Which of the following is true? Tangent is positive in Quadrant I. Sine is negative in Quadrant II. Cosine is positive in Quadrant III. Sine is positive in Quadrant IV.

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Which of the following is true? Tangent is positive in Quadrant I. Sine is negative in Quadrant II. Cosine is positive in Quadrant III. Sine is positive in Quadrant IV.

Mathematics
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@LexiLuvv2431 The true statement in that a tangent is positive in quadrant I.
tangent is positive in first quadrant
|dw:1439494300583:dw| remember the \(\huge\color{green}{\rm CAST }\) rule

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Thank you guys so much I have a few more can you guys help?
|dw:1439494347347:dw|
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@LexiLuvv2431 u have ta give me da medal because I answerd it 1 lol
Find the sine, cosine, and tangent of 120 degrees. and I did lol
use unit circle (x,y) where x -coordinate = cos y-coordinate = sin \[\rm tan \theta = \frac{ \sin \theta }{ \cos \theta }\]
|dw:1439494649288:dw|
so cos= ? sin=? tan = ?
Sin is \[\frac{ \sqrt{3} }{ 2 }\]
Idk the others though
hm yes that's right so if sin =sqrt{3}/2 cos = what look at the same parentheses where x-coordinate represents cos
waittt okay so then would the cos equal 1/2 and the tan equal \[\sqrt{3}\] ?
cos is not just 1/2 look at first drawing i posted above cos is negative in which quadrant ?
Quadrant S
|dw:1439495263173:dw| quadrant numbers i guess you don't understand why i wrote CAST C= cos S=sin T=tan A= all functions
|dw:1439495345327:dw| cos is negative in 2nd and 3rd quadrant
now look at the unit circle 120 degree is in 2nd quadrant where cos value should be negative |dw:1439495460867:dw|
Noo sorry this is soo confusing for me.
just focus on it it's not hard you don't have to solve for anything just read the unit circle
Ohhhh I seee. I get it nowww
One more similar to this, The point (−3, 1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of θ? Make sure to show all work.
|dw:1439495526308:dw| it's like simple (x,y) graph
(x,y) x= cos y=sin so what is sin and cos in that equation ?
-3 and 1??
yew which one is cos
-3
yes right so you have draw a right triangle cos =-3 sin =1 sow at which quadrant you should draw a right triangle ?
cos = negative (-3) and sin =positive (1) hint!
Quadrant 2
yep right !!
|dw:1439495963007:dw| apply Pythagorean theorem to find 3rd side of right triangle \[\huge\rm a^2+b^2=c^2\]
a=1 b=-3 solve for c
100?
hmm 100 no how did you get that ?
replace a with 1 and b with -3 then take square of both numbers and then add them after that take square root
ohhh I forgot the 3 was negative. So -64
show your work how did you get that ?
I did 1*1 and got one and the -3 * -3 and got -9 and I added that and got -8 and then multiplied -8 * -8 and got -64
okay \[(1)+(-3)^2=c^2\] when you take even power of negative you will get positive answer
and i just noticed you said 100 earlier
which is correct but i thought you r saying c =100 c^2=100 take square root both sides you need c not c^2
wait no nvm
Okay now I'm confused
\[\huge\rm (1)^2 + (-3)^2=c^2\]\[\rm 1+9=c^2\] (-3 times -3= 9) so now add 1+9\[\huge\rm 10=c^2\] now take square root both sides
10 squared is 100
square root not square
to cancel out square you should take square root both sides to cancel out square root you should take square both sides so there is c^2 to cancel out square take square root
Okay I have no idea how to do this part
\[\huge\rm \sqrt{c^2}= \sqrt{10}\]take square root both sides sqrt{c^2} = c so \[c =\sqrt{10}\]
|dw:1439497283978:dw|
Okayy I see
now remember definition of sin,cos and tan you need this \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]
|dw:1439497367283:dw| opp= opposite side of theta adj= adjacent of theta hyp=hypotenuse of theta
|dw:1439508247714:dw|
So is the answer finished?
yes right just write value of sin cosine and tan
How do I do that?
\[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\] just replace opposite , adjacent hyp words with their values
|dw:1439498158642:dw|
Thank youu so much
like sin = opposite over hyp so opposite side of theta is 1 and hyp = sqrt{10} write it as \[\sin \theta =\frac{ 1 }{ \sqrt{10} }\] you can't leave the radical sign at the denominator so multiply top and bottom by square root 10
my pleasure:)

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