Question

Find the following $ \sin \left[ {\dfrac{\pi }{2} - {{\sin }^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)} \right]$

Answer 1

Hint: In the question we can see both inverse and normal trigonometric identity. In case of inverse identity, we will get the angle as outcome and in case of trigonometric function we will get value as output:
Some formulas:$ \operatorname{Sin} \left( {\dfrac{\pi }{2} + \theta } \right) = \operatorname{Cos} \theta $

Complete step-by-step solution:
The given problem is related to trigonometric functions.
As we know,
The period of sin functions in $ 2\pi $ . This function is positive in the first and second quadrant.
The period of cos function is $ \pi $ . This function is positive in the first and fourth quadrant.
Sin is negative in third and fourth quadrant
$ = \sin (\pi + \dfrac{\pi }{6}) = - \dfrac{{\sqrt 3 }}{2} \\
   \Rightarrow \sin (\dfrac{{7\pi }}{6}) = - \dfrac{{\sqrt 3 }}{2} $
We can now rewrite the given equation as
$ = \sin \left[ {\dfrac{\pi }{2} - {{\sin }^{ - 1}}\left( {\dfrac{{7\pi }}{6}} \right)} \right] \\
   \Rightarrow \sin \left( {\dfrac{\pi }{2} - \dfrac{{7\pi }}{6}} \right) \\
   \Rightarrow \sin \left( {\dfrac{\pi }{2} - \pi - \dfrac{\pi }{6}} \right) $
By further solving we get,
$ = \sin \left[ { - \left( {\dfrac{\pi }{2} + \dfrac{\pi }{6}} \right)} \right] \\
  as,\,\sin ( - x) = - \sin (x) \\
 \Rightarrow - \sin \left[ {\left( {\dfrac{\pi }{2} + \dfrac{\pi }{6}} \right)} \right] $
Also,
$
  \because \sin (\dfrac{\pi }{2} + x) = \cos x \\
  = - \sin \left[ {\left( {\dfrac{\pi }{2} + \dfrac{\pi }{6}} \right)} \right] = - \cos \left( {\dfrac{\pi }{6}} \right) \\
 \Rightarrow - \dfrac{{\sqrt 3 }}{2} $
Hence, solution of problem $ \sin \left[ {\dfrac{\pi }{2} - {{\sin }^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)} \right]\,is\, - \dfrac{{\sqrt 3 }}{2}$

Note: In order to solve the problems we need to know the trigonometric functions very well.
Some trigonometric identities may come handy.
All functions positive in first quadrant
Sin in second, tan in third and cos in fourth quadrant respectively.