Answer
Verified
377.4k+ views
Hint: Equate the trigonometric function to a variable. And then find the ranges of the \[\cos \] functions and solve for them. Find the value that satisfies the condition. Substitute the angle of \[\pi \] and evaluate it.
Complete step-by-step solution:
Let us learn about inverse trigonometric functions. Inverse trigonometric functions are also known as anti trigonometric functions, arcus functions, and cyclometric functions. These inverse trigonometric functions formulas enable us to find out any angles with any of the trigonometry ratios. The inverse trigonometric function does exactly function the opposite way of the normal trigonometric functions. The -1 in the exponent of the trigonometric is not the exponent but it is the symbol for inverse function. The range of \[{{\cos }^{-1}}\] function is \[\left( 0,\pi \right)\].
Now let us find the value of the given trigonometric function \[{{\cos }^{-1}}\left( \cos \dfrac{13\pi }{6} \right)\]
Let \[y={{\cos }^{-1}}\left( \cos \dfrac{13\pi }{6} \right)\]
\[\dfrac{13\pi }{6}=\dfrac{13\times 180}{6}={{390}^{\circ }}\]
Multiply with \[\cos \] on both sides, we get
\[\Rightarrow \cos y=\cos \dfrac{13\pi }{6}\]
On substituting the value of \[\dfrac{\pi }{6}\], we get, i.e. \[\dfrac{13\pi }{6}=\dfrac{13\times 180}{6}=13\times 30={{390}^{\circ }}\]
\[\Rightarrow \cos y=\cos \left( {{390}^{\circ }} \right)\]
As we know that the range of principal values of \[\cos \] is \[\left(0, \pi \right)\].
\[\therefore y={{390}^{\circ }}\] is not possible.
Now,
\[\begin{align}
& \Rightarrow \cos y=\cos \left( {{390}^{\circ }} \right) \\
& \Rightarrow \cos y=\cos \left( {{360}^{\circ }}+{{30}^{\circ }} \right) \\
& \Rightarrow \cos y=\cos {{30}^{\circ }} \\
& \Rightarrow \cos y=\cos \left( {{30}^{\circ }}\times \frac{\pi }{180} \right) \\
& \Rightarrow \cos y=\cos \left( \frac{\pi }{6} \right) \\
& \therefore y=\frac{\pi }{6} \\
\end{align}\]
This is in the range of principle values of \[{{\cos }^{-1}}\] is \[\left( 0,\pi \right)\].
\[\therefore \] \[{{\cos }^{-1}}\left( \cos \dfrac{13\pi }{6} \right)\]=\[y=\dfrac{\pi }{6}\].
The graph of the given function is
Note: Here is the list of properties of inverse trigonometric functions.
\[\begin{align}
& \sin \left( {{\sin }^{-}}^{1}x \right)\text{ }=\text{ }x,\text{ }if\text{ }-1\text{ }\le \text{ }x\text{ }\le \text{ }1 \\
& \cos \left( {{\cos }^{-}}^{1}x \right)\text{ }=\text{ }x,\text{ }if\text{ }-1\text{ }\le \text{ }x\text{ }\le \text{ }1 \\
& \tan \left( {{\tan }^{-}}^{1}x \right)\text{ }=\text{ }x,\text{ }if\text{ }-\infty \text{ }\le \text{ }x\text{ }\le \infty \\
& \cot \left( {{\cot }^{-}}^{1}x \right)\text{ }=\text{ }x,\text{ }if\text{ }-\infty \le \text{ }x\text{ }\le \infty \\
& \sec \left( {{\sec }^{-}}^{1}x \right)\text{ }=\text{ }x,\text{ }if\text{ }-\infty \text{ }\le \text{ }x\text{ }\le \text{ }-1\text{ }or\text{ }1\text{ }\le \text{ }x\text{ }\le \text{ }\infty \\
& \csc \left( {{\csc }^{-}}^{1}x \right)\text{ }=\text{ }x,\text{ }if\text{ }-\infty \text{ }\le \text{ }x\text{ }\le \text{ }-1\text{ }or\text{ }1\text{ }\le \text{ }x\text{ }\le \text{ }\infty \\
\end{align}\]
The inverse trigonometric relations are not functions because for any given input there exists more than one output. That is, for a given number there exists more than one angle whose sine, cosine, etc., is that number.
