Question

For the principal values, evaluate ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)+2{{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)$.

Answer 1

Hint:We will use the trigonometric formulas which are used to find the angles. These are given by $\cos \left( x \right)=\cos \left( y \right)$ which results into $x=2n\pi \pm y$ and $\sin \left( x \right)=\sin \left( y \right)$ which results into $x=n\pi +{{\left( -1 \right)}^{n}}y$. We will also use the trigonometric values $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$ and the value of $\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}$ in the given expression in the question in order to solve the question further.

Complete step-by-step answer:
Now, we will first consider the expression ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)+2{{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)...(i)$. In this expression we will take the term ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$. Now, we will substitute ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=x$. By taking inverse sine to the right side of the equation we have $-\dfrac{1}{2}=\sin x$.
In trigonometry we are given the value of $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$ therefore, after substituting it in the term $-\dfrac{1}{2}=\sin x$ we will get $\sin x=-\sin \left( \dfrac{\pi }{6} \right)$.
Now, the trigonometric term sine is negative in the third and fourth quadrant. So, we will consider it in the fourth quadrant. This is for the reason here as the range of inverse sine is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ and $\dfrac{\pi }{6}$ belongs to $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ closed interval.
So, we will apply the formula of $\sin \left( x \right)=\sin \left( y \right)$ which results into $x=n\pi +{{\left( -1 \right)}^{n}}y$ we get $\sin x=\sin \left( -\dfrac{\pi }{6} \right)$ or, $x=n\pi +{{\left( -1 \right)}^{n}}\left( -\dfrac{\pi }{6} \right)$ or, $x=-\dfrac{\pi }{6}$. Since, ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=x$ therefore ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=-\dfrac{\pi }{6}$.
Now, we will consider the term ${{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)$. After substituting the trigonometric term ${{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)$ equal to y we will have ${{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)=x$. Now, we will place the inverse cosine term to the right side of the expression. Thus, we have a new equation as $-\dfrac{\sqrt{3}}{2}=\cos \left( x \right)$. Now, we will use the value of $\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}$ in $-\dfrac{\sqrt{3}}{2}=\cos \left( x \right)$ we have our equation converted into $\cos \left( x \right)=-\cos \left( \dfrac{\pi }{6} \right)$.
As cosine is negative only in the second and third quadrants so we will take it negative in the second quadrant. Since, the range of the inverse cosine is $\left[ 0,\pi \right]$. Thus, we get $\cos \left( x \right)=\cos \left( \pi -\dfrac{\pi }{6} \right)$ in the second quadrant. This results into $\cos \left( x \right)=\cos \left( \dfrac{6\pi -\pi }{6} \right)$ or, $\cos \left( x \right)=\cos \left( \dfrac{5\pi }{6} \right)$.
After applying the formula which is used to find the value given by $\cos \left( x \right)=\cos \left( y \right)$ results into the angle $x=2n\pi \pm y$. Therefore, we have $\cos \left( x \right)=\cos \left( \dfrac{5\pi }{6} \right)$ results into $x=2n\pi \pm \dfrac{5\pi }{6}$ or $x=\dfrac{5\pi }{6}$. Since, ${{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)=x$ thus we have ${{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)=\dfrac{5\pi }{6}$.
Now we will substitute the above inverse trigonometric values in the expression (i). Therefore, we have
$\begin{align}
  & {{\sin }^{-1}}\left( -\dfrac{1}{2} \right)+2{{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)=-\dfrac{\pi }{6}+2\times \dfrac{5\pi }{6} \\
 & \Rightarrow {{\sin }^{-1}}\left( -\dfrac{1}{2} \right)+2{{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)=-\dfrac{\pi }{6}+\dfrac{10\pi }{6} \\
 & \Rightarrow {{\sin }^{-1}}\left( -\dfrac{1}{2} \right)+2{{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)=\dfrac{-\pi +10\pi }{6} \\
 & \Rightarrow {{\sin }^{-1}}\left( -\dfrac{1}{2} \right)+2{{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)=\dfrac{9\pi }{6} \\
 & \Rightarrow {{\sin }^{-1}}\left( -\dfrac{1}{2} \right)+2{{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)=\dfrac{3\pi }{2} \\
\end{align}$
Hence, the value of the expression ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)+2{{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)=\dfrac{3\pi }{2}$.

Note: Be focused on placing any inverse trigonometric term to the right side of the equal sign. This is because when we do the placing we convert them as simple trigonometric terms which are without the inverse signs.
As we know that the trigonometric term sine is negative in the third and fourth quadrant. So, we will consider it in the fourth quadrant. This is because if we take the third quadrant we get $\begin{align}
  & \sin x=\sin \left( \pi +\dfrac{\pi }{6} \right) \\
 & \Rightarrow \sin x=\sin \left( \dfrac{7\pi }{6} \right) \\
\end{align}$
After applying the formula of $\sin \left( x \right)=\sin \left( y \right)$ which results into $x=n\pi +{{\left( -1 \right)}^{n}}y$ we get $\sin x=\sin \left( \dfrac{7\pi }{6} \right)$ or, $x=n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{7\pi }{6} \right)$ or, $x=\dfrac{7\pi }{6}$. Since, ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=x$ therefore ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{7\pi }{6}$ and $x=\dfrac{7\pi }{6}$ does not belongs to the closed interval $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ which is the range of inverse sine function. And in case of inverse cosine function we could have also used the fourth quadrant also instead of the second quadrant. This is because cosine is negative in both these quadrants. But since the range of inverse cosine is $\left[ 0,\pi \right]$ so, we will also take the value which belongs to this interval. As $\dfrac{5\pi }{6}$ belongs to the interval of inverse cosine therefore, we have selected this value. One should be aware that we are talking about the range of inverse cosine instead of cosine.