
How do you find the value of \[{\cos ^{ - 1}}\left( 0 \right)\]?
Answer
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Hint: In order to find the solution of a trigonometric equation start by taking the inverse trigonometric function (like inverse sin, inverse cosine, inverse tangent) of both sides of the equation and then set up reference angles to find the rest of the answers.
-For inverse sine function, the principal value branch is: $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$
-For inverse cosine function, the principal branch is: $\left[ {0,\pi } \right]$
-For inverse tangent function, the principal branch is: $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$
Complete step-by-step solution:
According to definition of inverse ratio,
If \[\cos x = 0\], then \[x = {\cos ^{ - 1}}\left( 0 \right)\] and the range is $\left[ {0,\pi } \right]$.
Now, we know that cosine function is zero for the odd multiples of $\left( {\dfrac{\pi }{2}} \right)$. So, odd multiple of $\left( {\dfrac{\pi }{2}} \right)$ lying in the principal branch of cosine $\left[ {0,\pi } \right]$ is $\dfrac{\pi }{2}$.
So, \[\cos \left( {\dfrac{\pi }{2}} \right) = 0\].
Hence, \[\dfrac{\pi }{2} = {\cos ^{ - 1}}\left( 0 \right)\]
Therefore, we have the value of \[{\cos ^{ - 1}}\left( 0 \right)\] as $\dfrac{\pi }{2}$.
Note: To find answers to such questions, we can also make use of reference triangles. Draw your triangle on the set of axes in the proper quadrant. Remember negative angles go in quadrant IV. The basic inverse trigonometric functions are used to find the missing angles in right triangles. While the regular trigonometric functions are used to determine the missing sides of the right-angled triangles, using the following formulae:
$\sin \theta = \dfrac{{{\text{Opposite Side}}}}{{{\text{Hypotenuse}}}}$
$\cos \theta = \dfrac{{{\text{Adjacent Side}}}}{{{\text{Hypotenuse}}}}$
$\tan \theta = \dfrac{{{\text{Opposite Side}}}}{{{\text{Adjacent Side}}}}$
Besides the trigonometric formulae and identity, we also have some trigonometric rules such as sine rule and cosine rule that are also of significance in solving questions involving trigonometry. Principal value branches of all the inverse trigonometric functions must be remembered.
-For inverse sine function, the principal value branch is: $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$
-For inverse cosine function, the principal branch is: $\left[ {0,\pi } \right]$
-For inverse tangent function, the principal branch is: $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$
Complete step-by-step solution:
According to definition of inverse ratio,
If \[\cos x = 0\], then \[x = {\cos ^{ - 1}}\left( 0 \right)\] and the range is $\left[ {0,\pi } \right]$.
Now, we know that cosine function is zero for the odd multiples of $\left( {\dfrac{\pi }{2}} \right)$. So, odd multiple of $\left( {\dfrac{\pi }{2}} \right)$ lying in the principal branch of cosine $\left[ {0,\pi } \right]$ is $\dfrac{\pi }{2}$.
So, \[\cos \left( {\dfrac{\pi }{2}} \right) = 0\].
Hence, \[\dfrac{\pi }{2} = {\cos ^{ - 1}}\left( 0 \right)\]
Therefore, we have the value of \[{\cos ^{ - 1}}\left( 0 \right)\] as $\dfrac{\pi }{2}$.
Note: To find answers to such questions, we can also make use of reference triangles. Draw your triangle on the set of axes in the proper quadrant. Remember negative angles go in quadrant IV. The basic inverse trigonometric functions are used to find the missing angles in right triangles. While the regular trigonometric functions are used to determine the missing sides of the right-angled triangles, using the following formulae:
$\sin \theta = \dfrac{{{\text{Opposite Side}}}}{{{\text{Hypotenuse}}}}$
$\cos \theta = \dfrac{{{\text{Adjacent Side}}}}{{{\text{Hypotenuse}}}}$
$\tan \theta = \dfrac{{{\text{Opposite Side}}}}{{{\text{Adjacent Side}}}}$
Besides the trigonometric formulae and identity, we also have some trigonometric rules such as sine rule and cosine rule that are also of significance in solving questions involving trigonometry. Principal value branches of all the inverse trigonometric functions must be remembered.
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