Question

Let \(k\) be a real number such that \(\sin \dfrac{3π}{14} \cos \dfrac{3π}{14} = k \cos \dfrac{π}{14}\).Then the value of \(4k\) is 

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)
• E. \(0\)

Answer 1

The Correct Option is B

Given: 

\(\sin \dfrac{3π}{14} \cos \dfrac{3π}{14} = k \cos \dfrac{π}{14}\)

\(⇒2 \sin\dfrac{3π}{14} \cos \dfrac{3π}{14} = 2k \cos \dfrac{π}{14}\)

 \(⇒\sin\dfrac{6π}{14} =2k \cos \dfrac{π}{14}\)

 \(\therefore( \dfrac{6π}{14} +\dfrac{π}{14} = \dfrac{7π}{14} = \dfrac{π}{2} )\)

\(\sin \dfrac{6π}{14} = \cos\dfrac{ π}{14}\)
1 = 2k
\(\therefore 4k=4×\dfrac{1}{2}\)
\(\therefore 4k=2\)
So, the correct option is (B) : 2.

Answer 2

We are given the equation:

\[ \sin\left(\frac{3\pi}{14}\right)\cos\left(\frac{3\pi}{14}\right) = k \cos\left(\frac{\pi}{14}\right) \]

We can use the double angle identity for sine: \( 2\sin(x)\cos(x) = \sin(2x) \). Therefore,

\[ \sin\left(\frac{3\pi}{14}\right)\cos\left(\frac{3\pi}{14}\right) = \frac{1}{2}\sin\left(\frac{6\pi}{14}\right) = \frac{1}{2}\sin\left(\frac{3\pi}{7}\right). \]

So the equation becomes:

\[ \frac{1}{2}\sin\left(\frac{3\pi}{7}\right) = k \cos\left(\frac{\pi}{14}\right). \]

We can use the triple angle identity for sine:

\[ \sin(3\theta) = 3\sin\theta - 4\sin^3\theta. \]

Let \( \theta = \frac{\pi}{14} \). Then \( 3\theta = \frac{3\pi}{14} \), and also \( 7\theta = \frac{\pi}{2} \).

\[ \sin\left(\frac{3\pi}{7}\right) = \sin\left(\pi - \frac{4\pi}{7}\right) = \sin\left(\frac{4\pi}{7}\right). \]

We have:

\[ \sin\left(\frac{3\pi}{7}\right) = 3\sin\left(\frac{\pi}{7}\right) - 4\sin^3\left(\frac{\pi}{7}\right). \]

This doesn't seem to lead to a simple solution using standard trigonometric identities. Let's try a different approach.

We know that:

\[ \sin(2x) = 2\sin(x)\cos(x). \]

Therefore:

\[ \sin\left(\frac{3\pi}{14}\right)\cos\left(\frac{3\pi}{14}\right) = \frac{1}{2}\sin\left(\frac{6\pi}{14}\right) = \frac{1}{2}\sin\left(\frac{3\pi}{7}\right). \]

Let's use the fact that \( \sin\left(\frac{3\pi}{7}\right) = \sin\left(\pi - \frac{4\pi}{7}\right) = \sin\left(\frac{4\pi}{7}\right) \).

We are given:

\[ \frac{1}{2}\sin\left(\frac{3\pi}{7}\right) = k \cos\left(\frac{\pi}{14}\right). \] \[ \sin\left(\frac{3\pi}{7}\right) = 2k \cos\left(\frac{\pi}{14}\right). \]

Using the appropriate trigonometric identities here is complex. Let's use a numerical approach:

\[ \sin\left(\frac{3\pi}{7}\right) \approx 0.9749, \quad \cos\left(\frac{\pi}{14}\right) \approx 0.9848. \]

Therefore:

\[ 0.9749 \approx 2k(0.9848). \] \[ k \approx \frac{0.9749}{2 \cdot 0.9848} \approx 0.4948. \] \[ 4k \approx 4 \cdot 0.4948 \approx 1.9792. \]

The value of \( 4k \) is approximately 2.