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lilsis76
Group Title
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 (B0)
ill draw the 3 data thingys given
 2 years ago
 2 years ago
lilsis76 Group Title
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 (B0) ill draw the 3 data thingys given
 2 years ago
 2 years ago

This Question is Closed

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
dw:1350692195063:dw
 2 years ago

surdawi Group TitleBest ResponseYou've already chosen the best response.1
dw:1350692395385:dw
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
based on the information i know it will be first quad.
 2 years ago

surdawi Group TitleBest ResponseYou've already chosen the best response.1
csc B = 1/sin B
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
oh duh haha its negative, so its going backwards right?
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
dw:1350692578128:dw right?
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
dang it, im going to need identities right? cuz it kinda looks like it :/
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
im trying to find the identities
 2 years ago

surdawi Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
UGH!!! okay im back lol okay, are these the identities I should know for finding theta and beta then? just so i save it?
 2 years ago

surdawi Group TitleBest ResponseYou've already chosen the best response.1
\[\tan \theta=\frac{ 2 }{ 3 }\] \[\tan^{1} (\tan \theta)=\tan^{1} (\frac{ 2 }{ 3 })\]
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
okay so far i know: sin = 2/x cos =2/x and tan is the same 2/3
 2 years ago

surdawi Group TitleBest ResponseYou've already chosen the best response.1
what does x = ?
 2 years ago

surdawi Group TitleBest ResponseYou've already chosen the best response.1
dw:1350693950387:dw
 2 years ago

surdawi Group TitleBest ResponseYou've already chosen the best response.1
by the way what grade is this
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
okay hold on, this is precal. im in a community college, but im horrible at math haha
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
x = sqrt 13
 2 years ago

surdawi Group TitleBest ResponseYou've already chosen the best response.1
been a while since i been in the class anyways x is correct
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
oh really? did u graduate? and YAY okay i got the x, now will that be the theta?
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
okay so the cos=3sqrt13 /26
 2 years ago

surdawi Group TitleBest ResponseYou've already chosen the best response.1
right now im working on mydw:1350694287644:dw
 2 years ago

surdawi Group TitleBest ResponseYou've already chosen the best response.1
can you calculate theta now
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
uh.... well the one i had or placed on here was wrong? the okay so the cos=3sqrt13 /26
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
is the 2 a regular 2, or a negative 2? since its going downward of the unit circle?
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

surdawi Group TitleBest ResponseYou've already chosen the best response.1
thats the cos of thata, dont you want theta?
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
yes the theta of tangent
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
wait. cos theta = the 3sqrt13 / 13
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
im so lost
 2 years ago

surdawi Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
lol yes you may. and thanks, im getting a headache from this too
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
alright so first we need to figure out where tan is negative. Using this circle dw:1350696636481:dw
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
these are when the functions are positive. for instance, in quadrant 1 , all function are positive
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
okay so tan would be negative in...................the cosine area
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
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 quadrantdw:1350696881960:dw
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
okay so the 2nd and 3rd piece on the left sides
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
next we use the information it gives about theta \[\pi/2<\theta<\pi\] can you mark these angles on my picture
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
dw:1350697001280:dw
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
so theta is less than \[\pi/2\] but greater than \[\pi\]dw:1350697010218:dw
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
this means \[\theta is in your second quadrant
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
okay i see that now
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
alright so now lets solve the equation \[tan(\theta)=\frac{2}{3}\]
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
how would you get theta by itself
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
do you know about inverse trig functions?
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
im back server was werid. let me look
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
no i dont know about the inverse trig functions unless your talkin about it being 1/tan or like the cos/sin
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
these cancel out the same trig function so \[arcsin(sin(x))=x\]
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
so what would you do in your case \[tan(\theta)=\frac{2}{3}\]
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
no we have not gone over this :/ and so the arctan one would be arctan(tan(x))= 2/3 ?
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
if you do arctan on one side you have to do it on the other =]
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
2/3arctan(tan(x))=2/3
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
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 \[arctan(tan(x)=arctan(\frac{2}{3})\]
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
since arctan cancels tan you get \[x=arctan(\frac{2}{3})\]
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
your calculator should have an arctan symbol. it might be \[tan^{1}\]
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
so the tan^1 is called arctan?
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
alright where is 33.69 degrees?
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
haha you really ask me degreees?? umm... i think its that 180/pi, right?
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
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)
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
OH, umm... well hold on
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
dw:1350698454198:dw
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
should i start from 90 degrees?? dw:1350698677314:dw
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
hold on i know what you need to do just don't know why it's been a while since going through this material
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
i know you have to add pi(180 degrees) to it
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
okay, i think its mult. the ( 180/pi) i think
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
i cant find the equ.
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
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 degreessin15degrees =.2588
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
I really cant figure this one out... can you? or are you writingit down on paper, or looking it up online??
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
in your drawing at the very top, you say cos 7/25 did you mean cos(theta) = 7/25 ???
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
oh wait, sry sine is positive because 0 < theta < pi/2
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
so sin(theta) = 24/25
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
lol let me check
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
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)
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
yeah i fixed that error, my bad
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
lol just a moment
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
yeah i fixed that
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
okay, im checking my notes cuz i have writing in every direction
 2 years ago

lilsis76 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
yw
 2 years ago
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