Find the value of ${{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$.
Answer
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Hint:As we know that the values of $\tan \left( \dfrac{\pi }{4} \right)=1,\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$ and $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$. So, we will use these values in order to solve the question. In this question we have also applied the formulas given as $\tan \left( x \right)=\tan \left( y \right)$ which results into $x=n\pi +y$ and $\cos \left( x \right)=\cos \left( y \right)$ which results into $x=2n\pi \pm y$ and $\sin \left( x \right)=\sin \left( y \right)$ which results into $x=n\pi +{{\left( -1 \right)}^{n}}y$. Here n is the number that belongs to integers.
Complete step-by-step answer:
Now we will consider the expression ${{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)...(i)$.
We will solve it by taking one term at a time. So, first we take ${{\tan }^{-1}}\left( 1 \right)$ and we will substitute ${{\tan }^{-1}}\left( 1 \right)$ equal to x. This can be written as ${{\tan }^{-1}}\left( 1 \right)=x$.
Now, we will take the inverse tangent expression to the right side of the equal sign. This results in $\tan x=1$.
Now, at this step we will use the value of $\tan \left( \dfrac{\pi }{4} \right)=1$ and substitute it into $\tan x=\tan \left( \dfrac{\pi }{4} \right)$. Therefore, we have $\tan x=\tan \left( \dfrac{\pi }{4} \right)$.
The tangent trigonometric operation is positive in first and third quadrants only. Thus, in the first quadrant the equation becomes $\tan x=\tan \left( \dfrac{\pi }{4} \right)$. By using the formula given by $\tan \left( x \right)=\tan \left( y \right)$ which results in $x=n\pi +y$. Thus, we get $x=\left( \dfrac{\pi }{4} \right)$. As we know that the range of inverse tangents which is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$. So, clearly $x=\left( \dfrac{\pi }{4} \right)$ belongs to this open interval.
So, $x=\left( \dfrac{\pi }{4} \right)$ is considered here. As ${{\tan }^{-1}}\left( 1 \right)=x$ thus, we have ${{\tan }^{-1}}\left( 1 \right)=\left( \dfrac{\pi }{4} \right)$.
Now we will consider ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)$. Now, we will put this value equal to y. Therefore, we will have ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=y$.
Now, we will place the inverse cosine term to the right side of the expression. Thus, we get $\cos y=-\dfrac{1}{2}$.
As we know that the value of $\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$. Thus, we get $\cos y=-\cos \left( \dfrac{\pi }{3} \right)$.
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 $\cos \left( y \right)=\cos \left( \theta \right)$ which results into $y=2n\pi \pm \theta $. Therefore, we have $\cos \left( y \right)=\cos \left( \dfrac{2\pi }{3} \right)$ results into $y=2n\pi \pm \dfrac{2\pi }{3}$ or $y=\dfrac{2\pi }{3}$. Since, ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=y$ thus we have ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{2\pi }{3}$.
Hence, the required principal value of ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{2\pi }{3}$.
Now, we will focus on the term ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$. 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 $-\dfrac{1}{2}=\sin z$.
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 substitute these values in expression (i). Therefore, we have
${{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\left( \dfrac{\pi }{4} \right)+\dfrac{2\pi }{3}-\dfrac{\pi }{6}$
As the l.c.m. of 4, 3 and 6 is 12. Thus, we now have
$\begin{align}
& {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{\pi }{4}+\dfrac{2\pi }{3}-\dfrac{\pi }{6} \\
& \Rightarrow {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{3\pi +8\pi -2\pi }{12} \\
& \Rightarrow {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{11\pi -2\pi }{12} \\
& \Rightarrow {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{9\pi }{12} \\
& \Rightarrow {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{3\pi }{4} \\
\end{align}$
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: In the inverse tangent term we have not selected the angle in the third quadrant as the equation $\tan x=\tan \left( \dfrac{\pi }{4} \right)$ changes into $\tan x=\tan \left( \pi +\dfrac{\pi }{4} \right)$ or,
$\begin{align}
& \tan x=\tan \left( \dfrac{4\pi +\pi }{4} \right) \\
& \Rightarrow \tan x=\tan \left( \dfrac{5\pi }{4} \right) \\
\end{align}$
By using the formula given by $\tan \left( x \right)=\tan \left( y \right)$ which results in $x=n\pi +y$. Thus, we get $x=\left( \dfrac{5\pi }{4} \right)$. As the range of inverse tangent which is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ therefore, $x=\left( \dfrac{5\pi }{4} \right)$ does not belong to this interval.
