Animated Solution for Mathematics - Trigonometry: If $\sin \theta +\cos \theta =\frac{1}{2}$, then $16(\sin (2\theta )+\cos (4\theta )+\sin (6\theta ))$ is equal to:

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of multiple angles. You see $\sin \theta +\cos \theta =\frac{1}{2}$ and then a target expression: $16(\sin 2\theta +\cos 4\theta +\sin 6\theta )$.

Your brain might immediately jump to trying to find $\theta$. Stop. Take a breath.

In JEE Advanced, the goal is rarely to find the variable itself; the goal is to find the structure of the expression. We are going to treat this not as a calculation, but as a construction project.

The Foundation (The Squaring Trick)

We start with our given: $\sin \theta +\cos \theta =\frac{1}{2}$.

Why do we square this? Because of the beautiful, fundamental identity of trigonometry: ${\sin }^2\theta +{\cos }^2\theta =1$. When we square the left side, we get $(\sin \theta +\cos \theta {)}^2={\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta$.

Notice what happens? The ${\sin }^2\theta +{\cos }^2\theta$ collapses into a simple $1$, and the $2\sin \theta \cos \theta$ becomes our golden ticket: $\sin 2\theta$.

So, we have:

With a simple subtraction, we find our foundational value: $\sin 2\theta =\frac{1}{4}-1=-\frac{3}{4}$. This value is the heartbeat of our entire solution. Everything else we do is just building upon this single, solid number.

The Bridge to Higher Angles

Now, look at the target expression: $16(\sin 2\theta +\cos 4\theta +\sin 6\theta )$. We have $\sin 2\theta$, but we need $\cos 4\theta$ and $\sin 6\theta$. This is where your toolkit of identities comes into play.

We need to bridge the gap between $2\theta$ and the higher multiples. For $\cos 4\theta$, we use the double-angle formula. Recall that $\cos 2x=1-2{\sin }^{2}x$.

If we let $x=2\theta$, then $\cos 4\theta =1-2{\sin }^{2}2\theta$. This is perfect! We already know $\sin 2\theta =-\frac{3}{4}$.

Let's calculate:

See how the math flows? We aren't guessing; we are simply translating the higher-order angles into the language of our foundational value, $\sin 2\theta$.

The Triple Angle Challenge

Now, for the final piece: $\sin 6\theta$. This looks intimidating, but it is just a triple angle in disguise. The formula for $\sin 3x$ is $3\sin x-4{\sin }^{3}x$.

If we set $x=2\theta$, we get $\sin 6\theta =3\sin 2\theta -4{\sin }^{3}2\theta$. Let's substitute our value, $\sin 2\theta =-\frac{3}{4}$, with extreme care.

Precision is the hallmark of an elite student:

To combine these, we need a common denominator of $16$. So, $-\frac{9}{4}$ becomes $-\frac{36}{16}$.

The Final Assembly

We have arrived at the final stage. We have all our components: 1. $\sin 2\theta =-\frac{3}{4}$ 2. $\cos 4\theta =-\frac{1}{8}$ 3. $\sin 6\theta =-\frac{9}{16}$

Now, we substitute these into our target expression: $16(\sin 2\theta +\cos 4\theta +\sin 6\theta )$.

Instead of finding a common denominator inside the bracket, let's distribute the $16$. This is the "pro move" that saves time and reduces calculation errors:

And there it is. The complexity collapses into a clean, integer result. This is the beauty of mathematics—no matter how tangled the expression appears, if you follow the logical threads of identities and substitutions, the path clears before you.

You didn't just solve a problem; you navigated a system. Keep this mindset, and no JEE problem will ever be too daunting. The final answer is -23.