katieb
  • katieb
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anonymous
  • anonymous
Nnesha
  • Nnesha
you can use *unit circle * where cos is x-coordinate and sin - is y-coordinate and tan = sin over cos
Nnesha
  • Nnesha
without calculator you need to apply 45-45-90 triangle theorem or 30-60-90

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jdoe0001
  • jdoe0001
|dw:1438299380960:dw|
anonymous
  • anonymous
ok im lost
jdoe0001
  • jdoe0001
check your Unit Circle as Nnesha suggested
anonymous
  • anonymous
ok. what do i look for
Nnesha
  • Nnesha
thats why just use unit circle instead 45-45-90 theorem
Nnesha
  • Nnesha
360 degree
Nnesha
  • Nnesha
look 't jdoe's drawing
anonymous
  • anonymous
ok
Nnesha
  • Nnesha
do you have unit circle sheet ? if not then google it
anonymous
  • anonymous
got one
anonymous
  • anonymous
i look for 360 on the unit circle?
anonymous
  • anonymous
there is none because 360 IS 0 and vice versa right?
Nnesha
  • Nnesha
0 doesn't mean *nothing* :=)
Nnesha
  • Nnesha
there is an order pair right (x,y) where x=cos y =sin
anonymous
  • anonymous
yes. (1,0)
Nnesha
  • Nnesha
|dw:1438299090911:dw|
Nnesha
  • Nnesha
yes so sin 360 = ? and cos 360 = ?
anonymous
  • anonymous
sin 360 = 0 cos 360 = 1
anonymous
  • anonymous
and tan 360 = 0
anonymous
  • anonymous
ok thx. what about this one. The point (1, −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.
anonymous
  • anonymous
i find where the point 1 and -1 are at
Nnesha
  • Nnesha
you have to draw a right triangle
anonymous
  • anonymous
ok
anonymous
  • anonymous
from quadrant 1?
Nnesha
  • Nnesha
wait a sec i have to refresh the page *lag is real*
Nnesha
  • Nnesha
given point is (1,-1) so cos =1 (positive ) sin = -1 ( negative ) so at which quadrant cos is positive but sin is negative remember the CAST rule ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
4th quadrent
Nnesha
  • Nnesha
yep right
Nnesha
  • Nnesha
|dw:1438300117250:dw| now you need to find hyp apply Pythagorean theorem and find 3rd side of right triangle
Nnesha
  • Nnesha
Pythagorean theorem a^2 +b^2 = c^2 where c is hypotenuse longest side of right triangle
Nnesha
  • Nnesha
ello ?
Nnesha
  • Nnesha
if you have question , let me know plz :=)
Nnesha
  • Nnesha
|dw:1438301856130:dw| a=1 b=-1 c= ? use that formula \[\huge\rm a^2+b^2=c^2\] substitute a for 1 substitute b for -1 solve for c
anonymous
  • anonymous
c = 0
anonymous
  • anonymous
thats the answer???
Nnesha
  • Nnesha
how is it c ?
Nnesha
  • Nnesha
nope
Nnesha
  • Nnesha
first of all take square of a and b
Nnesha
  • Nnesha
\[\huge\rm 1^2 +(-1)^2=c^2\] solve for c
Nnesha
  • Nnesha
and then add them
anonymous
  • anonymous
1+ 1 =2
Nnesha
  • Nnesha
yes right so c^2=2 now take square root both sides bec you need c not c^2
anonymous
  • anonymous
root2
Nnesha
  • Nnesha
yep right |dw:1438302263294:dw| so what is sin theta = ?
Nnesha
  • Nnesha
\[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\] remember this
anonymous
  • anonymous
so sin theta = -1/root 2
Nnesha
  • Nnesha
yes right you can't leave the radical at the denominator so multiply both denoimnator and numerator by sqr root 2
anonymous
  • anonymous
root 2 sin theta = 1?
anonymous
  • anonymous
-1/2
Nnesha
  • Nnesha
correction:\[\frac{ -1 }{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} }\]
anonymous
  • anonymous
so thats answer to cos?
Nnesha
  • Nnesha
nope sin theta = -1 over square root 2 you can't have the radical at the denominator thats why multiply numerator and denominator by square root 2
anonymous
  • anonymous
shoot this is hard
anonymous
  • anonymous
idk i gtg
anonymous
  • anonymous
thx tho
Nnesha
  • Nnesha
\[\frac{ -1 }{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} }\] multiply
Nnesha
  • Nnesha
alright i have to go,too we will continue later
anonymous
  • anonymous
ok
Nnesha
  • Nnesha
:\[\frac{ -1 }{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} }\] here multiply numerator by numerator and denominator by denominator\[\frac{ -1 \times \sqrt{2} }{ \sqrt{2} \times \sqrt{2}}\] sqrt{2} times sqrt{2} = sqrt{4} which is equal to 2 you will get \[\frac{ -\sqrt{2} }{ 2 }\] now at what degrree sin equal to -sqrt{2} over 2 ?
Nnesha
  • Nnesha
same way to find value of cos that's it!
Nnesha
  • Nnesha
if you have any question let me know tomorrow. :=)

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