anonymous
  • anonymous
evaluate the six trigonometric functions of the angle (see below for picture)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
anonymous
  • anonymous
the point is (4, -3) see the picture above
anonymous
  • anonymous
im just confused because the arrow goes all the way around so i dont know how to find answers

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myininaya
  • myininaya
|dw:1328392741044:dw|
anonymous
  • anonymous
is that all i have to do
myininaya
  • myininaya
\[\sin(\theta)=\frac{opp}{hyp}=\frac{-3}{5}\]
anonymous
  • anonymous
and do i list negatives in the answers
anonymous
  • anonymous
oh wait
myininaya
  • myininaya
adj=4 opp=-3 hyp=5
anonymous
  • anonymous
i forgot thats not the angle. here is a new picture
anonymous
  • anonymous
anonymous
  • anonymous
feta is on the outside its the big one going around
anonymous
  • anonymous
??????
myininaya
  • myininaya
I'm confused Do you want to find theta or do you want us to evaluate theta using the 6 trig ratios?
anonymous
  • anonymous
Do the operation myinaya suggested, then add 360 to get the angle that is there
anonymous
  • anonymous
evalute using ratios. but im confused because the theta is on the outside and not the inside
myininaya
  • myininaya
let me look at your pic again one sec
anonymous
  • anonymous
ok look at the second one
myininaya
  • myininaya
ok
myininaya
  • myininaya
|dw:1328393266943:dw|
myininaya
  • myininaya
\[\sin(360^o-\theta^o)=\frac{opp}{hyp}=\frac{-3}{5}\] \[\sin(360)\cos(\theta)-\sin(\theta) \cos(360)=\frac{-3}{5}\]
myininaya
  • myininaya
\[(0) \cos(\theta)-\sin(\theta)(1)=\frac{-3}{5}\]
myininaya
  • myininaya
\[\sin(\theta)=\frac{3}{5}\]
myininaya
  • myininaya
Does that make since?
myininaya
  • myininaya
sense*?
anonymous
  • anonymous
no sry i dont get what u just said
myininaya
  • myininaya
which part?
myininaya
  • myininaya
There are 360 degrees in circle we have theta already making up part of that 360 degrees 360-theta is left to make a full rotation
myininaya
  • myininaya
Recall your sum and difference formulas
anonymous
  • anonymous
ok but how do i solve for sin cos tan etc if its on the outside
myininaya
  • myininaya
you will need those here
myininaya
  • myininaya
I just did sin(theta) for you
myininaya
  • myininaya
\[\cos(360-\theta)=\frac{4}{5}\] \[\cos(360)\cos(\theta)+\sin(360 )\sin(\theta)=\frac{4}{5}\]
myininaya
  • myininaya
\[1\cos(\theta)+0 \sin(\theta)=\frac{4}{5} => \cos(\theta)=\frac{4}{5}\]
anonymous
  • anonymous
so is it the same as being on teh inside
myininaya
  • myininaya
well sin(theta) does not equal sin(360-theta) but cos(theta) equals cos(360-theta) so it is not the same
myininaya
  • myininaya
it is not the same because it is not always true and i showed this by showing sin(theta) does not equal sin(360-theta)
anonymous
  • anonymous
but when i do it using theta on teh inside i get the same answers (sin is -3/5, cos is 4/5, tan is -3/4, etc
myininaya
  • myininaya
no sin(theta)=3/5 sin(360-theta)=-3/5 3/5 does not equal -3/5
anonymous
  • anonymous
ohh ok....thanks!!!
anonymous
  • anonymous
can i ask you another question
anonymous
  • anonymous
its kinda silly...but how do u simplify down (-4/\[\sqrt{52}\])
myininaya
  • myininaya
what is 13*4
anonymous
  • anonymous
myininaya
  • myininaya
\[\sqrt{52}=\sqrt{4} \sqrt{13}=2 \sqrt{13}\]
anonymous
  • anonymous
see pic above...i got it down to there but then stuck
myininaya
  • myininaya
\[\frac{-4}{2 \sqrt{13}}=\frac{-2}{\sqrt{13}}=\frac{-2 \sqrt{13}}{13}\]

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