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

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

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Other answers:

csc B = 1/sin B
oh duh haha its negative, so its going backwards right?
yes
|dw:1350692578128:dw| right?
yes, next
dang it, im going to need identities right? cuz it kinda looks like it :/
im trying to find the identities
1st find theta and beta sin (0 + B) = sin 0 cos B + cos 0 sin B cos (0 – B) = cos 0 cos B + sin 0 sin B
UGH!!! okay im back lol okay, are these the identities I should know for finding theta and beta then? just so i save it?
\[\tan \theta=\frac{ -2 }{ 3 }\] \[\tan^{-1} (\tan \theta)=\tan^{-1} (\frac{ -2 }{ 3 })\]
okay so far i know: sin = 2/x cos =2/x and tan is the same -2/3
what does x = ?
|dw:1350693950387:dw|
by the way what grade is this
okay hold on, this is precal. im in a community college, but im horrible at math haha
x = sqrt 13
been a while since i been in the class anyways x is correct
oh really? did u graduate? and YAY okay i got the x, now will that be the theta?
okay so the cos=3sqrt13 /26
right now im working on my|dw:1350694287644:dw|
can you calculate theta now
uh.... well the one i had or placed on here was wrong? the okay so the cos=3sqrt13 /26
is the 2 a regular 2, or a negative 2? since its going downward of the unit circle?
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?
thats the cos of thata, dont you want theta?
yes the theta of tangent
wait. cos theta = the 3sqrt13 / 13
im so lost
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
lol yes you may. and thanks, im getting a headache from this too
alright so first we need to figure out where tan is negative. Using this circle |dw:1350696636481:dw|
these are when the functions are positive. for instance, in quadrant 1 , all function are positive
okay so tan would be negative in...................the cosine area
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|
okay so the 2nd and 3rd piece on the left sides
next we use the information it gives about theta \[\pi/2<\theta<\pi\] can you mark these angles on my picture
|dw:1350697001280:dw|
so theta is less than \[\pi/2\] but greater than \[\pi\]|dw:1350697010218:dw|
this means \[\theta is in your second quadrant
okay i see that now
alright so now lets solve the equation \[tan(\theta)=\frac{-2}{3}\]
how would you get theta by itself
do you know about inverse trig functions?
im back server was werid. let me look
no i dont know about the inverse trig functions unless your talkin about it being 1/tan or like the cos/sin
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
these cancel out the same trig function so \[arcsin(sin(x))=x\]
so what would you do in your case \[tan(\theta)=\frac{-2}{3}\]
no we have not gone over this :/ and so the arctan one would be arctan(tan(x))= -2/3 ?
if you do arctan on one side you have to do it on the other =]
-2/3arctan(tan(x))=-2/3
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 \[arctan(tan(x)=arctan(\frac{-2}{3})\]
since arctan cancels tan you get \[x=arctan(\frac{-2}{3})\]
your calculator should have an arctan symbol. it might be \[tan^{-1}\]
so the tan^-1 is called arctan?
yes =]
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
alright where is -33.69 degrees?
haha you really ask me degreees?? umm... i think its that 180/pi, right?
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)
OH, umm... well hold on
|dw:1350698454198:dw|
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
should i start from 90 degrees?? |dw:1350698677314:dw|
hold on i know what you need to do just don't know why it's been a while since going through this material
i know you have to add pi(180 degrees) to it
okay, i think its mult. the ( 180/pi) i think
i cant find the equ.
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
I really cant figure this one out... can you? or are you writingit down on paper, or looking it up online??
in your drawing at the very top, you say cos 7/25 did you mean cos(theta) = 7/25 ???
yes
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
oh wait, sry sine is positive because 0 < theta < pi/2
so sin(theta) = 24/25
lol let me check
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
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)
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
yeah i fixed that error, my bad
lol just a moment
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
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
yeah i fixed that
okay, im checking my notes cuz i have writing in every direction
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
yw

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