Animated Solution for Mathematics - Trigonometry: Let $\frac{\pi }{2}<\theta <\pi$ and $\cot \theta =-\frac{1}{2\sqrt{2}}$. Then the value of $\sin \left(\frac{15\theta }{2}\right)(\cos 8\theta +\sin 8\theta )+\cos \left(\frac{15\theta }{2}\right)(\cos 8\theta -\sin 8\theta )$ is equal to

Welcome, future engineers! Today, we are going to dismantle a trigonometric beast. When you first look at the expression:

It is natural to feel a surge of anxiety. It looks like a chaotic mess of angles and functions. But here is the secret: in JEE Advanced, complexity is often just a mask for elegance.

Our first step is to strip away that mask. We begin by expanding the brackets. By distributing the terms, we get:

Now, look at what we have created. We have four distinct terms. This is not chaos; this is a pattern waiting to be grouped.

We now regroup these terms to form the building blocks of trigonometry. Let us pair the 'cosine-cosine' and 'sine-sine' terms together, and the 'sine-cosine' and 'cosine-sine' terms together. This gives us:

Do you see it now? The first bracket is the classic expansion of $\cos (A-B)$, and the second is the expansion of $\sin (A-B)$. By substituting $A=8\theta$ and $B=\frac{15\theta }{2}$, the entire expression collapses into:

A quick subtraction reveals the beauty: $\cos (\frac{\theta }{2})+\sin (-\frac{\theta }{2})$. Since $\sin$ is an odd function, this simplifies to $\cos (\frac{\theta }{2})-\sin (\frac{\theta }{2})$. The beast has been tamed!

Now, we face a new challenge. We have the expression $\cos (\frac{\theta }{2})-\sin (\frac{\theta }{2})$, but we only know the value of $\cot \theta$. How do we connect these? We use the 'squaring bridge'.

Let $X=\cos (\frac{\theta }{2})-\sin (\frac{\theta }{2})$. Then:

Using the Pythagorean identity and the double-angle formula, this becomes $1-\sin \theta$. We are given $\cot \theta =-\frac{1}{2\sqrt{2}}$ in the second quadrant.

Using ${\csc }^2\theta =1+{\cot }^2\theta$, we find ${\csc }^2\theta =1+\frac{1}{8}=\frac{9}{8}$, which means ${\sin }^2\theta =\frac{8}{9}$. Since $\theta$ is in the second quadrant, $\sin \theta =\frac{2\sqrt{2}}{3}$.

Thus, $X^2=1-\frac{2\sqrt{2}}{3}=\frac{3-2\sqrt{2}}{3}$. This numerator, $3-2\sqrt{2}$, is a perfect square: $(\sqrt{2}-1{)}^2$. So:

We are almost there, but we must be careful. Taking the square root of $X^2$ gives us two possibilities: $\pm \frac{\sqrt{2}-1}{\sqrt{3}}$. Which one is correct?

We must check the quadrant of $\frac{\theta }{2}$. We know $\frac{\pi }{2}<\theta <\pi$, so $\frac{\pi }{4}<\frac{\theta }{2}<\frac{\pi }{2}$. In this interval, the sine function is greater than the cosine function.

Therefore, $\cos (\frac{\theta }{2})-\sin (\frac{\theta }{2})$ must be negative. We choose the negative root: $-\frac{\sqrt{2}-1}{\sqrt{3}}$, which simplifies to:

And there it is! A perfect, elegant solution. You have navigated the complexity, used the identities, bridged the gap, and avoided the quadrant trap. Well done!