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lilsis76
Given tan theta = -2/3, where pi/2 < 0 < pi and csc B=2 where 0<B<pi/2 Find sin (0+B) and cos (B-0) ill draw the 3 data thingys given
based on the information i know it will be first quad.
oh duh haha its negative, so its going backwards right?
|dw:1350692578128:dw| right?
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
by the way what grade is this
okay hold on, this is precal. im in a community college, but im horrible at math haha
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
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
so theta is less than \[\pi/2\] but greater than \[\pi\]|dw:1350697010218:dw|
this means \[\theta is in your second quadrant
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?
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
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
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 ???
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
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
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