Question

If \(\sin\theta \sin(60^\circ-\theta)\sin(60^\circ+\theta)=\dfrac18\), then \(\cos6\theta=\)

• A. \(\dfrac{\sqrt3}{2}\)
• B. \(\dfrac12\)
• C. \(\dfrac1{\sqrt2}\)
• D. \(0\)

Hint:

Memorize: \[ \sin3x=4\sin x\sin(60^\circ+x)\sin(60^\circ-x). \] It appears frequently in competitive examinations.

Answer 1

The Correct Option is B

Concept: A standard trigonometric identity is \[ \sin3x=4\sin x\sin(60^\circ+x)\sin(60^\circ-x). \] This identity directly connects the given product with a multiple-angle expression.

Step 1: Apply the standard identity. Given, \[ \sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta) = \frac18. \] Using \[ 4\sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta) = \sin3\theta, \] we obtain \[ \sin3\theta = 4\cdot\frac18 = \frac12. \]

Step 2: Determine \(\cos6\theta\). Using \[ \cos6\theta = 1-2\sin^23\theta, \] \[ = 1-2\left(\frac12\right)^2. \] \[ = 1-\frac12. \] \[ = \frac12. \] Conclusion: \[ \boxed{\cos6\theta=\frac12} \]