JEE Main 2026 — Mathematics Trigonometry

Expanding and regrouping:

$$\sin\frac{15\theta}{2}(\cos 8\theta + \sin 8\theta) + \cos\frac{15\theta}{2}(\cos 8\theta - \sin 8\theta)$$sin215θ​(cos8θ+sin8θ)+cos215θ​(cos8θ−sin8θ)

$$= \left(\cos\frac{15\theta}{2}\cos 8\theta + \sin\frac{15\theta}{2}\sin 8\theta\right) + \left(\sin\frac{15\theta}{2}\cos 8\theta - \cos\frac{15\theta}{2}\sin 8\theta\right)$$=(cos215θ​cos8θ+sin215θ​sin8θ)+(sin215θ​cos8θ−cos215θ​sin8θ)

$= \cos\left(8\theta - \frac{15\theta}{2}\right) + \sin\left(\frac{15\theta}{2} - 8\theta\right) = \cos\frac{\theta}{2} - \sin\frac{\theta}{2}$=cos(8θ−215θ​)+sin(215θ​−8θ)=cos2θ​−sin2θ​.

From $\cot\theta = -\frac{1}{2\sqrt{2}}$cotθ=−22​1​: $\sin\theta = \frac{2\sqrt{2}}{3}$sinθ=322​​, $\cos\theta = -\frac{1}{3}$cosθ=−31​.

$\left(\cos\frac{\theta}{2} - \sin\frac{\theta}{2}\right)^2 = 1 - \sin\theta = 1 - \frac{2\sqrt{2}}{3} = \frac{(\sqrt{2}-1)^2}{3}$(cos2θ​−sin2θ​)2=1−sinθ=1−322​​=3(2​−1)2​.

Since $\frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{2}$4π​<2θ​<2π​, $\sin\frac{\theta}{2} > \cos\frac{\theta}{2}$sin2θ​>cos2θ​, so the value is $\frac{1-\sqrt{2}}{\sqrt{3}}$3​1−2​​.