=arctan(\frac{-2}{3})\]

- anonymous

your calculator should have an arctan symbol. it might be \[tan^{-1}\]

- lilsis76

so the tan^-1 is called arctan?

- anonymous

yes =]

- lilsis76

weird....okay, ya i didnt know any inverse ones i couldnt find the identities either :/
but i did tan^-1(-2/3) and it gave me.............-33.69

- anonymous

alright where is -33.69 degrees?

- lilsis76

haha you really ask me degreees?? umm... i think its that 180/pi, right?

- anonymous

theta is in degrees unless you put in radians but keep it in degrees , all i'm asking is where on a circle would -33.69 degrees be( quadrant wise)

- lilsis76

OH, umm... well hold on

- lilsis76

|dw:1350698454198:dw|

- anonymous

well uh oh that isn't in the second quadrant we need to find an angle that is in the second quadrant with the same value as -33.69 degrees

- lilsis76

should i start from 90 degrees?? |dw:1350698677314:dw|

- anonymous

hold on i know what you need to do just don't know why it's been a while since going through this material

- anonymous

i know you have to add pi(180 degrees) to it

- lilsis76

okay, i think its mult. the ( 180/pi) i think

- lilsis76

i cant find the equ.

- lilsis76

WAIT!!! to change to degrees you multiply 180 deg/ pi. but i would need the point for the other one to do it..... here is an example from my notes:
pi/12(180/pi) cross reduce to--->180 deg./12 = 15 degrees---sin15degrees =.2588

- lilsis76

I really cant figure this one out... can you? or are you writingit down on paper, or looking it up online??

- jim_thompson5910

in your drawing at the very top, you say cos 7/25
did you mean cos(theta) = 7/25 ???

- lilsis76

yes

- jim_thompson5910

so because
sin^2(x) + cos^2(x) = 1
you can find sin(theta) like so
sin^2(theta) + cos^2(theta) = 1
sin^2(theta) + (7/25)^2 = 1
sin^2(theta) + 49/625 = 1
sin^2(theta) = 1 - 49/625
sin^2(theta) = 576/625
sin(theta) = -sqrt(576/625)
sin(theta) = -24/25

- jim_thompson5910

oh wait, sry sine is positive because 0 < theta < pi/2

- jim_thompson5910

so sin(theta) = 24/25

- lilsis76

lol let me check

- jim_thompson5910

sin^2(B) + cos^2(B) = 1
sin^2(B) + (-3/5)^2 = 1
sin^2(B) + 9/25 = 1
sin^2(B) = 1 - 9/25
sin^2(B) = 16/25
sin(B) = -sqrt(16/25) ... sin(B) is negative for pi < B < 3pi/2
sin(B) = -4/5

- jim_thompson5910

So after those two steps, we now know these four pieces of info
sin(theta) = 24/25
cos(theta) = 7/25
sin(B) = -4/5
cos(B) = -3/25
You can now use them in the identities
sin(theta + B) = sin(theta)cos(B) + cos(theta)sin(B)
cos(theta - B) = cos(theta)cos(B) + sin(theta)sin(B)

- lilsis76

wait sin B is negative?? because..... its in the ..3rd quardinate and sin is negative
and
24/25 is positive because it falls between the |dw:1350700800280:dw| 0 < theta < pi/2

- jim_thompson5910

yeah i fixed that error, my bad

- lilsis76

lol just a moment

- lilsis76

- lilsis76

Sin B is negative??? because its in the 3rd quar. and that sin is negative
and
24/25 is positive because it falls between the 0 < theta < pi/2

- jim_thompson5910

yeah i fixed that

- lilsis76

okay, im checking my notes cuz i have writing in every direction

- lilsis76

okay, thank you, lol i have to write it neater and go over it again to get things down. thank you for helping me. ill give u a medal

- jim_thompson5910

yw

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