Jee Main
Trigonometric Ratio And Identites
JEE Main 2026 (Online) 23rd January Evening Shift
MCQ+4 / -12026
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 :
