anonymous
  • anonymous
Need some Trig HElp
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
what do you need help with?
anonymous
  • anonymous
i will attach the file, give me a second
anonymous
  • anonymous

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anonymous
  • anonymous
anonymous
  • anonymous
DanJS
  • DanJS
hmm, try replacing cos^2 with 1 - sin^2
anonymous
  • anonymous
which attachment are you on
DanJS
  • DanJS
the first one...
DanJS
  • DanJS
\[3\sin(\theta)+3 = 2\cos^2(\theta)\]
anonymous
  • anonymous
ok
anonymous
  • anonymous
i did that
DanJS
  • DanJS
ok, it looks like you can factor that now if you put it in ax^2+bx+c form , with the x being sin
DanJS
  • DanJS
\[2\sin^2(\theta)+3\sin(\theta)+1=0\]
anonymous
  • anonymous
wait but this is what my equation looks like so far... is it right
anonymous
  • anonymous
DanJS
  • DanJS
right, from the unit circle , there is the property, sin^2 + cos^2 = 1, that is what we did
DanJS
  • DanJS
then just simplify that down to
DanJS
  • DanJS
\[2\sin^2(\theta)+3\sin(\theta)+1=0\]
anonymous
  • anonymous
ok got it
DanJS
  • DanJS
so you can factor that like usual quadratics , let v=sin(theta) 2v^2 + 3v + 1 = 0
anonymous
  • anonymous
ok give me a sec
DanJS
  • DanJS
\[(v+1)(2v+1)=0\]
DanJS
  • DanJS
replace v back with sin(theta) \[(\sin \theta + 1)(2\sin \theta +1) = 0\]
DanJS
  • DanJS
good so far?
anonymous
  • anonymous
yeah
DanJS
  • DanJS
that is true if either of those parenthesis terms are zero, so you have \[\sin(\theta) = -1~~~or~~~\sin(\theta) = -1/2\]
DanJS
  • DanJS
do you know how to figure those from the unit circle?
anonymous
  • anonymous
yeah, but im kind of weak in my unit circle so i might get it wrong
DanJS
  • DanJS
Ill tell ya, soon as you get that picture, everything clicks good...
anonymous
  • anonymous
ok should I gust google unit circle
DanJS
  • DanJS
http://www.embeddedmath.com/downloads/files/unitcircle/unitcircle-letter.pdf
DanJS
  • DanJS
do you want a quick rundown of it, how to determine angles
anonymous
  • anonymous
no i got it
anonymous
  • anonymous
i just needed a brush up
anonymous
  • anonymous
the picture is fine
DanJS
  • DanJS
aight, what are the answers for this one
anonymous
  • anonymous
the answers are -pi/2, 3pi/2, -pi/6, 7p/6
DanJS
  • DanJS
read the problem one more time through...
anonymous
  • anonymous
oh so it is just 3pi/2 and 7pi/6
DanJS
  • DanJS
not quite... lets look at the first one... sin(theta) = -1 Each point on the circle is the coordinate (cos , sin) , the y coordinate is the sin of the angle... y is -1 at 270 or 3pi/2
DanJS
  • DanJS
for sin(theta) = -1/2 there are 2 angle values where the y coordinate is -1/2
DanJS
  • DanJS
|dw:1438716059055:dw|
anonymous
  • anonymous
ahhhh
DanJS
  • DanJS
i remember taking trig and never got how angles were figured till this circle pic , after the class was over
DanJS
  • DanJS
radius is 1, so sin = opp/hyp = opp/1 = opp , or the y coordinate
anonymous
  • anonymous
ok that makes sense
anonymous
  • anonymous
thats helpful thanks
DanJS
  • DanJS
welcome
anonymous
  • anonymous
so the solutions to the equation are?
anonymous
  • anonymous
anonymous
  • anonymous
@Loser66 can you help me with my other questions

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