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

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

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

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

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

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