Question

Find the values of \[\sin \left( {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)} \right)\]
a). \[\dfrac{1}{2}\]
b). \[\dfrac{1}{3}\]
c). \[\dfrac{1}{4}\]
d). \[1\]

Answer 1

Hint: To solve this question first we have to assume the value of the given expression. Now we have to find the value of that variable. In order to find the value of that variable first we solve the inverse trigonometry function and find that answer in radian. Then solve the angle given in the trigonometry function. And then find the value of sin function at that angle.

Complete step-by-step solution:
We have to find the value of \[\sin \left( {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)} \right)\]
Let, \[x = \sin \left( {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)} \right)\]……(i)
To solve further we solve inverse trigonometry function
The value of \[{\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right) = - \dfrac{\pi }{6}\]
On putting this value in the equation (i)
\[x = \sin \left( {\dfrac{\pi }{3} - \left( { - \dfrac{\pi }{6}} \right)} \right)\]
On multiplying the negative sign with negative \[\dfrac{\pi }{6}\] then the answer is positive \[\dfrac{\pi }{6}\]
\[x = \sin \left( {\dfrac{\pi }{3} + \dfrac{\pi }{6}} \right)\]
Now solving the angle of the trigonometric function
\[x = \sin \left( {\dfrac{\pi }{2}} \right)\]
Now on putting the value of \[\sin \dfrac{\pi }{2} = 1\]
\[x = 1\]
Final answer:
The value of the given expression is
\[\sin \left( {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)} \right) = 1\]
According to the obtained answer option d is the correct answer

Note: If the value becomes negative on \[\sin \] inverse then the value is also negative this is the property of \[\sin \] function. Example If \[{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6}\] and the value of value is negative in \[\sin \] inverse function the output is also negative \[{\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = - \dfrac{\pi }{6}\]. This condition is also applicable for normal trigonometry functions. If the value in \[\cos \] inverse function is negative then the answer is not affected on the basis of sign. Example If \[{\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{3}\] and the value of value is negative in \[\cos \] inverse function the output is also not affected \[{\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \dfrac{\pi }{3}\]. This condition is also applicable for normal trigonometry functions. Like these conditions are also applicable for all the trigonometric functions.