on the unit circle, x^2 + y^2 = 1
so
(-3/5)^2 + y^2 = 1 solve for y, you'll get two y values and you have to pick the one that would put it in the quadrant II
Vocaloid:
if it wasn't clear before, you just need to solve
(-3/5)^2 + y^2 = 1
for y ^^
zarkam21:
4/5
Vocaloid:
awesome, since it's qII we choose the positive y-value 4/5
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first, what would cos(theta) be based on that info?
zarkam21:
um ?/5?
Vocaloid:
try sketching a right triangle in quadrant II with sin(theta) = 4/5
zarkam21:
4/3 for tan
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Vocaloid:
nope, not 4/3, it can't be positive
Vocaloid:
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Vocaloid:
this is really important to know/be able to do, this is how you would sketch a triangle in QII, now add the values for sin(theta) = 4/5 then calculate cos(theta)
zarkam21:
and i know
adj=x=3
opp=y=4
hyp=r=5
zarkam21:
okay
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Vocaloid:
check the sign of x again
Vocaloid:
we're on the left side of the y-axis so x must be negative
then calculate cos(theta), then calculate cot(theta) given sin and cos
zarkam21:
cot=cos/sin
cot=(-3/5)/(4/5)=-3/4
zarkam21:
I'm sorry im jumping ahead of myself but i know how to do this. just was trying to refresh my memory
Vocaloid:
awesome, -3/4 is it
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