_____________________________________________________________
Calculus I                                         Quiz 3                       September 21, 2004

Name _________________        R. Hammack                              Score_____
_____________________________________________________________

(1) Find the equation of the line that passes through the point (5, 9) and has x-intercept 7.
Please express your answer in slope-intercept form.

We know 2 points on the line.  (5, 9) and (7, 0).  Thus its slope is (0 - 9)/(7 - 5) = -9/2

Using the point-slope form of the equation for a line, we get
y - 0 = -9/2 (x - 7)  y = -9/2x + 63/2


(2)
(a)   sin(-(3π)/4) =-1/2^(1/2) (by looking at the unit circle)
[Graphics:HTMLFiles/Q3F04A_5.gif]

(b)   cos ( 99π) = -1
(Because any radian measure that's an odd multiple of π corresponds to the point (-1, 0) on the unit circle.)


(3)  Find all values of x (in radians) for which tan(x) = -1.

From the unit circle, you can see that if x = (3π)/4, then tan(x) = -1.

[Graphics:HTMLFiles/Q3F04A_10.gif]
But note also you can add any integer multiple of π to this value, and still  tan(x) = -1.

SOLUTION:  x = (3π)/4 + n π, where n = 0,  1,  2,  3, ...


(4)  Suppose x is a number for which sin(x) = 1/3.  What are the possible values for tan(x)?

From the identity sin^2(x) + cos^2(x) = 1, we get  (1/3)^2 + cos^2(x) = 1, or  1/9 + cos^2(x) = 1.
Thus cos^2(x) = 1 - 1/9 = 8/9,    so  cos(x) =  (8/9)^(1/2) =  (22^(1/2))/3.

Now, tan(x) = sin(x)/cos(x) = (1/3)/ (22^(1/2))/3 =  1/(22^(1/2))