What is the \[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)\] \[?\]
Answer
599.4k+ views
Hint: The inverse of the trigonometric function must be used to determine the measure of the angle. The inverse of the tangent function is read tangent inverse and is also called the arctangent relation. The inverse of the cosine function is read cosine inverse and is also called the arccosine relation. The inverse of the sine function is read sine inverse and is also called the arcsine relation.
Complete step by step solution:
Given \[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)\] -----(1)
We know that \[\sin \left( {\dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}\] ------(2)
Taking \[{\sin ^{ - 1}}\] on both sides of the equation (2). Then the equation (2) becomes
\[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = \dfrac{\pi }{3}\] -------(3)
Since \[\sin x\] is a periodic function with period \[\pi \] . By definition of a periodic function, there exist any integer \[n\] , such that
\[\sin \left( {2n\pi + \dfrac{\pi }{3}} \right) = \sin \left( {\dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}\] --(4) for any integer \[n\] .
Since the range of \[{\sin ^{ - 1}}\left( x \right)\] lie in the range \[\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] \] .
From the equation (4) only \[\dfrac{\pi }{3}\] lies in the closed interval \[\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] \] .
Hence, the value of \[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)\] is \[\dfrac{{\sqrt 3 }}{2}\] .
So, the correct answer is “\[\dfrac{\pi }{3}\]”.
Note: Note that the domain of \[{\sin ^{ - 1}}\left( x \right)\] is \[\left[ { - 1,1} \right] \] . The principal value denoted \[{\tan ^{ - 1}}\] is chosen to lie in the range \[\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)\] . Hence the exact value of \[{\tan ^{ - 1}}\left( { - x} \right)\] for any value of \[x\] lies in the \[\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)\] .Also note that \[\sin ( - x) = - \sin (x)\] , \[\cos ( - x) = \cos (x)\] and \[\tan ( - x) = - \tan (x)\] .
Complete step by step solution:
Given \[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)\] -----(1)
We know that \[\sin \left( {\dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}\] ------(2)
Taking \[{\sin ^{ - 1}}\] on both sides of the equation (2). Then the equation (2) becomes
\[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = \dfrac{\pi }{3}\] -------(3)
Since \[\sin x\] is a periodic function with period \[\pi \] . By definition of a periodic function, there exist any integer \[n\] , such that
\[\sin \left( {2n\pi + \dfrac{\pi }{3}} \right) = \sin \left( {\dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}\] --(4) for any integer \[n\] .
Since the range of \[{\sin ^{ - 1}}\left( x \right)\] lie in the range \[\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] \] .
From the equation (4) only \[\dfrac{\pi }{3}\] lies in the closed interval \[\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] \] .
Hence, the value of \[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)\] is \[\dfrac{{\sqrt 3 }}{2}\] .
So, the correct answer is “\[\dfrac{\pi }{3}\]”.
Note: Note that the domain of \[{\sin ^{ - 1}}\left( x \right)\] is \[\left[ { - 1,1} \right] \] . The principal value denoted \[{\tan ^{ - 1}}\] is chosen to lie in the range \[\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)\] . Hence the exact value of \[{\tan ^{ - 1}}\left( { - x} \right)\] for any value of \[x\] lies in the \[\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)\] .Also note that \[\sin ( - x) = - \sin (x)\] , \[\cos ( - x) = \cos (x)\] and \[\tan ( - x) = - \tan (x)\] .
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

Draw a labelled diagram of the neuron and describe class 11 biology CBSE

