Question

Find the value of the trigonometric expression given below
${{\cos }^{-1}}(\cos 6)$

Answer 1

Hint: Here we will use the value of inverse trigonometric cosine function of the form ${{\cos }^{-1}}(\cos x)=x$, when x lies in the interval $\left[ 0,\pi \right]$. So we will check for the quadrant in which the argument of the cosine function lies to solve these kind of problems.

Complete step-by-step solution -
In the question we have to find the value of the expression ${{\cos }^{-1}}(\cos 6)$.
Now, we know that when we have ${{\cos }^{-1}}(\cos x)=x$ then it means that x lies in the interval $\left[ 0,\pi \right]$.
But when x lies in the interval $\dfrac{3\pi }{2}<\,x\,<\,2\pi$, then ${{\cos }^{-1}}(\cos x)=2\pi -x$. So, this is the important concept that is to be used here.
Now, in the problem we have $\cos 6$ which has the argument as 6 which lies in $\dfrac{3\pi }{2}<\,\text{6}\,<\,2\pi$
So to bring that in the required interval of $\left[ 0,\pi \right]$ we can write 6 as $2\pi -6$.
Also we know that $(\cos (2\pi -6))=(\cos 6)\,$
So, we can write ${{\cos }^{-1}}(\cos 6)={{\cos }^{-1}}(\cos (2\pi -6))$
Now, finally we have the expression ${{\cos }^{-1}}(\cos (2\pi -6))$ for the given expression ${{\cos }^{-1}}(\cos 6)$
Here, $(2\pi -6)$ lies in the interval $\left[ 0,\pi \right]$ and that can be shown below:

& \Rightarrow \dfrac{3\pi }{2}<\,\text{6}\,<\,2\pi \\\ & \Rightarrow -\dfrac{3\pi }{2}>\,-\text{6}\,>\,\text{}-2\pi \\\ & \Rightarrow 2\pi -\dfrac{3\pi }{2}>2\pi -\,\text{6}\,>\,2\pi -2\pi \\\ & \Rightarrow \dfrac{\pi }{2}>2\pi -\,\text{6}\,>\,0 \\\ \end{aligned}$$ And thus, now we can write the value of $$\begin{aligned} & \Rightarrow {{\cos }^{-1}}(\cos (2\pi -6))=(2\pi -6)\,\,\,\,\,\,\,\, [ \because {{\cos }^{-1}}(\cos x)=x ] \\\ & \Rightarrow {{\cos }^{-1}}(\cos (6))=(2\pi -6)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, [ \because {{\cos }^{-1}}(\cos (2\pi -6))={{\cos }^{-1}}(\cos (6)) ]\\\ \end{aligned}$$ So finally we have the value of the expression $${{\cos }^{-1}}(\cos 6)$$ as $$(2\pi -6)$$. Note: We have to be careful in finding the value of the inverse trigonometric function. It is important to check the quadrant in which the argument of the trigonometric function lies. So $${{\cos }^{-1}}(\cos x)=x$$ is not true for all x, but this is only true if the argument x lies in the interval $$\left[ 0,\pi \right]$$.