Animated Solution for Mathematics - Trigonometry: In this questions there are entries in columns 1 and 2. Each entry in column 1 is related to exactly one entry in column 2. $\frac{\sin 3\alpha }{\cos 2\alpha }$ is

Mastering Trigonometric Signs

Welcome, future engineers! Today, we are going to demystify a problem that often trips up even the brightest students. We are looking at the expression:

We aim to determine its sign across different intervals of $\alpha$. It might look like a simple algebra problem, but it is actually a beautiful exercise in visualization and the geometry of the unit circle.

Imagine the unit circle as a clock. As $\alpha$ increases, the angle sweeps around the circle. $3\alpha$ and $2\alpha$ move at different speeds, like two runners on a track, and we must track their positions at any given moment.

The Unit Circle

Before we dive into the math, let us ground ourselves. The unit circle is our map.

In the first quadrant ($0$ to $\frac{\pi }{2}$), everything is positive. In the second quadrant ($\frac{\pi }{2}$ to $\pi$), sine is positive, but cosine is negative.

In the third quadrant ($\pi$ to $\frac{3\pi }{2}$), both are negative. In the fourth quadrant ($\frac{3\pi }{2}$ to $2\pi$), cosine is positive while sine is negative. If you know the quadrant, you know the sign.

A Guided Walkthrough

Let us take the first interval: $\alpha \in (\frac{13\pi }{48},\frac{14\pi }{48})$. We need to find the sign of $f(\alpha )=\frac{\sin 3\alpha }{\cos 2\alpha }$.

First, let us look at the numerator, $\sin 3\alpha$. If $\alpha \in (\frac{13\pi }{48},\frac{14\pi }{48})$, then $3\alpha \in (\frac{39\pi }{48},\frac{42\pi }{48})$, which simplifies to $3\alpha \in (\frac{13\pi }{16},\frac{14\pi }{16})$.

Since $\frac{\pi }{2}<\frac{13\pi }{16}$ and $\frac{14\pi }{16}<\pi$, we know $3\alpha$ is in the second quadrant. In the second quadrant, sine is positive.

Now, for the denominator, $\cos 2\alpha$. Multiplying our interval by $2$, we get $2\alpha \in (\frac{26\pi }{48},\frac{28\pi }{48})$, which simplifies to $2\alpha \in (\frac{13\pi }{24},\frac{14\pi }{24})$.

These values are between $\frac{\pi }{2}$ and $\pi$, meaning $2\alpha$ is also in the second quadrant. Since cosine is negative in the second quadrant, we have a positive numerator divided by a negative denominator. The result is negative.

Reinforcing the Logic

Now, let us tackle the second interval: $\alpha \in (\frac{18\pi }{48},\frac{23\pi }{48})$. We repeat the process.

For the numerator, $3\alpha \in (\frac{54\pi }{48},\frac{69\pi }{48})$, which simplifies to $3\alpha \in (\frac{9\pi }{8},\frac{23\pi }{16})$. Since $\pi <\frac{9\pi }{8}$ and $\frac{23\pi }{16}<\frac{3\pi }{2}$, $3\alpha$ is in the third quadrant, where sine is negative.

For the denominator, $2\alpha \in (\frac{36\pi }{48},\frac{46\pi }{48})$, which simplifies to $2\alpha \in (\frac{3\pi }{4},\frac{23\pi }{24})$. Both these values are between $\frac{\pi }{2}$ and $\pi$, placing $2\alpha$ in the second quadrant, where cosine is negative.

We have a negative numerator divided by a negative denominator. A negative divided by a negative is positive.

It is not about memorizing formulas; it is about tracking the position of the angle. By mapping the transformed angles back to the unit circle, you can confidently determine the sign of any trigonometric expression.

Keep practicing this visualization, and you will find that even the most complex problems become clear and manageable. You have got this!