The cosine function is the trigonometric function, so it is known as the periodic function.
The period of the given function is π/5. The option 4 is the correct option.
What is the period of the cosine function?
The cosine function is the trigonometric function, hence it is known as the periodic function.
The period of a function is the total interval in which the function repeats itself over and over again the of the function for a complete cycle. If,
[tex]f(x)=A\cos(Bx+C)+D[/tex]
Then the period of the cosine function can be given as,
[tex]I=\dfrac{2\pi}{|B|}[/tex]
Given information-
The given function in the problem is,
[tex]f(x)=\cos(10x)[/tex]
The above function is the cosine function. Let a cosine function of in with values of [tex]x[/tex] 0 to 2π.
Now the period of the periodic function is the interval of [tex]x[/tex] values in which the the cycle of graph repeats itself in both directions.
Compare the given function with above equation we get,
[tex]B=10[/tex]
Thus the period of the [tex]\cos (10x)[/tex] is,
[tex]I=\dfrac{2\pi}{|10|}\\I=\dfrac{\pi}{|5|}\\I=\dfrac{\pi}{5}\\[/tex]
Hence the period of the given function is π/5. Thus the option 4 is the correct option.
Learn more about the period of the function here;
https://brainly.com/question/4599903
Answer:
○ [tex]\displaystyle \frac{\pi}{5}[/tex]
Explanation:
[tex]\displaystyle \boxed{f(x) = sin\:(10x + \frac{\pi}{2})} \\ \\ y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{-\frac{\pi}{20}} \hookrightarrow \frac{-\frac{\pi}{2}}{10} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\frac{\pi}{5}} \hookrightarrow \frac{2}{10}\pi \\ Amplitude \hookrightarrow 1[/tex]
OR
[tex]\displaystyle y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\frac{\pi}{5}} \hookrightarrow \frac{2}{10}\pi \\ Amplitude \hookrightarrow 1[/tex]
From the information above, you have your answer.
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