Question

Evaluate ${{\cos }^{-1}}\left( \cos 3 \right)$?

Answer 1

To evaluate the above cosine inverse function i.e. ${{\cos }^{-1}}\left( \cos 3 \right)$. Now, we are equating this expression to $\theta$. After that, we are going to take cosine on both sides of the equation. Then, we are going to use this cosine property which is equal to: $\cos \alpha =\cos \theta$. The general solution for this equation is $\theta =2n\pi \pm \alpha$.

Complete step by step answer:
The inverse expression given in the above problem which we have to evaluate is as follows:
${{\cos }^{-1}}\left( \cos 3 \right)$
Let us equate the above expression to $\theta$ and then the above expression will look like:
${{\cos }^{-1}}\left( \cos 3 \right)=\theta$
Now, we are going to take cosine $\left( or\cos \right)$ on both the sides of the above equation and we get,
$\cos \left( {{\cos }^{-1}}\left( \cos 3 \right) \right)=\cos \theta$
We know from the algebraic properties that if we multiply a number by its inverse then we will get 1 as the result of the multiplication. Then in the L.H.S of the above equation $\cos \left( {{\cos }^{-1}} \right)$ will become 1 and the above equation will look as follows:
$\cos 3=\cos \theta$
We know the general solution for the cosine in the following form as follows:
If $\cos \alpha =\cos \theta$ then:
$\theta =2n\pi \pm \alpha$
Then the solution of the above cosine equation is as follows:
$\theta =2n\pi \pm 3$
In the above solution, n can take any values from 0, 1, 2, 3……
Hence, we have evaluated the above expression to $2n\pi \pm 3$. Here 3 is given in radian and it is lying between 0 to 180 so we can write 3 as an answer.

Note: The common mistake which could be possible in the above problem is that you might have written the answer of the above solution as 3. Because the property in the algebra which says that if we multiply a number with its inverse then answer will be 1 but when in the trigonometric functions, we multiply the inverse of trigonometric function with its inverse then the answer is not same as multiplying trigonometric function with its inverse.
In the above problem, we have given:
${{\cos }^{-1}}\left( \cos 3 \right)$
In the above type of inverse multiplication, we cannot write just 3 but if we have given the trigonometric function as:
$\cos \left( {{\cos }^{-1}}3 \right)$
The answer of the above evaluation is 3.