Question

If $\cot x=\dfrac{5}{12}$ for some $x\in\left(\pi,\dfrac{3\pi}{2}\right)$, then \[ \sin 7x\left(\cos\frac{13x}{2}+\sin\frac{13x}{2}\right) +\cos 7x\left(\cos\frac{13x}{2}-\sin\frac{13x}{2}\right) \] is equal to:

• A. $\dfrac{1}{\sqrt{13}}$
• B. $\dfrac{5}{\sqrt{13}}$
• C. $-\dfrac{1}{\sqrt{13}}$
• D. $\dfrac{8}{\sqrt{13}}$

Hint:

In trigonometric problems, always simplify long expressions first using sum-to-product or phase-shift identities.

Answer 1

The Correct Option is A

To solve this problem, we begin by analyzing the given condition: \(\cot x = \dfrac{5}{12}\) for some \(x \in \left(\pi, \dfrac{3\pi}{2}\right)\). Since \(\cot x = \dfrac{\cos x}{\sin x}\), we have:

• \(\cos x = 5k\)
• \(\sin x = 12k\)

Using the Pythagorean identity, \(\cos^2 x + \sin^2 x = 1\), we get:

• \((5k)^2 + (12k)^2 = 1\)
• \(25k^2 + 144k^2 = 1\)
• \(169k^2 = 1\)
• \(k = \pm \dfrac{1}{13}\)

Since \(x \in \left(\pi, \dfrac{3\pi}{2}\right)\), \(\sin x\) is negative and \(\cos x\) is also negative in the third quadrant. Therefore, \(k = -\dfrac{1}{13}\), and thus:

• \(\cos x = -\dfrac{5}{13}\)
• \(\sin x = -\dfrac{12}{13}\)

We need to evaluate:

• \(\sin 7x\left(\cos\frac{13x}{2}+\sin\frac{13x}{2}\right) + \cos 7x\left(\cos\frac{13x}{2}-\sin\frac{13x}{2}\right)\)

Using the identities:

• \(\cos A + \sin A = \sqrt{2} \sin\left(A + \dfrac{\pi}{4}\right)\)
• \(\cos A - \sin A = \sqrt{2} \cos\left(A + \dfrac{\pi}{4}\right)\)

Substituting these identities, the expression simplifies to:

• \(\sqrt{2} \left(\sin 7x \sin \left(\dfrac{13x}{2} + \dfrac{\pi}{4}\right) + \cos 7x \cos \left(\dfrac{13x}{2} + \dfrac{\pi}{4}\right)\right)\)

Using the identity \(\cos A \cos B + \sin A \sin B = \cos(A-B)\), this becomes:

• \(\sqrt{2} \cos \left(7x - \left(\dfrac{13x}{2} + \dfrac{\pi}{4}\right)\right)\)
• \(\sqrt{2} \cos \left(\dfrac{x}{2} - \dfrac{\pi}{4}\right)\)

Now, compute \(\cos \left(\dfrac{x}{2} - \dfrac{\pi}{4}\right)\):

Using \(\cos\left(a - b\right) = \cos a \cos b + \sin a \sin b\), we plug in the known values:

• \(\cos \dfrac{x}{2} = \sqrt{\dfrac{13}{2}}\)
• \(\sin \dfrac{x}{2} = \sqrt{\dfrac{-13}{2}}\)

Substituting we get:

• \(\sqrt{2} \left(\cos \left(\dfrac{x}{2}\right) \cos \dfrac{\pi}{4} + \sin \left(\dfrac{x}{2}\right) \sin \dfrac{\pi}{4}\right)\)

This results in value \(\dfrac{1}{\sqrt{13}}\)