Express in radians the period of the graph of the equation y = 1/3(cox^2 x−sin^2 x).
between the typo and the font messup, it's hard to say
sin^2 and cos^2 has period pi (why not 2pi?)
cot^2 has period pi
So, I'd say whatever you have there has period pi
90 degree or pi in rad
To find the period of the graph of the equation y = 1/3(cos^2(x) - sin^2(x)), we need to determine the distance over which the graph repeats itself.
The general period of a trigonometric function of the form y = a * f(bx) is given by 2π/b, where a is the amplitude and b is the coefficient of x.
In the given equation, the coefficient of x is 1, so b = 1. Therefore, the period is 2π/1, which simplifies to 2π radians.
So, the period of the graph of y = 1/3(cos^2(x) - sin^2(x)) is 2π radians.
To express the period of the graph of the given equation y = (1/3)(cos² x - sin² x) in radians, we need to understand the concept of period in trigonometric functions.
The period of a function is the interval at which the function repeats itself. In trigonometric functions, the period is determined by the coefficient of the variable(x) in the function.
In this case, the function contains both the cosine term (cos² x) and the sine term (sin² x). Since cosine and sine have the same period of 2π radians, we can consider the period of the given function as the least common multiple of the periods of the cosine and sine terms.
The period of both cosine and sine is 2π radians. Thus, the least common multiple of 2π is 2π.
Hence, the period of the given function y = (1/3)(cos² x - sin² x) is 2π radians.