HI,
Function: y=5cot(-2x)
can someone can please help me find the coordinates of the two halfway points of the principal cycles. Here are some answers that might help.
-I found the interval: (0,pi/2)
-The equation of the left vertical asymptote: x=0, and right asy: x=pi/2
-The coordinate of the center point of the principal cycle: (pi/4,0)
Now, I have to find the coordinate of the two halfway points of the principal cycle. I found the x coordinates of the question, which are(pi/6, ?),(pi/3,?). I'm trying to figure out the y coordinates. Please show step by step process + answer.
THANKS!
Excuse me?? Can you not just plug in the x-value?
y = 5cot(-2x)
so, at x = π/6,
y = 5cot(-2 * π/6)
= 5cot(-π/3)
= -5cot(π/3)
= -5/√3
To find the y-coordinates of the halfway points of the principal cycle, we can substitute the x-coordinates you provided into the function y = 5cot(-2x).
Let's start with the first x-coordinate, pi/6.
Step 1: Substitute pi/6 into the function:
y = 5cot(-2(pi/6))
y = 5cot(-pi/3)
Step 2: Find the cotangent of -pi/3:
Recall that cot(-x) = 1/tan(-x)
Since tangent is an odd function, tan(-x) = -tan(x), so cot(-pi/3) = 1/(-tan(pi/3))
Step 3: Find the tangent of pi/3:
The tangent of pi/3 is sqrt(3).
Step 4: Substitute the tangent value into the equation:
cot(-pi/3) = 1/(-sqrt(3))
y = 5 / (-sqrt(3))
Therefore, the y-coordinate for the first halfway point at x = pi/6 is -5/sqrt(3).
Now, let's find the y-coordinate for the second x-coordinate, pi/3.
Step 1: Substitute pi/3 into the function:
y = 5cot(-2(pi/3))
y = 5cot(-2pi/3)
Step 2: Find the cotangent of -2pi/3:
Similar to before, cot(-2x) = 1/tan(-2x)
But since tangent is an odd function, tan(-2x) = -tan(2x), so cot(-2pi/3) = 1/(-tan(2pi/3))
Step 3: Find the tangent of 2pi/3:
The tangent of 2pi/3 is -sqrt(3).
Step 4: Substitute the tangent value into the equation:
cot(-2pi/3) = 1 / (-(-sqrt(3)))
y = 5 / sqrt(3)
Therefore, the y-coordinate for the second halfway point at x = pi/3 is 5/sqrt(3).
To summarize, the coordinates of the two halfway points of the principal cycle are:
- (pi/6, -5/sqrt(3))
- (pi/3, 5/sqrt(3))