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Question: The value of \[\sin \left[ \left( \dfrac{\pi }{2} \right)-{{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} ...

The value of sin⁡[(π2)−sin⁡−1(−32)]=\sin \left[ \left( \dfrac{\pi }{2} \right)-{{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right) \right]=
A.32\dfrac{\sqrt{3}}{2}
B.−32\dfrac{-\sqrt{3}}{2}
C.12\dfrac{1}{2}
D.−12\dfrac{-1}{2}

Explanation

Solution

Hint : An equation which involves trigonometric ratio of any angle is said to be a trigonometric identity if it is satisfied for all values for which the given trigonometric ratios are defined. Trigonometric functions are periodic functions and all trigonometric functions are not bijections. Consequently their inverse does not exist. If no branch of an inverse trigonometric function is given, then it means that the principal value branch of the function. An inverse function reverses the direction of the original function.

Complete step-by-step answer :
sin⁡[(π2)−sin⁡−1(−32)]\sin \left[ \left( \dfrac{\pi }{2} \right)-{{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right) \right]
Applying the inverse trigonometric identities to solve the problem
As we know that sin⁡−1(−x)=−sin⁡−1x{{\sin }^{-1}}(-x)=-{{\sin }^{-1}}x
⇒sin⁡[(π2)+sin⁡−1(32)]\Rightarrow \sin \left[ \left( \dfrac{\pi }{2} \right)+{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) \right]
Using another trigonometric identity [sin⁡−1(32)]=π3\left[ {{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) \right]=\dfrac{\pi }{3} we get
⇒sin⁡(π2+π3)\Rightarrow \sin \left( \dfrac{\pi }{2}+\dfrac{\pi }{3} \right)
An trigonometric identity represents a relationship that is always true. A conditional relationship represents an equation which is sometimes true.
The domain of sin⁡−1x{{\sin }^{-1}}x is [−1,1]\left[ -1,1 \right] and range is [−π2,π2]\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right].
⇒sin⁡(5π6)\Rightarrow \sin \left( \dfrac{5\pi }{6} \right)
We can rewrite the above function as
⇒sin⁡(π−π6)\Rightarrow \sin \left( \pi -\dfrac{\pi }{6} \right)
Using the trigonometric identity we get
⇒sin⁡π6\Rightarrow \sin \dfrac{\pi }{6}
Further using the trigonometric identity we get
⇒sin⁡π6=32\Rightarrow \sin \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}
Hence we conclude that sin⁡[(π2)−sin⁡−1(−32)]=32\sin \left[ \left( \dfrac{\pi }{2} \right)-{{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right) \right]=\dfrac{\sqrt{3}}{2}
Therefore, option AA is the correct answer.
So, the correct answer is “Option A”.

Note : Before solving the trigonometric problems, one must be familiar with the trigonometric ratios, trigonometric identities, inverse trigonometric functions and trigonometric applications. The word ‘trigonometry’ is derived from the Greek words ‘tri’ which means three, ‘gon’ (means sides) and ‘metron’ (means measure). Trigonometry is the study of relationships between the sides and angles of a triangle. Some ratios of the sides with respect to its acute angles, called trigonometric ratios of the angle. If one of the trigonometric ratios of an acute angle is known, the remaining trigonometric ratios of the angle can be easily determined.