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Question: Find the values of \[\sin \left( {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \righ...

Find the values of sin⁡(π3−sin⁡−1(−12))\sin \left( {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)} \right)
a). 12\dfrac{1}{2}
b). 13\dfrac{1}{3}
c). 14\dfrac{1}{4}
d). 11

Explanation

Solution

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⁡(π3−sin⁡−1(−12))\sin \left( {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)} \right)
Let, x=sin⁡(π3−sin⁡−1(−12))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(−12)=−π6{\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right) = - \dfrac{\pi }{6}
On putting this value in the equation (i)
x=sin⁡(π3−(−π6))x = \sin \left( {\dfrac{\pi }{3} - \left( { - \dfrac{\pi }{6}} \right)} \right)
On multiplying the negative sign with negative π6\dfrac{\pi }{6} then the answer is positive π6\dfrac{\pi }{6}
x=sin⁡(π3+π6)x = \sin \left( {\dfrac{\pi }{3} + \dfrac{\pi }{6}} \right)
Now solving the angle of the trigonometric function
x=sin⁡(π2)x = \sin \left( {\dfrac{\pi }{2}} \right)
Now on putting the value of sin⁡π2=1\sin \dfrac{\pi }{2} = 1
x=1x = 1
Final answer:
The value of the given expression is
sin⁡(π3−sin⁡−1(−12))=1\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⁡\sin inverse then the value is also negative this is the property of sin⁡\sin function. Example If sin⁡−1(12)=π6{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6} and the value of value is negative in sin⁡\sin inverse function the output is also negative sin⁡−1(−12)=−π6{\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = - \dfrac{\pi }{6}. This condition is also applicable for normal trigonometry functions. If the value in cos⁡\cos inverse function is negative then the answer is not affected on the basis of sign. Example If cos⁡−1(12)=π3{\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{3} and the value of value is negative in cos⁡\cos inverse function the output is also not affected cos⁡−1(−12)=π3{\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.