Q.Find $\sin\frac{x}{2}$, $\cos\frac{x}{2}$ and $\tan\frac{x}{2}$ if $\tan x = -\frac{4}{3}$, $x$ in quadrant II.

Concept: Trigonometric half-angle formulas with quadrant-based sign determination.

Step 1 – Find $\sin x$ and $\cos x$ from $\tan x$.

In quadrant II, $\sin x > 0$, $\cos x < 0$.

Given $\tan x = -\frac{4}{3}$, construct a right triangle: opposite $=4$, adjacent $=-3$, hypotenuse $=5$.

Thus $\sin x = \frac{4}{5}$, $\cos x = -\frac{3}{5}$.

Step 2 – Determine the quadrant for $\frac{x}{2}$.

$x$ in quadrant II means $90^\circ < x < 180^\circ$, so $45^\circ < \frac{x}{2} < 90^\circ$.

Hence $\frac{x}{2}$ is in quadrant I: all trigonometric ratios are positive.

Step 3 – Apply half-angle formulas.

$\sin\frac{x}{2} = \sqrt{\frac{1-\cos x}{2}} = \sqrt{\frac{1-(-\frac{3}{5})}{2}} = \sqrt{\frac{1+\frac{3}{5}}{2}} = \sqrt{\frac{8/5}{2}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$ …

Concept and Intuition

When a problem gives you a trigonometric ratio and the quadrant of the angle, the real challenge isn't the algebra — it's the signs. Every trigonometric function has a specific sign pattern in each quadrant, and half-angle formulas introduce an extra layer: you must first figure out which quadrant $\frac{x}{2}$ lies in, because that determines the signs of $\sin\frac{x}{2}$, $\cos\frac{x}{2}$, and $\tan\frac{x}{2}$.

Here, $x$ is in quadrant II. That means $90^\circ < x < 180^\circ$ (or $\frac{\pi}{2} < x < \pi$). If you halve that range, you get $45^\circ < \frac{x}{2} < 90^\circ$ (or $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$). So $\frac{x}{2}$ lies in quadrant I, where all trigonometric ratios are positive. That's the key insight — it saves you from guessing signs later.

Now, we also need $\cos x$ and $\sin x$ from $\tan x$. Since $\tan x = -\frac{4}{3}$ and $x$ is in quadrant II, $\sin x$ is positive and $\cos x$ is negative. We can find them using a right triangle or the identity $\sec^2 x = 1 + \tan^2 x$.

Step-by-Step Solution

1. Find $\cos x$ and $\sin x$ from $\tan x$.

We know $\tan x = \frac{\sin x}{\cos x} = -\frac{4}{3}$. In quadrant II, $\sin x > 0$ and $\cos x < 0$.

Using $\sec^2 x = 1 + \tan^2 x$:

So $\sec x = \pm \frac{5}{3}$. Since $\cos x$ is negative in quadrant II, $\sec x$ is also negative:

Now $\sin x$ can be found from $\sin^2 x = 1 - \cos^2 x$:

Since $\sin x > 0$ in quadrant II:

2. Determine the quadrant of $\frac{x}{2}$ and its sign implications.

$x$ in quadrant II means $90^\circ < x < 180^\circ$. Halving:

So $\frac{x}{2}$ is in quadrant I. Therefore:

• $\sin\frac{x}{2} > 0$
• $\cos\frac{x}{2} > 0$
• $\tan\frac{x}{2} > 0$

3. Apply the half-angle formulas.

The standard half-angle formulas are:

The $\pm$ sign is chosen based on the quadrant of $\frac{x}{2}$.

Since $\frac{x}{2}$ is in quadrant I, we take the positive sign for all three.

4. Compute $\sin\frac{x}{2}$. …