Question

\[\cos x=-\dfrac{1}{3}\], $x$ in third quadrant. Find the value of $\sin \dfrac{x}{2},\cos \dfrac{x}{2},\tan \dfrac{x}{2}$.

Answer 1

Hint: For solving such types of questions first we have to find out the range of the given function. After that we have to find out the range and values of the other functions. For this we have to use the following formula:
$\sin \dfrac{x}{2}=\pm \sqrt{\dfrac{1-\cos x}{2}}$

Complete step by step answer:
Given that
$\cos x=-\dfrac{1}{3}$
The range of this function is
$\pi Means $x$lies in the third quadrant.
Now we know that
$2{{\sin }^{2}}\dfrac{x}{2}=1-\cos x$
Further rearranging the terms of above expression
$\sin \dfrac{x}{2}=\pm \sqrt{\dfrac{1-\cos x}{2}}$
Substitute the value of $\cos x$in the above expression we get
$\sin \dfrac{x}{2}=\pm \sqrt{\dfrac{1-\left( -\dfrac{1}{3} \right)}{2}}$
Further solving
$\sin \dfrac{x}{2}=\pm \sqrt{\dfrac{4}{6}}$
The above function lies in the following range
$\pi Divide the above expression by $2$ we get
$\dfrac{\pi }{2}<\dfrac{x}{2}<\dfrac{3\pi }{4}$
Since $\sin x$ is positive in the second quadrant. So,
$\sin \dfrac{x}{2}=\dfrac{\sqrt{2}}{\sqrt{5}}$
We know that
$\cos x+1=2{{\cos }^{2}}\dfrac{x}{2}$
Further rearranging the terms of the above expression
$\cos \dfrac{x}{2}=\pm \sqrt{\dfrac{1+\cos x}{2}}$
Substitute the value of $\cos x$in the above expression we get
$\cos \dfrac{x}{2}=\pm \sqrt{\dfrac{1+\left( -\dfrac{1}{3} \right)}{2}}$
Further solving
$\cos \dfrac{x}{2}=\pm \sqrt{\dfrac{1}{3}}$
The above function lies in the following range
$\pi Divide the above expression by $2$ we get
$\dfrac{\pi }{2}<\dfrac{x}{2}<\dfrac{3\pi }{4}$
The function $\cos x$ is negative in the second quadrant.
$\cos \dfrac{x}{2}=-\dfrac{1}{\sqrt{3}}$
Now we also know that
$\cos x=\dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}$
Further rearranging the terms of the above expression
$\tan \dfrac{x}{2}=\pm \sqrt{\dfrac{1-\cos x}{1+\cos x}}$
Substitute the value of $\cos x$in the above expression we get
$\tan \dfrac{x}{2}=\pm \sqrt{\dfrac{1-\left( -\dfrac{1}{3} \right)}{1+\left( -\dfrac{1}{3} \right)}}$
By solving the above expression
$\tan \dfrac{x}{2}=\pm \sqrt{2}$
The above function lies in the following range
$\pi Divide the above expression by $2$ we get
$\dfrac{\pi }{2}<\dfrac{x}{2}<\dfrac{3\pi }{4}$
We know that $\tan x$ is negative in the second quadrant. Therefore,
$\tan \dfrac{x}{2}=-\sqrt{2}$

Therefore,
$\sin \dfrac{x}{2}=\dfrac{\sqrt{2}}{\sqrt{5}}$ and $\cos \dfrac{x}{2}=-\dfrac{1}{\sqrt{3}}$ and $\tan \dfrac{x}{2}=-\sqrt{2}$

Note:
Here, in these types of questions we have to be careful while finding the range of the functions. Because the whole answer totally depends upon the range and the quadrant. So it is necessary to find the range of the functions correctly.