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
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- lilsis76

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

Mathematics
- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- lilsis76

|dw:1350692195063:dw|

- anonymous

|dw:1350692395385:dw|

- lilsis76

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.