Question

Find the values of the following:
${{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$

Answer 1

Hint: We will apply the formula of trigonometry here. There are three of them which we will use. These are $\tan \left( x \right)=\tan \left( y \right)$ which results into $x=n\pi +y$, $\cos \left( x \right)=\cos \left( y \right)$ which results into $x=2n\pi \pm y$ and $\sin \left( x \right)=\sin \left( y \right)$ which results into $x=n\pi +{{\left( -1 \right)}^{n}}y$.

Complete step-by-step answer:
We first consider the expression ${{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)...(i)$. Now we will find the value of ${{\tan }^{-1}}\left( 1 \right)$. This is done as by substituting ${{\tan }^{-1}}\left( 1 \right)=x$. After placing inverse tan to the right side of the equation we get $1=\tan \left( x \right)$.
As we know that the value of $\tan \left( \dfrac{\pi }{4} \right)=1$. Therefore, we have $\tan \left( \dfrac{\pi }{4} \right)=\tan \left( x \right)$.
Now we will apply the formula $\tan \left( x \right)=\tan \left( y \right)$ which results in $x=n\pi +y$. Thus, we get $x=n\pi +\dfrac{\pi }{4}$. Here, n belongs to integers. So we will substitute n = 0. Therefore we have $x=\dfrac{\pi }{4}$. Since, ${{\tan }^{-1}}\left( 1 \right)=x$ this results into ${{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}$.
Now, we will consider the next trigonometric term ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)$. We will substitute it as ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=y$. By placing the inverse cosine to the right side of the equation we get $\cos \left( y \right)=-\dfrac{1}{2}$.
As we know that the value of $\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$ therefore, we have $\cos \left( y \right)=-\cos \left( \dfrac{\pi }{3} \right)$. 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 $\cos \left( y \right)=\cos \left( \pi -\dfrac{\pi }{3} \right)$ in the second quadrant. This results into $\cos \left( y \right)=\cos \left( \dfrac{3\pi -\pi }{3} \right)$ or, $\cos \left( y \right)=\cos \left( \dfrac{2\pi }{3} \right)$.
Now we will use the formula $\cos \left( x \right)=\cos \left( y \right)$ which results in $x=2n\pi \pm y$. By taking x as y and y as $\dfrac{2\pi }{3}$. Therefore, we have $y=2n\pi \pm \dfrac{2\pi }{3}$ or, $y=\dfrac{2\pi }{3}$. Since, ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=y$ thus we have ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{2\pi }{3}$.
Now we will substitute ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=z$. By taking inverse sine to the right side of the equation we have $-\dfrac{1}{2}=\sin z$.
As we know that the value of $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$ therefore, we get $\sin z=-\sin \left( \dfrac{\pi }{6} \right)$. Since, sine is negative in the third and fourth quadrant. We will consider it in the fourth quadrant. This is for the reason here as the range of inverse sine is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$.
So, by the formula $\sin \left( x \right)=\sin \left( y \right)$ which results into $x=n\pi +{{\left( -1 \right)}^{n}}y$ we get $\sin z=\sin \left( -\dfrac{\pi }{6} \right)$ or, $z=n\pi +{{\left( -1 \right)}^{n}}\left( -\dfrac{\pi }{6} \right)$ or, $z=-\dfrac{\pi }{6}$. Since, ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=z$ therefore ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=-\dfrac{\pi }{6}$.
Now we will substitute all the values in the expression (i). Therefore, we get
${{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{\pi }{4}+\dfrac{2\pi }{3}-\dfrac{\pi }{6}$
As we know that the lcm of 4, 3 and 6 is 12. Thus we have
$\begin{aligned} & {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{3\pi +8\pi -2\pi }{12} \\\ & \Rightarrow {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{9\pi }{12} \\\ & \Rightarrow {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{3\pi }{4} \\\ \end{aligned}$
Hence, the value of the expression ${{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{3\pi }{4}$.

Note: While placing ${{\cos }^{-1}}$ to the right side of the equation we will convert it as $\cos$. This is applied to all the inverse functions. We could have used the fourth quadrant also since cos is negative only in the second and third quadrants. But here the range matters. Since, the range of the inverse cosine is $\left[ 0,\pi \right]$. Thus, we get $\cos \left( y \right)=\cos \left( \pi -\dfrac{\pi }{3} \right)$ in the second quadrant only. Similarly, sine is negative in the third and fourth quadrant. We will consider it in the fourth quadrant. This is for the reason here as the range of inverse sine is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$.