Expert: Sherry Wallin - 12/15/2008

Question
QUESTION: I have 2 questions on my review for my final that I am
stumped on. 1. Verify the identity: (1/1-cos x) -
(1/1+cos x)= 2cot x csc x.

2. convert to rectangular form r-sin t=-7cos t(solve for 0)




ANSWER: Clarissa~
    For the first question just find a common denominator for the left hand side and then look at what cot x is and csc x is and you will see you have it in about 2 steps...

[1(1+cosx)-1(1-cosx)]/[(1-cosx)(1+cosx)]
=(1+cosx-1+cosx)/(1-cos^2x)
= 2cosx/sin^2x
= 2(cosx/sinx)(1/sinx)
= 2cotxcscx

For the 2nd part I am not certain what you mean by solve for 0. Are you setting  r - sint + 7cos t = 0? Or are you talking about the angle 0? Are you sure you wrote the question correctly?

Above for problem 1 you want to write it as: (1/(1+cos x))-(1/(1-cos x))
or else you are saying one minus cos x minus one plus cos x.

Math Prof

---------- FOLLOW-UP ----------

QUESTION: For the second part I'm setting it equal to 0. not finding
the angle.

Answer
Hi Clarissa~
    So you have r - sint = -7cost multiply everything by r getting
r^2 -rsint = -7rcost recall y = rsint and x = rcost so we now have
r^2 = rsint -7rcost -> x^2+y^2 = y - 7x, since r^2 = x^2+y^2,

now move everything to one side
x^2 + 7x + y^2 - y = 0.

Complete the square on x and y
x^2 + 7x +(7/2)^2 + y^2 - y + (-1/2)^2 = (7/2)^2 +(-1/2)^2
(x + (7/2)^2) + (y - (1/2)^2)= 49/4 +1/4 = 50/ 4 = 25/2

The center of the circle is (-7/2,1/2) and the radius is 5/sqrt2

Is this what you were looking for?

Math Prof