what is the period of these functions y=1/2 cos 3pi/2
y=2 cos piX/2
To find the period of a trigonometric function, we need to determine the length of one complete cycle of the function. In other words, we need to identify the distance between two consecutive maximum or minimum points.
Let's start with the first function: y = (1/2)cos(3π/2).
The general form of the cosine function is y = A cos(Bx + C), where A, B, and C are constants. The value of B determines the period of the function.
In this case, B is 3π/2. The period of the cosine function is given by the formula: T = (2π)/|B|
So, for the first function, the period will be T = (2π)/|(3π/2)|.
To simplify the expression, we divide the numerator and denominator of the fraction by π/2:
T = (2π)/(3π/2) = (2π) * (2/3π) = 4/3
Therefore, the period of the first function y = (1/2)cos(3π/2) is 4/3.
Now let's move on to the second function: y = 2cos((πx)/2).
Again, we analyze the value of B, which is (π/2) in this case.
Using the formula for the period, T = (2π)/|B|, we can calculate:
T = (2π)/|π/2| = (2π) * (2/π) = 4
Hence, the period of the second function y = 2cos((πx)/2) is 4.