Super Confused...Any help is appreciated. Its Geometry btw

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Super Confused...Any help is appreciated. Its Geometry btw

Mathematics
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wat im posting it
i only need help with last 2 pages. The others are for references

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Other answers:

@jim_thompson5910 please help
@jim_thompson5910 please help
what do you need, the sine and cosine questions?
i just dont understand what im supposed to do on the last 2 pages.
what you need is a good trig cheat sheet look at the unit circle on the last page of the attached
lets say you want \[\cos(\frac{\pi}{6})\] find the angle \(\frac{\pi}{6}\) and look at the corresponding point on the unit circle the first coordinate of that point is \(\cos(\frac{\pi}{6})\) and the second coordinate is \(\sin(\frac{\pi}{6})\)
hmmm
let me know when you see it, then we can do one more
i see it, whats confusing me is that, im not sure what my teacher is wanting me to put in the blanks on the last 2 pages.
i think the values of sine and cosine asked for
as for the first question, since the radius is 1, instead of \[\cos(\theta)=\frac{opposite}{adjacent}\] it is just \[\cos(\theta)\] is the first coordinate of the point on the unit circle (since the hypotenuse is 1)
i still dont understand. sorry
|dw:1441853953843:dw|
yes that's the point on the unit circle when the angle is 30 degrees the x coordinate is equal to cos(30 degrees) the y coordinate is equal to sin(30 degrees) |dw:1441854248061:dw|
Is this the answer?
So that means \[\Large \cos(30^{\circ}) = \frac{\sqrt{3}}{2}\] \[\Large \sin(30^{\circ}) = \frac{1}{2}\]
|dw:1441854362314:dw|
so i would put |dw:1441854545172:dw|
30 degrees is NOT in quadrant 2
would this be right for quadrant 1??
yeah 30 degrees is in quadrant 1
change the 30 degrees to 120??
why do that?
because 120 degrees is in quadrant II
what would cos(120) be equal to
-1/2?
yes
and sin(120) ?
square root of 3 /2
good
Do i just do that for the rest of them???
yes
OK thanks a ton!
no problem

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