Complete step-by-step solution:
Let us learn about inverse trigonometric functions. Inverse trigonometric functions are also known as anti trigonometric functions, arcus functions, and cyclometric functions. These inverse trigonometric functions formulas enable us to find out any angles with any of the trigonometry ratios. The inverse trigonometric function does exactly function the opposite way of the normal trigonometric functions. The -1 in the exponent of the trigonometric is not the exponent but it is the symbol for inverse function. The range of \[{{\cos }^{-1}}\] function is \[\left( 0,\pi \right)\].
Now let us find the value of the given trigonometric function \[{{\cos }^{-1}}\left( \cos \dfrac{13\pi }{6} \right)\]
Let \[y={{\cos }^{-1}}\left( \cos \dfrac{13\pi }{6} \right)\]
\[\dfrac{13\pi }{6}=\dfrac{13\times 180}{6}={{390}^{\circ }}\]
Multiply with \[\cos \] on both sides, we get
\[\Rightarrow \cos y=\cos \dfrac{13\pi }{6}\]
On substituting the value of \[\dfrac{\pi }{6}\], we get, i.e. \[\dfrac{13\pi }{6}=\dfrac{13\times 180}{6}=13\times 30={{390}^{\circ }}\]
\[\Rightarrow \cos y=\cos \left( {{390}^{\circ }} \right)\]
As we know that the range of principal values of \[\cos \] is \[\left(0, \pi \right)\].
\[\therefore y={{390}^{\circ }}\] is not possible.
Now,
\[\begin{align}
& \Rightarrow \cos y=\cos \left( {{390}^{\circ }} \right) \\
& \Rightarrow \cos y=\cos \left( {{360}^{\circ }}+{{30}^{\circ }} \right) \\
& \Rightarrow \cos y=\cos {{30}^{\circ }} \\
& \Rightarrow \cos y=\cos \left( {{30}^{\circ }}\times \frac{\pi }{180} \right) \\
& \Rightarrow \cos y=\cos \left( \frac{\pi }{6} \right) \\
& \therefore y=\frac{\pi }{6} \\
\end{align}\]
This is in the range of principle values of \[{{\cos }^{-1}}\] is \[\left( 0,\pi \right)\].
\[\therefore \] \[{{\cos }^{-1}}\left( \cos \dfrac{13\pi }{6} \right)\]=\[y=\dfrac{\pi }{6}\].
The graph of the given function is
Note: Here is the list of properties of inverse trigonometric functions.
\[\begin{align}
& \sin \left( {{\sin }^{-}}^{1}x \right)\text{ }=\text{ }x,\text{ }if\text{ }-1\text{ }\le \text{ }x\text{ }\le \text{ }1 \\
& \cos \left( {{\cos }^{-}}^{1}x \right)\text{ }=\text{ }x,\text{ }if\text{ }-1\text{ }\le \text{ }x\text{ }\le \text{ }1 \\
& \tan \left( {{\tan }^{-}}^{1}x \right)\text{ }=\text{ }x,\text{ }if\text{ }-\infty \text{ }\le \text{ }x\text{ }\le \infty \\
& \cot \left( {{\cot }^{-}}^{1}x \right)\text{ }=\text{ }x,\text{ }if\text{ }-\infty \le \text{ }x\text{ }\le \infty \\
& \sec \left( {{\sec }^{-}}^{1}x \right)\text{ }=\text{ }x,\text{ }if\text{ }-\infty \text{ }\le \text{ }x\text{ }\le \text{ }-1\text{ }or\text{ }1\text{ }\le \text{ }x\text{ }\le \text{ }\infty \\
& \csc \left( {{\csc }^{-}}^{1}x \right)\text{ }=\text{ }x,\text{ }if\text{ }-\infty \text{ }\le \text{ }x\text{ }\le \text{ }-1\text{ }or\text{ }1\text{ }\le \text{ }x\text{ }\le \text{ }\infty \\
\end{align}\]
The inverse trigonometric relations are not functions because for any given input there exists more than one output. That is, for a given number there exists more than one angle whose sine, cosine, etc., is that number.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE