
How do you find the exact value for \[\sec \left( 18\pi \right)\]?
Answer
544.8k+ views
Hint: This type of problem is based on the concept of trigonometry. First, we have to convert \[\sec \left( 18\pi \right)\] into ‘cos’ function, that is, \[\sec \theta =\dfrac{1}{\cos \theta }\]. We find that \[\cos 0=1,\cos \pi =-1,\cos 2\pi =1\] and so on. Therefore, we need to find a sequence to obtain the value of \[\cos n\pi \]. Then, substitute n=18 and find the value of \[\cos 18\pi \]. And convert this into a sec function to obtain the final answer.
Complete step-by-step answer:
According to the question, we are asked to find the value of the given function \[\sec \left( 18\pi \right)\].
We have given the function is \[\sec \left( 18\pi \right)\]. -----(1)
We first have to convert \[\sec \left( 18\pi \right)\] into a ‘cos’ function.
We know that \[\sec \theta =\dfrac{1}{\cos \theta }\]. Substituting this in (1), we get,
\[\Rightarrow \sec \left( 18\pi \right)=\dfrac{1}{\cos \left( 18\pi \right)}\] -------(2)
We know that
\[\cos 0=1,\cos \pi =-1,\cos \left( 2\pi \right)=1,\cos \left( 3\pi \right)=-1\] and so on.
Here we observe 1 and -1 repeating consecutively in addition to \[\pi \] to the ‘cos’ function.
Therefore, we get
\[\cos \left( n\pi \right)={{\left( -1 \right)}^{n}}\]
Where n=0,1,2,3,……
We need to find \[\cos \left( 18\pi \right)\].
\[\therefore \cos \left( 18\pi \right)={{\left( -1 \right)}^{18}}\]
Since 18 is an even number, \[{{\left( -1 \right)}^{18}}=1\].
Therefore,
\[\cos \left( 18\pi \right)=1\] -----(3)
Now, let us substitute (3) in (2). We get,
\[\sec \left( 18\pi \right)=\dfrac{1}{1}\]
\[\therefore \sec \left( 18\pi \right)=1\]
Hence, the exact value of \[\sec \left( 18\pi \right)\] is 1.
Note: Whenever you get this type of problems, we should always try to make the necessary changes in the given function to get the final of the function which will be the required answer. We should avoid calculation mistakes based on sign conventions. We should be thorough with the trigonometric identities and use then, if needed. Similarly, we can find the solution of \[\sec \left( 7\pi \right)\] the exact same way.
Complete step-by-step answer:
According to the question, we are asked to find the value of the given function \[\sec \left( 18\pi \right)\].
We have given the function is \[\sec \left( 18\pi \right)\]. -----(1)
We first have to convert \[\sec \left( 18\pi \right)\] into a ‘cos’ function.
We know that \[\sec \theta =\dfrac{1}{\cos \theta }\]. Substituting this in (1), we get,
\[\Rightarrow \sec \left( 18\pi \right)=\dfrac{1}{\cos \left( 18\pi \right)}\] -------(2)
We know that
\[\cos 0=1,\cos \pi =-1,\cos \left( 2\pi \right)=1,\cos \left( 3\pi \right)=-1\] and so on.
Here we observe 1 and -1 repeating consecutively in addition to \[\pi \] to the ‘cos’ function.
Therefore, we get
\[\cos \left( n\pi \right)={{\left( -1 \right)}^{n}}\]
Where n=0,1,2,3,……
We need to find \[\cos \left( 18\pi \right)\].
\[\therefore \cos \left( 18\pi \right)={{\left( -1 \right)}^{18}}\]
Since 18 is an even number, \[{{\left( -1 \right)}^{18}}=1\].
Therefore,
\[\cos \left( 18\pi \right)=1\] -----(3)
Now, let us substitute (3) in (2). We get,
\[\sec \left( 18\pi \right)=\dfrac{1}{1}\]
\[\therefore \sec \left( 18\pi \right)=1\]
Hence, the exact value of \[\sec \left( 18\pi \right)\] is 1.
Note: Whenever you get this type of problems, we should always try to make the necessary changes in the given function to get the final of the function which will be the required answer. We should avoid calculation mistakes based on sign conventions. We should be thorough with the trigonometric identities and use then, if needed. Similarly, we can find the solution of \[\sec \left( 7\pi \right)\] the exact same way.
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