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 $\dfrac{2\pi }{3}$ 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.
Complete step-by-step answer:
Now we will consider the expression ${{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)...(i)$.
We will solve it by taking one term at a time. So, first we take ${{\tan }^{-1}}\left( 1 \right)$ and we will substitute ${{\tan }^{-1}}\left( 1 \right)$ equal to x. This can be written as ${{\tan }^{-1}}\left( 1 \right)=x$.
Now, we will take the inverse tangent expression to the right side of the equal sign. This results in $\tan x=1$.
Now, at this step we will use the value of $\tan \left( \dfrac{\pi }{4} \right)=1$ and substitute it into $\tan x=\tan \left( \dfrac{\pi }{4} \right)$. Therefore, we have $\tan x=\tan \left( \dfrac{\pi }{4} \right)$.
The tangent trigonometric operation is positive in first and third quadrants only. Thus, in the first quadrant the equation becomes $\tan x=\tan \left( \dfrac{\pi }{4} \right)$. By using the formula given by $\tan \left( x \right)=\tan \left( y \right)$ which results in $x=n\pi +y$. Thus, we get $x=\left( \dfrac{\pi }{4} \right)$. As we know that the range of inverse tangents which is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$. So, clearly $x=\left( \dfrac{\pi }{4} \right)$ belongs to this open interval.
So, $x=\left( \dfrac{\pi }{4} \right)$ is considered here. As ${{\tan }^{-1}}\left( 1 \right)=x$ thus, we have ${{\tan }^{-1}}\left( 1 \right)=\left( \dfrac{\pi }{4} \right)$.
Now we will consider ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)$. Now, we will put this value equal to y. Therefore, we will have ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=y$.
Now, we will place the inverse cosine term to the right side of the expression. Thus, we get $\cos y=-\dfrac{1}{2}$.
As we know that the value of $\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$. Thus, we get $\cos y=-\cos \left( \dfrac{\pi }{3} \right)$.
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 $\cos \left( y \right)=\cos \left( \theta \right)$ which results into $y=2n\pi \pm \theta $. Therefore, we have $\cos \left( y \right)=\cos \left( \dfrac{2\pi }{3} \right)$ results into $y=2n\pi \pm \dfrac{2\pi }{3}$ or $y=\dfrac{2\pi }{3}$. Since, ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=y$ thus we have ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{2\pi }{3}$.
Hence, the required principal value of ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{2\pi }{3}$.
Now, we will focus on the term ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$. 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 $-\dfrac{1}{2}=\sin z$.
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 substitute these values in expression (i). Therefore, we have
${{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\left( \dfrac{\pi }{4} \right)+\dfrac{2\pi }{3}-\dfrac{\pi }{6}$
As the l.c.m. of 4, 3 and 6 is 12. Thus, we now have
$\begin{align}
& {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{\pi }{4}+\dfrac{2\pi }{3}-\dfrac{\pi }{6} \\
& \Rightarrow {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{3\pi +8\pi -2\pi }{12} \\
& \Rightarrow {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{11\pi -2\pi }{12} \\
& \Rightarrow {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{9\pi }{12} \\
& \Rightarrow {{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{3\pi }{4} \\
\end{align}$
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: In the inverse tangent term we have not selected the angle in the third quadrant as the equation $\tan x=\tan \left( \dfrac{\pi }{4} \right)$ changes into $\tan x=\tan \left( \pi +\dfrac{\pi }{4} \right)$ or,
$\begin{align}
& \tan x=\tan \left( \dfrac{4\pi +\pi }{4} \right) \\
& \Rightarrow \tan x=\tan \left( \dfrac{5\pi }{4} \right) \\
\end{align}$
By using the formula given by $\tan \left( x \right)=\tan \left( y \right)$ which results in $x=n\pi +y$. Thus, we get $x=\left( \dfrac{5\pi }{4} \right)$. As the range of inverse tangent which is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ therefore, $x=\left( \dfrac{5\pi }{4} \right)$ does not belong to this interval.
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 $\dfrac{2\pi }{3}$ 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.
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