Question

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

Answer 1

Hint: 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 $g\left( -x \right)=g\left( x \right).$ A function is odd, if $g\left( -x \right)=-g\left( x \right).$

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 $\cos \left( \cos x \right)+\cos \left( \sin x \right).$
Now, let us suppose that $f\left( x \right)=\cos \left( \cos x \right)+\cos \left( \sin x \right).$
Also, we know that the cosine function is even and the sine function is odd.
We have already learnt that, for an even function $g\left( x \right), g\left( -x \right)=g\left( x \right).$
Similarly, for an odd function $h\left( x \right), h\left( -x \right)=-h\left( x \right).$
So, we will get $\cos \left( -x \right)=\cos x$ and $\sin \left( -x \right)=-\sin x.$
Now we use these identities to find the period of the given function $\Rightarrow f\left( x \right)=\cos \left( \cos x \right)+\cos \left( \sin x \right).$
 Also, we know that $\cos \left( \dfrac{\pi }{2}+x \right)=-\sin x.$
Similarly, $\sin \left( \dfrac{\pi }{2}+x \right)=\cos x.$
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
$\Rightarrow f\left( \dfrac{\pi }{2}+x \right)=\cos \left( \cos \left( \dfrac{\pi }{2}+x \right) \right)+\cos \left( \sin \left( \dfrac{\pi }{2}+x \right) \right).$
From the above, we will get the following
$\Rightarrow f\left( \dfrac{\pi }{2}+x \right)=\cos \left( -\sin x \right)+\cos \left( \cos x \right).$
Now we can use the properties of even functions to get,
$\Rightarrow f\left( \dfrac{\pi }{2}+x \right)=\cos \left( \sin x \right)+\cos \left( \cos x \right).$
Because, we will get $\cos \left( -\sin x \right)=\cos \left( \sin x \right)$ and $\cos \left( \cos x \right)=\cos \left( \cos x \right).$
And now we can see that
$\Rightarrow f\left( \dfrac{\pi }{2}+x \right)=\cos \left( \cos x \right)+\cos \left( \sin x \right).$
And this will give us the following
$\Rightarrow f\left( \dfrac{\pi }{2}+x \right)=\cos \left( \cos x \right)+\cos \left( \sin x \right)=f\left( x \right).$
Thus, we will get $f\left( \dfrac{\pi }{2}+x \right)=f\left( x \right).$
Hence, we can conclude that the period of the given function $f\left( x \right)=\cos \left( \cos x \right)+\cos \left( \sin x \right)$ is $\dfrac{\pi }{2}.$

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