Question 95528: Hi,
I have 2 questions, Im so coonfused.
1) Let P be the point on the unit circle U that corresponds to an angle t. Find the exact coordinates(x,y) of P and the exactvalues of the trigonometric functions at angle t. Use sqrt (2) for the square root of 2 and if the value of a trigonometric function is undefined, enter und. t=(-pi)
x=? y=?
sin(t)=?,cos(t)=?,tan(t)=?,csc(t)=?, sec(t)=?, cot(t)=?
2) Use the fundmental identities to find the values of the trigonometric functions(as decimal values within 0.001) for the give conditions.
sin(t)=0.6, cos(t)<0
sin(t)=?,cos(t)=?,tan(t)=?,csc(t)=?, sec(t)=?, cot(t)=?
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! the circumference of any circle is 2pi radians.
An angle of pi (-1,0)is 180 degrees from the standard position (1,0) on the circumference going counterclockwise.
-pi (-1,0) has the same coordinates but arrives there from the standard position going clockwise.
In the unit circle sin=y cos=x & tan=y/x
In this case sin=0 cos=-1 tan=0/-1=0 csc=1/0=und. sec=1/-1=-1 cot=-1/0=und
We derived all of these above formulas from (-1,0).
Easy, huh!
2nd question:
sin(t)=0.6, cos(t)<0
sin is + in quadrants 1 and 2
cos is <0 (ie. negative) in quadrants 2 and 3
So the terminal side of the angle is in quadrant 2
y=.6
If we draw a right triangle in quadrant 2 using the suppliment of the angle as our reference angle then y=.6 is the opposite side and the hypotenuse=1. In the unit circle the hypotenuse is always 1 and is always positive!
Now we can find x which is negative (quadrant 2).
y^2+x^2=c^2
.36+x^2=1
x^2=1-.36=.64
x=-.8
cos=-.8 tan=-.6/.8=-.75 cot=-.8/.6 csc=1/.6 sec=-1/.8=-1.25
EdJones
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