Question

Write the principal value of the expression${{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)$.

Answer 1

Hint:We will apply here the formulas by $\tan \left( p \right)=\tan \left( y \right)$ which results into $p=n\pi +y$ so that we can find the value of the angle p. Also, we will use the formula $\cos \left( q \right)=\cos \left( \theta \right)$ which results into $q=2n\pi \pm \theta$ and with the help of this w will find the value of angle q.

Complete step-by-step answer:
We will first consider the trigonometric expression that is given to us as ${{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)...(i)$. Here we will consider the first term that is ${{\tan }^{-1}}\left( 1 \right)$. After this we will put ${{\tan }^{-1}}\left( 1 \right)$ equal to p which results into the expression ${{\tan }^{-1}}\left( 1 \right)=p$. Now, we will take the inverse tangent expression to the right side of the equal sign so that we can have $\tan p=1$. As we know that the value of $\tan \left( \dfrac{\pi }{4} \right)=1$ therefore we will substitute it into $\tan p=1$ to get $\tan p=\tan \left( \dfrac{\pi }{4} \right)$. Therefore, we have $\tan p=\tan \left( \dfrac{\pi }{4} \right)$. The positivity of the inverse tangent function is in two quadrants. These are namely first and third quadrants only. Thus, in the first quadrant the equation becomes $\tan p=\tan \left( \dfrac{\pi }{4} \right)$. To find the angle here we will use the formula given by $\tan \left( p \right)=\tan \left( y \right)$ which results in $p=n\pi +y$. Thus, we get $p=\left( \dfrac{\pi }{4} \right)$. As we know that the range of inverse tangents which is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$. So, clearly $p=\left( \dfrac{\pi }{4} \right)$ belongs to this open interval.
And, in the third quadrant the equation $\tan p=\tan \left( \dfrac{\pi }{4} \right)$ changes into $\tan p=\tan \left( \pi +\dfrac{\pi }{4} \right)$. Therefore, we have
$\begin{aligned} & \tan p=\tan \left( \dfrac{4\pi +\pi }{4} \right) \\\ & \Rightarrow \tan p=\tan \left( \dfrac{5\pi }{4} \right) \\\ \end{aligned}$
By using the formula given by $\tan \left( p \right)=\tan \left( y \right)$ which results in $p=n\pi +y$. Thus, we get $p=\left( \dfrac{5\pi }{4} \right)$. As the range of inverse tangent which is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ therefore, $p=\left( \dfrac{5\pi }{4} \right)$ does not belong to this interval.
So, the value of $p=\left( \dfrac{\pi }{4} \right)$ is considered here. As ${{\tan }^{-1}}\left( 1 \right)=p$ thus, we have ${{\tan }^{-1}}\left( 1 \right)=\left( \dfrac{\pi }{4} \right)$.
Now, we will consider the expression ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)$. Now, we will put this value equal to q. Therefore, we will have ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=q$. Now, we will place the inverse cosine term to the right side of the expression. Thus, we get $\cos q=-\dfrac{1}{2}$. As we know that the value of $\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$. Thus, we get $\cos q=-\cos \left( \dfrac{\pi }{3} \right)$. 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 $\cos \left( q \right)=\cos \left( \pi -\dfrac{\pi }{3} \right)$ in the second quadrant. This results into $\cos \left( q \right)=\cos \left( \dfrac{3\pi -\pi }{3} \right)$. Therefore, we have $\cos \left( q \right)=\cos \left( \dfrac{2\pi }{3} \right)$.
Now, we will use the formula which is given by $\cos \left( q \right)=\cos \left( \theta \right)$ which results in $q=2n\pi \pm \theta$. Therefore, we have $\cos \left( q \right)=\cos \left( \dfrac{2\pi }{3} \right)$ results into $q=2n\pi \pm \dfrac{2\pi }{3}$ or $q=\dfrac{2\pi }{3}$. Since, ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=q$ thus we have ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{2\pi }{3}$.
Hence, the required principal value of ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{2\pi }{3}$.
Now we will substitute the values in the expression ${{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)$. Therefore, we get
${{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{\pi }{4}+\dfrac{2\pi }{3}$. As we know that the l.c.m. of 3 and 4 is 12. Thus, we get
$\begin{aligned} & {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{\pi }{4}+\dfrac{2\pi }{3} \\\ & \Rightarrow {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{3\pi +8\pi }{12} \\\ & \Rightarrow {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{11\pi }{12} \\\ \end{aligned}$
Hence, the value of the expression ${{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{11\pi }{12}$.

Note: If we had given a chance to find any angle which may or may not belongs to the range of inverse tangent the, we would have also considered the angle in the third quadrant as the equation $\tan p=\tan \left( \dfrac{\pi }{4} \right)$ changes into
$\begin{aligned} & \tan p=\tan \left( \pi +\dfrac{\pi }{4} \right) \\\ & \Rightarrow \tan p=\tan \left( \dfrac{4\pi +\pi }{4} \right) \\\ & \Rightarrow \tan p=\tan \left( \dfrac{5\pi }{4} \right) \\\ \end{aligned}$
By using the formula given by $\tan \left( p \right)=\tan \left( y \right)$ which results in $p=n\pi +y$. Thus, we get $p=\left( \dfrac{5\pi }{4} \right)$.
And, we could have also used the fourth quadrant also instead of the second quadrant while solving inverse cosine function. 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 $\dfrac{2\pi }{3}$ 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.