Trigonometric Identities and Values
DAILY_CHALLENGE
Grade None
Question:
Let $\dfrac{\pi}{2}<x<\pi$ be such that $\cot x=\dfrac{-5}{\sqrt{11}}$. Then
$$\left(\sin\dfrac{11x}{2}\right)(\sin 6x-\cos 6x)+\left(\cos\dfrac{11x}{2}\right)(\sin 6x+\cos 6x)$$
is equal to
$\dfrac{\sqrt{11}-1}{2\sqrt{3}}$
$\dfrac{\sqrt{11}+1}{2\sqrt{3}}$
$\dfrac{\sqrt{11}+1}{3\sqrt{2}}$
$\dfrac{\sqrt{11}-1}{3\sqrt{2}}$
Step-by-Step Solution
Key Concept: Group sine and cosine product terms using sum-to-product; then use half-angle formulas
Simplify the expression:
$$\sin\tfrac{11x}{2}\sin 6x - \sin\tfrac{11x}{2}\cos 6x + \cos\tfrac{11x}{2}\sin 6x + \cos\tfrac{11x}{2}\cos 6x$$
$$= \cos\!\left(\tfrac{11x}{2}-6x\right)+\sin\!\left(6x-\tfrac{11x}{2}\right) = \cos\!\left(-\tfrac{x}{2}\right)+\sin\!\left(\tfrac{x}{2}\right) = \cos\tfrac{x}{2}+\sin\tfrac{x}{2}$$
With $\cot x=-5/\sqrt{11}$ and $x\in(\pi/2,\pi)$: $\sin x>0$, $\cos x<0$.
$\csc^2 x=1+\cot^2 x=1+25/11=36/11\Rightarrow\sin x=\sqrt{11}/6$, $\cos x=-5/6$.
Half-angle ($x/2\in(\pi/4,\pi/2)$, so $\cos(x/2)>0$):
$$\cos\tfrac{x}{2}=\sqrt{\tfrac{1+\cos x}{2}}=\sqrt{\tfrac{1/6}{2}}=\dfrac{1}{2\sqrt{3}},\quad\sin\tfrac{x}{2}=\sqrt{\tfrac{1-\cos x}{2}}=\sqrt{\tfrac{11/6}{2}}=\dfrac{\sqrt{11}}{2\sqrt{3}}$$
$$\cos\tfrac{x}{2}+\sin\tfrac{x}{2}=\dfrac{\sqrt{11}+1}{2\sqrt{3}}$$
Correct Answer: B