Animated Solution for Mathematics - Trigonometry: Which of the following number(s) is/are rational?

Welcome, future engineers! Today, we are going to embark on a journey through the elegant world of trigonometry and number theory. We are tasked with identifying which of the given trigonometric expressions represents a rational number.

Before we dive into the calculations, let's ground ourselves. A rational number is any number that can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q eq 0$. If a number cannot be written in this form—like the square root of two or the square root of three—it is irrational.

Our goal is to find the expression that, when simplified, results in a clean, rational value.

Let's start by evaluating our first option, $\sin 15^{\circ }$. Fifteen degrees is not a standard angle, but we can express it as a difference of two standard angles: $45^{\circ }-30^{\circ }$.

Using the compound angle formula, $\sin (A-B)=\sin A\cos B-\cos A\sin B$, we substitute $A=45^{\circ }$ and $B=30^{\circ }$. This gives us:

Substituting the standard values, we get:

Since $\sqrt{3}$ and $\sqrt{2}$ are irrational, this entire expression is clearly irrational. Similarly, for $\cos 15^{\circ }$, using $\cos (A-B)=\cos A\cos B+\sin A\sin B$, we find:

This is also irrational. These options are traps; they look like simple trigonometric values, but they hide irrational roots within them.

Now, let's examine the third option: $\sin 15^{\circ }\cos 15^{\circ }$. At first glance, you might be tempted to multiply the two irrational values we just found. But wait! This is where the JEE mindset kicks in.

Instead of brute-forcing the multiplication, look for a pattern. We have a product of sine and cosine of the same angle. This screams for the double angle identity: $\sin 2\theta =2\sin \theta \cos \theta$.

Rearranging this, we get $\sin \theta \cos \theta =\frac{1}{2}\sin 2\theta$. By setting $\theta =15^{\circ }$, our expression becomes:

Since $\sin 30^{\circ }=\frac{1}{2}$, the expression simplifies to:

Look at that! The irrational terms have completely vanished, leaving us with a perfect rational number. This is the beauty of trigonometry—identities are your best friends.

To be absolutely thorough, let's check the fourth option: $\sin 15^{\circ }\cos 75^{\circ }$. Using the co-function identity, $\cos 75^{\circ }=\cos (90^{\circ }-15^{\circ })=\sin 15^{\circ }$. Thus, the expression becomes ${\sin }^{2}15^{\circ }$.

Using the power-reducing formula ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$, we get:

Again, the presence of $\sqrt{3}$ confirms this is irrational. We have successfully navigated the traps and found our answer.

Remember, in JEE, it's not just about calculating; it's about choosing the most elegant path. The rational expression is $\sin 15^{\circ }\cos 15^{\circ }=\frac{1}{4}$. Keep practicing, and keep falling in love with the logic!