Q.If $\tan x = \frac{3}{4}$, $\pi < x < \frac{3\pi}{2}$, find the value of $\sin\frac{x}{2}$, $\cos\frac{x}{2}$ and $\tan\frac{x}{2}$.

Concept: Half-angle formulas with quadrant analysis.

Since $\pi < x < \frac{3\pi}{2}$, the angle $x$ lies in the third quadrant where both sine and cosine are negative.

From $\tan x = \frac{3}{4}$ in Q3, we construct a reference triangle: opposite = 3, adjacent = 4, so hypotenuse = 5. Thus:

For the half-angle $\frac{x}{2}$: since $\pi < x < \frac{3\pi}{2}$, we have $\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$ (second quadrant), where sine is positive and cosine is negative.

Applying half-angle formulas:

$\sin\frac{x}{2} = \sqrt{\frac{1-\cos x}{2}} = \sqrt{\frac{1-(-\frac{4}{5})}{2}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10}$ …

The heart of this problem is understanding how trigonometric values behave across quadrants and how half-angle formulas inherit their signs from the quadrant in which the half-angle lies.

When $\tan x = \frac{3}{4}$ and $\pi < x < \frac{3\pi}{2}$, we know $x$ is in the third quadrant where both sine and cosine are negative. The half-angle $\frac{x}{2}$ will satisfy $\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$, placing it in the second quadrant where sine is positive but cosine is negative.

Finding $\sin x$ and $\cos x$

1. Use the Pythagorean identity to find $\sin x$ and $\cos x$

Since $\tan x = \frac{\sin x}{\cos x} = \frac{3}{4}$, we can write $\sin x = \frac{3}{4}\cos x$.

Substituting into $\sin^2 x + \cos^2 x = 1$:

Since $x$ is in the third quadrant, $\cos x < 0$, so $\cos x = -\frac{4}{5}$.

Then $\sin x = \frac{3}{4} \cdot \left(-\frac{4}{5}\right) = -\frac{3}{5}$.

Applying Half-Angle Formulas

2. Determine the quadrant of $\frac{x}{2}$

Since $\pi < x < \frac{3\pi}{2}$, dividing by 2 gives:

This places $\frac{x}{2}$ in the second quadrant, where $\sin\frac{x}{2} > 0$, $\cos\frac{x}{2} < 0$, and $\tan\frac{x}{2} < 0$.

3. Calculate $\sin\frac{x}{2}$ using the half-angle formula

Since $\frac{x}{2}$ is in the second quadrant, $\sin\frac{x}{2} > 0$:

$\sin\frac{x}{2} = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10}$ …