
What is the fundamental period of $ f\left( x \right) = \dfrac{{\sin x + \sin 3x}}{{\cos x + \cos 3x}} $ ?
A. $ \dfrac{\pi }{2} $
B. $ \pi $
C. $ 2\pi $
D. $ 3\pi $
Answer
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Hint: The fundamental period of a function is the minimum time period after which a function repeats its value. So clearly, in this question we need to find the minimum time period after which $ f\left( x \right) $ will repeat its value. To find the fundamental period of $ f\left( x \right) $ you must know that the period of $ \sin x $ and $ \cos x $ is $ 2\pi $ .
Complete step-by-step answer:
Given function: $ f\left( x \right) = \dfrac{{\sin x + \sin 3x}}{{\cos x + \cos 3x}} $
In the given function Putting $ x \to \pi + x $ , we get
$ \Rightarrow f\left( {\pi + x} \right) = \dfrac{{\sin \left( {\pi + x} \right) + \sin 3\left( {\pi + x} \right)}}{{\cos \left( {\pi + x} \right) + \cos 3\left( {\pi + x} \right)}} $
We know that $ \sin (\pi + x) = - \sin (x) $ and $ \cos (\pi + x) = - \cos x $ . Putting these values, we get
$ \Rightarrow f\left( {\pi + x} \right) = \dfrac{{ - \sin x + \sin \left( {3\pi + 3x} \right)}}{{ - \cos x + \cos \left( {3\pi + 3x} \right)}} $
We also know that $ \sin (3\pi + 3x) = - \sin (3x) $ and $ \cos (3\pi + 3x) = - \cos 3x $ . Putting these values, we get
$ \Rightarrow f\left( {\pi + x} \right) = \dfrac{{ - \sin x - \sin 3x}}{{ - \cos x - \cos 3x}} $
Multiplying numerator and denominator by (-1), we get
$ \Rightarrow f\left( {\pi + x} \right) = \dfrac{{\sin x + \sin 3x}}{{\cos x + \cos 3x}} $
The obtained function in the right hand side is equal to $ f(x) $ , which means $ f(x) $ repeats its value after every $ \pi $ interval. Hence, $ \pi $ will be the fundamental period of a given function.
So, the correct answer is “ $ \pi $ ”.
Note: : The period of addition of two periodic functions is the L.C.M. of periods of those two functions.
Alternatively, in these types of questions we can also use the hit and trial method by putting given options. After putting which option the obtained function is equal to the original function, that would be the fundamental period of the function.
Complete step-by-step answer:
Given function: $ f\left( x \right) = \dfrac{{\sin x + \sin 3x}}{{\cos x + \cos 3x}} $
In the given function Putting $ x \to \pi + x $ , we get
$ \Rightarrow f\left( {\pi + x} \right) = \dfrac{{\sin \left( {\pi + x} \right) + \sin 3\left( {\pi + x} \right)}}{{\cos \left( {\pi + x} \right) + \cos 3\left( {\pi + x} \right)}} $
We know that $ \sin (\pi + x) = - \sin (x) $ and $ \cos (\pi + x) = - \cos x $ . Putting these values, we get
$ \Rightarrow f\left( {\pi + x} \right) = \dfrac{{ - \sin x + \sin \left( {3\pi + 3x} \right)}}{{ - \cos x + \cos \left( {3\pi + 3x} \right)}} $
We also know that $ \sin (3\pi + 3x) = - \sin (3x) $ and $ \cos (3\pi + 3x) = - \cos 3x $ . Putting these values, we get
$ \Rightarrow f\left( {\pi + x} \right) = \dfrac{{ - \sin x - \sin 3x}}{{ - \cos x - \cos 3x}} $
Multiplying numerator and denominator by (-1), we get
$ \Rightarrow f\left( {\pi + x} \right) = \dfrac{{\sin x + \sin 3x}}{{\cos x + \cos 3x}} $
The obtained function in the right hand side is equal to $ f(x) $ , which means $ f(x) $ repeats its value after every $ \pi $ interval. Hence, $ \pi $ will be the fundamental period of a given function.
So, the correct answer is “ $ \pi $ ”.
Note: : The period of addition of two periodic functions is the L.C.M. of periods of those two functions.
Alternatively, in these types of questions we can also use the hit and trial method by putting given options. After putting which option the obtained function is equal to the original function, that would be the fundamental period of the function.
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