Question

sin6θ+sin4θ−sin2θ=4cosθ⋅sin2θ⋅cos3θsin⁡6θ+sin⁡4θ−sin⁡2θ=4cos⁡θ⋅sin⁡2θ⋅cos⁡3θ

• A. True
• B. False

Answer 1

The correct option is A True

Given

$\sin 6\theta +\sin 4\theta -\sin 2\theta$

$⟹\sin 6\theta +2\cos 3\theta \sin \theta$

$(\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right))$

$⟹2\sin 3\theta \cos 3\theta +2\cos 3\theta \sin \theta$

$⟹2\cos 3\theta (\sin 3\theta +\sin \theta )$

$⟹2\cos 3\theta \left(2\sin 2\theta \cos \theta \right)$

$(\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right))$

$⟹4\cos \theta \sin 2\theta \cos 3\theta$