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Question: Find the period of \(\cos \left( \cos x \right)+\cos \left( \sin x \right)?\)...

Find the period of cos⁡(cos⁡x)+cos⁡(sin⁡x)?\cos \left( \cos x \right)+\cos \left( \sin x \right)?

Explanation

Solution

The length of one complete cycle is called a period of a trigonometric function. The cosine function is even and the sine function is odd. A function is even, if g(−x)=g(x).g\left( -x \right)=g\left( x \right). A function is odd, if g(−x)=−g(x).g\left( -x \right)=-g\left( x \right).

Complete step-by-step solution:
We know that the period of a trigonometric function is the length of one complete cycle.
Consider the given trigonometric function cos⁡(cos⁡x)+cos⁡(sin⁡x).\cos \left( \cos x \right)+\cos \left( \sin x \right).
Now, let us suppose that f(x)=cos⁡(cos⁡x)+cos⁡(sin⁡x).f\left( x \right)=\cos \left( \cos x \right)+\cos \left( \sin x \right).
Also, we know that the cosine function is even and the sine function is odd.
We have already learnt that, for an even function g(x),g(−x)=g(x).g\left( x \right), g\left( -x \right)=g\left( x \right).
Similarly, for an odd function h(x),h(−x)=−h(x).h\left( x \right), h\left( -x \right)=-h\left( x \right).
So, we will get cos⁡(−x)=cos⁡x\cos \left( -x \right)=\cos x and sin⁡(−x)=−sin⁡x.\sin \left( -x \right)=-\sin x.
Now we use these identities to find the period of the given function ⇒f(x)=cos⁡(cos⁡x)+cos⁡(sin⁡x).\Rightarrow f\left( x \right)=\cos \left( \cos x \right)+\cos \left( \sin x \right).
Also, we know that cos⁡(π2+x)=−sin⁡x.\cos \left( \dfrac{\pi }{2}+x \right)=-\sin x.
Similarly, sin⁡(π2+x)=cos⁡x.\sin \left( \dfrac{\pi }{2}+x \right)=\cos x.
Because, in the second quadrant, the sine function is positive and the cosine function is negative.
Since we have these identities, we can find that
⇒f(π2+x)=cos⁡(cos⁡(π2+x))+cos⁡(sin⁡(π2+x)).\Rightarrow f\left( \dfrac{\pi }{2}+x \right)=\cos \left( \cos \left( \dfrac{\pi }{2}+x \right) \right)+\cos \left( \sin \left( \dfrac{\pi }{2}+x \right) \right).
From the above, we will get the following
⇒f(π2+x)=cos⁡(−sin⁡x)+cos⁡(cos⁡x).\Rightarrow f\left( \dfrac{\pi }{2}+x \right)=\cos \left( -\sin x \right)+\cos \left( \cos x \right).
Now we can use the properties of even functions to get,
⇒f(π2+x)=cos⁡(sin⁡x)+cos⁡(cos⁡x).\Rightarrow f\left( \dfrac{\pi }{2}+x \right)=\cos \left( \sin x \right)+\cos \left( \cos x \right).
Because, we will get cos⁡(−sin⁡x)=cos⁡(sin⁡x)\cos \left( -\sin x \right)=\cos \left( \sin x \right) and cos⁡(cos⁡x)=cos⁡(cos⁡x).\cos \left( \cos x \right)=\cos \left( \cos x \right).
And now we can see that
⇒f(π2+x)=cos⁡(cos⁡x)+cos⁡(sin⁡x).\Rightarrow f\left( \dfrac{\pi }{2}+x \right)=\cos \left( \cos x \right)+\cos \left( \sin x \right).
And this will give us the following
⇒f(π2+x)=cos⁡(cos⁡x)+cos⁡(sin⁡x)=f(x).\Rightarrow f\left( \dfrac{\pi }{2}+x \right)=\cos \left( \cos x \right)+\cos \left( \sin x \right)=f\left( x \right).
Thus, we will get f(π2+x)=f(x).f\left( \dfrac{\pi }{2}+x \right)=f\left( x \right).
Hence, we can conclude that the period of the given function f(x)=cos⁡(cos⁡x)+cos⁡(sin⁡x)f\left( x \right)=\cos \left( \cos x \right)+\cos \left( \sin x \right) is π2.\dfrac{\pi }{2}.

Note: The period of the cosine function is 2π.2\pi . That is, the period of cos⁡x=2π.\cos x=2\pi . The period of the sine function is also 2π.2\pi . That is, the period of sin⁡x=2π.\sin x=2\pi . The period can be found as follows:
Period of cos⁡(cos⁡x)=π\cos \left( \cos x \right)=\pi and cos⁡(sin⁡x)=π.\cos \left( \sin x \right)=\pi .
Now the period of f(x)=12(LCM ofπ and π)f\left( x \right)=\dfrac{1}{2}\left( \text{LCM of} \pi \text{ and } \pi \right)
That is, f(x)=π2,f\left( x \right)=\dfrac{\pi }{2}, Since LCM of π and π=π.\text{LCM of }\pi \text{ and }\pi =\pi .