Trigonometry
Trigonometric expression with cotangent condition
MJAT_TS7_P1
Grade 12
Question:
Let $\dfrac{\pi}{2}<x<\pi$ such that $\cot x = -\dfrac{5}{\sqrt{11}}$. Then $\left(\sin\dfrac{11x}{2}\right)(\sin^6 x-\cos^6 x)+\left(\cos\dfrac{11x}{2}\right)(\sin^6 x+\cos^6 x)$ equals:
A) $\dfrac{\sqrt{11}-1}{2\sqrt{3}}$
B) $\dfrac{\sqrt{11}+1}{2\sqrt{3}}$
C) $\dfrac{\sqrt{11}+1}{3\sqrt{2}}$
D) $\dfrac{\sqrt{11}-1}{3\sqrt{2}}$
Step-by-Step Solution
Key Concept: Use $\sin^6x-\cos^6x=(\sin^2x-\cos^2x)(\sin^4x+\sin^2x\cos^2x+\cos^4x)=-\cos2x(1-\sin^2x\cos^2x)$ and $\sin^6x+\cos^6x=1-3\sin^2x\cos^2x$. Combine with the half-angle expressions.
Answer: **B**.
Correct Answer: B