The equation y= 2 - 3sin (pi/ 4) (x-1) has a fundemental period of?
2pi / (pi/4) = 8
To determine the fundamental period of the given equation y = 2 - 3sin(pi/4)(x-1), we need to understand the concept of the fundamental period.
The fundamental period refers to the smallest positive value of "T" for which the trigonometric function repeats itself. In this case, it represents the length of one complete cycle of the function.
To find the fundamental period, we must examine the coefficient of the independent variable (x) within the sine function, as well as any additional transformations applied to it.
In the equation y = 2 - 3sin(pi/4)(x-1), the coefficient of (x-1) is 1. This indicates that the function will complete one full cycle for every x-increment of 1 unit.
Additionally, the coefficient within the sine function is pi/4. The coefficient affects the frequency or "squishing/stretching" of the sine wave. A coefficient of pi/4 means that the wave will complete four cycles within an interval of 2π (1 full cycle).
Since we already established that the function completes one full cycle for every x-increment of 1 unit, we need to determine how many total cycles the function completes within a 2π interval.
Thus, we can calculate the fundamental period by dividing the interval 2π by the number of cycles within this interval.
Number of cycles within 2π = 2π / (pi/4)
Simplifying, we can multiply by the reciprocal:
Number of cycles within 2π = 2π * (4/pi) = 8
Hence, the fundamental period of the equation y = 2 - 3sin(pi/4)(x-1) is 8 units. This means the function repeats itself every 8 units along the x-axis.