Question

The principal value of ${{\cos }^{-1}}\left[ \sin \left( {{\cos }^{-1}}\dfrac{1}{2} \right) \right]$ is
(a) $\dfrac{\pi }{5}$
(b) $\dfrac{\pi }{4}$
(c) $\dfrac{\pi }{6}$
(d) $\dfrac{\pi }{12}$

Answer 1

We know that if $\cos \theta =b$, then the inverse cosine could be written as $\theta ={{\cos }^{-1}}\left( b \right)$. We also know the principal solution for cosine inverse lies in the interval $\left[ 0,\pi \right]$. Also, we all know very well that $\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$, $\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}$ and $\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}$.

Complete step-by-step answer:
We all know very well about the 6 trigonometric identities, sine (sin), cosine (cos), tangent (tan), cosecant (cosec), secant (sec) and cotangent (cot). We also know that if any of these trigonometric identities is related with an angle and a value, we can always write the inverse of that.
For example, if $\sin \theta =a$, then we can write $\theta ={{\sin }^{-1}}\left( a \right)$, which is called an inverse trigonometric identity.
Similarly, if $\cos \theta =b$, then we can write $\theta ={{\cos }^{-1}}\left( b \right)$.
We need to find ${{\cos }^{-1}}\left[ \sin \left( {{\cos }^{-1}}\dfrac{1}{2} \right) \right]$. Let us start from the innermost function.
We know that $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$. Also, we are aware that the principal solution for cos inverse lies in the interval $\left[ 0,\pi \right]$.
So, the principal solution of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{3}$.
So, we can say that we need to find ${{\cos }^{-1}}\left[ \sin \left( \dfrac{\pi }{3} \right) \right]$.
We all know very well that the value of $\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}$.
Thus, putting the value of $\sin \left( \dfrac{\pi }{3} \right)$, we can say that we need to evaluate ${{\cos }^{-1}}\left[ \dfrac{\sqrt{3}}{2} \right]$.
We know that $\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$. Also, we are aware that the principal solution for cos inverse lies in the interval $\left[ 0,\pi \right]$.
So, the principal solution of ${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{6}$.
Thus, the principal solution of ${{\cos }^{-1}}\left[ \sin \left( {{\cos }^{-1}}\dfrac{1}{2} \right) \right]$ is $\dfrac{\pi }{6}$.

So, the correct answer is “Option (c)”.

Note: We must remember that the principal solution for sine inverse lies in the interval $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ and that of cosine inverse lies in the interval $\left[ 0,\pi \right]$. We must note that all the angles in this solution is in radians and not degrees.