If $\tan x=\dfrac{3}{4}, \pi < x < \dfrac{3 \pi}{2}$, then find the values of $\sin \dfrac{x}{2}, \cos \dfrac{x}{2}$ and $\tan \dfrac{x}{2}$.
Verified
From the above figure we can say that the angle $\dfrac{x}{2}$ lies in second quadrant.
The signs of the trigonometric identities for the angle lies in different quadrants is demonstrated in below image.
Hence the signs of the different trigonometric identities of an angel $\dfrac{x}{2}$ are $\sin \left(\dfrac{x}{2}\right) \rightarrow+v e$
$\cos \left(\dfrac{x}{2}\right) \rightarrow-v e$
$\tan \left(\dfrac{x}{2}\right) \rightarrow-v e$
Now we have $\tan x=\dfrac{3}{4}$
Using the half angle formula $\tan x=\dfrac{2 \tan \left(\dfrac{x}{2}\right)}{1-\tan ^{2}\left(\dfrac{x}{2}\right)}$, then
$\dfrac{2 \tan \left(\dfrac{x}{2}\right)}{1-\tan ^{2}\left(\dfrac{x}{2}\right)}=\dfrac{3}{4}$
$8 \tan \left(\dfrac{x}{2}\right)=3-3 \tan ^{2}\left(\dfrac{x}{2}\right)$
$3 \tan ^{2}\left(\dfrac{x}{2}\right)+8 \tan \left(\dfrac{x}{2}\right)-3=0$
Solving the above quadratic equation. $3 \tan ^{2}\left(\dfrac{x}{2}\right)+8 \tan \left(\dfrac{x}{2}\right)-3=0$
$3 \tan ^{2}\left(\dfrac{x}{2}\right)+9 \tan \left(\dfrac{x}{2}\right)-\tan \left(\dfrac{x}{2}\right)-3=0$
$3 \tan \left(\dfrac{x}{2}\right)\left[\tan \left(\dfrac{x}{2}\right)+3\right]-1\left[\tan \left(\dfrac{x}{2}\right)+3\right]=0$
$\left(3 \tan \left(\dfrac{x}{2}\right)-1\right)\left(\tan \left(\dfrac{x}{2}\right)+3\right)=0$
Here we get the values of $\tan \left(\dfrac{x}{2}\right)$ as $\dfrac{1}{3}$ which is not acceptable because the angle $\dfrac{x}{2}$ lies in second quadrant then the value of $\tan \left(\dfrac{x}{2}\right)$ should be $-v e .$ So, the value of $\tan \left(\dfrac{x}{2}\right)$ is taken as $-3$. $\therefore \tan \left(\dfrac{x}{2}\right)=-3$ for $\dfrac{\pi}{2}<\dfrac{x}{2}<\dfrac{3 \pi}{4}$
Now from the formulas $\cos x=\dfrac{1}{\sec x}$ and $\sec ^{2} x-\tan ^{2} x=1$ we are finding the value of $\cos \left(\dfrac{x}{2}\right)$. So,
$\cos \left(\dfrac{x}{2}\right)=\dfrac{1}{\sec \left(\dfrac{x}{2}\right)}$
From the formula $\sec ^{2} x-\tan ^{2} x=1$ substituting $\sec x=\sqrt{1+\tan ^{2} x}$ in the above equation, then we have
$\cos \left(\dfrac{x}{2}\right)=\dfrac{1}{\sqrt{1+\tan ^{2}\left(\dfrac{x}{2}\right)}}$
$=\dfrac{1}{\sqrt{1+(-3)^{2}}}$
$=\dfrac{1}{\sqrt{10}}$
But $\cos \left(\dfrac{x}{2}\right)=-\dfrac{1}{\sqrt{10}}$ for $\dfrac{\pi}{2}<\dfrac{x}{2}<\dfrac{3 \pi}{4}$
We have the relation between $\sin x, \cos x, \tan x$ as $\sin x=\tan x \times \cos x$, so
$\sin \left(\dfrac{x}{2}\right)=\tan \left(\dfrac{x}{2}\right) \times \cos \left(\dfrac{x}{2}\right)$
$=-3 \times\left(-\dfrac{1}{\sqrt{10}}\right)$
$=\dfrac{3}{\sqrt{10}}$
Hence $\sin \left(\dfrac{x}{2}\right)=\dfrac{3}{\sqrt{10}}$ for $\dfrac{\pi}{2}<\dfrac{x}{2}<\dfrac{3 \pi}{4}$
<b>Note:</b> Make sure that you have found the range of the angle $\dfrac{x}{2}$ and its quadrant. Based on the quadrant of the angle $\dfrac{x}{2}$ find the signs of the all trigonometric ratios. Based on the sign conventions, take the value of $\tan \left(\dfrac{x}{2}\right)$ carefully.
