Animated Solution for Mathematics - Trigonometry: The value of $2\sin (12^{\circ })-\sin (72^{\circ })$ is :

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a problem that, at first glance, seems to defy the standard rules of trigonometry. We are faced with the expression $2\sin (12^{\circ })-\sin (72^{\circ })$.

You might look at these angles, $12^{\circ }$ and $72^{\circ }$, and feel a sense of unease. They are not the familiar $30^{\circ }$, $45^{\circ }$, or $60^{\circ }$ that we love. But here is the secret: in the world of JEE, unfamiliar angles are just invitations to use identities.

The first hurdle is that coefficient '$2$' in front of $\sin (12^{\circ })$. It feels like a roadblock, but we can view it as an opportunity. We rewrite $2\sin (12^{\circ })$ as $\sin (12^{\circ })+\sin (12^{\circ })$.

This allows us to isolate one $\sin (12^{\circ })$ and pair the other with the $-\sin (72^{\circ })$ term. Our expression now looks like this:

This grouping is the key. It is a strategic move, setting the stage for the identity we need.

Now, look at the expression inside the parentheses: $\sin (72^{\circ })-\sin (12^{\circ })$. This follows the classic $\sin C-\sin D$ pattern. The identity is:

With $C=72^{\circ }$ and $D=12^{\circ }$, we substitute these values. The calculation is straightforward:

Suddenly, the magic happens. We get $2\cos (42^{\circ })\sin (30^{\circ })$. Since $\sin (30^{\circ })=\frac{1}{2}$, the expression simplifies to $\cos (42^{\circ })$. Our original expression has now transformed into $\sin (12^{\circ })-\cos (42^{\circ })$.

We are almost there, but we have a mismatch: a sine and a cosine. To use our identity again, we need them to be the same function. We use the complementary angle identity: $\cos (\theta )=\sin (90^{\circ }-\theta )$.

Thus, $\cos (42^{\circ })=\sin (90^{\circ }-42^{\circ })=\sin (48^{\circ })$. Now we have $\sin (12^{\circ })-\sin (48^{\circ })$.

We apply the $\sin C-\sin D$ identity one last time, with $C=12^{\circ }$ and $D=48^{\circ }$. This gives us:

The angles become $30^{\circ }$ and $-18^{\circ }$. We know $\cos (30^{\circ })=\frac{\sqrt{3}}{2}$ and $\sin (-18^{\circ })=-\sin (18^{\circ })=-\frac{\sqrt{5}-1}{4}$. Substituting these values, we get:

After the final cancellation, we arrive at the elegant result:

This is the beauty of trigonometry—taking a complex, seemingly impossible expression and, through a series of logical steps, revealing its simple, underlying truth. Keep practicing, keep questioning, and most importantly, keep falling in love with the process.