Find the value of the trigonometric expression given below
\[{{\cos }^{-1}}(\cos 6)\]
Answer
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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:
\[\begin{align}
& \Rightarrow \dfrac{3\pi }{2}<\,\text{6}\,<\,2\pi \\
& \Rightarrow -\dfrac{3\pi }{2}>\,-\text{6}\,>\,\text{}-2\pi \\
& \Rightarrow 2\pi -\dfrac{3\pi }{2}>2\pi -\,\text{6}\,>\,2\pi -2\pi \\
& \Rightarrow \dfrac{\pi }{2}>2\pi -\,\text{6}\,>\,0 \\
\end{align}\]
And thus, now we can write the value of
\[\begin{align}
& \Rightarrow {{\cos }^{-1}}(\cos (2\pi -6))=(2\pi -6)\,\,\,\,\,\,\,\, [ \because {{\cos }^{-1}}(\cos x)=x ] \\
& \Rightarrow {{\cos }^{-1}}(\cos (6))=(2\pi -6)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, [ \because {{\cos }^{-1}}(\cos (2\pi -6))={{\cos }^{-1}}(\cos (6)) ]\\
\end{align}\]
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 \[\left[ 0,\pi \right]\].
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:
\[\begin{align}
& \Rightarrow \dfrac{3\pi }{2}<\,\text{6}\,<\,2\pi \\
& \Rightarrow -\dfrac{3\pi }{2}>\,-\text{6}\,>\,\text{}-2\pi \\
& \Rightarrow 2\pi -\dfrac{3\pi }{2}>2\pi -\,\text{6}\,>\,2\pi -2\pi \\
& \Rightarrow \dfrac{\pi }{2}>2\pi -\,\text{6}\,>\,0 \\
\end{align}\]
And thus, now we can write the value of
\[\begin{align}
& \Rightarrow {{\cos }^{-1}}(\cos (2\pi -6))=(2\pi -6)\,\,\,\,\,\,\,\, [ \because {{\cos }^{-1}}(\cos x)=x ] \\
& \Rightarrow {{\cos }^{-1}}(\cos (6))=(2\pi -6)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, [ \because {{\cos }^{-1}}(\cos (2\pi -6))={{\cos }^{-1}}(\cos (6)) ]\\
\end{align}\]
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 \[\left[ 0,\pi \right]\].
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