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Question: How do you find the period of \[y = - 10\cos \left( {\dfrac{{\pi x}}{6}} \right)\]...

How do you find the period of y=10cos(πx6)y = - 10\cos \left( {\dfrac{{\pi x}}{6}} \right)

Explanation

Solution

Normalize the function and find out the coefficient and substitute in formula.
From the given equation, we need to compare it to the normalized cosine function equation of f(x) = Acos(Bx + C) + Df\left( x \right){\text{ }} = {\text{ }}Acos\left( {Bx{\text{ }} + {\text{ }}C} \right){\text{ }} + {\text{ }}D. If it's in the required form, no need for simplification. Then later, we identify the value of B which is a coefficient of x, which will be used in the period formula for the given cosine function, for which the formula isPeriod=2πBPeriod = \dfrac{{2\pi }}{{\left| B \right|}}.

Complete step by step solution:
The given equation we have is,
y=10cos(πx6)y = - 10\cos \left( {\dfrac{{\pi x}}{6}} \right)
We have to make sure that the equation is in the required and normalized form and from the given equation, we see that it is in the normalized form.
Since the formula for period isPeriod=2πBPeriod = \dfrac{{2\pi }}{{\left| B \right|}}, we need to find B, which is the coefficient of x.
So, we compare the given equation with the normal equation to find the value of B, from which the value of B is π6\dfrac{\pi }{6}.
The next step is to replace “B” with its value in the formula, which gives us
Period=2ππ6Period = \dfrac{{2\pi }}{{\left| {\dfrac{\pi }{6}} \right|}}
The 6 in the denominator gets shifted to the numerator and the π\pi in the numerator and denominator get cancelled out and finally giving the answer 12.

Period=2π×6πPeriod = \dfrac{{2\pi \times 6}}{\pi }
Period=2×6Period = 2 \times 6
Period=12Period = 12

Note: It is necessary to have the equation be normalized first, as it makes it easy for identifying the coefficient of A, B, C and so on. And also , we should be careful not to miss out the coefficients which are in fraction form, in which we can forget to consider the denominator as a part of the coefficient value.