Question

I: $\sin \theta .\sin ({60}^0-\theta ).\sin ({60}^0+\theta )=\frac{1}{4}\sin 3\theta$
II: $\cos \theta \cos ({120}^0-\theta )\cos ({120}^0+\theta )=$$\frac{1}{4}\cos 3\theta$

• A. only I is true
• B. only II is true
• C. Both I and II are true
• D. Neither I nor II are true

Answer 1

Answer: (C)

Both I and II are true

$a=\sin \theta \sin (60-\theta )\sin (60+\theta )$
$\Rightarrow a=\frac{\sin \theta \left[2\sin (60-\theta )\sin (60+\theta )\right]}{2}$
$\Rightarrow a=\frac{\sin \theta [\cos (60-\theta -60-\theta )-\cos (60-\theta +60+\theta )]}{2}\left\{2\sin A\sin B=\cos (A-B)-\cos (A+B)\right\}$
$\Rightarrow a=\frac{\sin \theta [\cos 2\theta -\cos 120]}{2}=\frac{\sin \theta [1-2{\sin }^2\theta +\frac{1}{2}]}{2}$
$\Rightarrow a=\frac{3\sin \theta -4{\sin }^3\theta }{4}=\frac{1}{4}\sin 3\theta$
$b=\cos \theta \cos (120-\theta )\cos (120+\theta )$
$\Rightarrow b=\frac{\cos \theta \left[2\cos (120-\theta )\cos (120+\theta )\right]}{2}$
$\Rightarrow b=\frac{\cos \theta [\cos (120-\theta -60-\theta )+\cos (120-\theta +120+\theta )]}{2}\{2\cos A\cos B=\cos (A-B)+\cos (A+B)\}$
$\Rightarrow b=\frac{\cos \theta [\cos 2\theta +\cos 240]}{2}=\frac{\cos \theta [2{\cos }^2\theta -1-\frac{1}{2}]}{2}$
$\Rightarrow b=\frac{4{\cos }^3\theta -3\cos \theta }{4}=\frac{1}{4}\cos 3\theta$
Hence, option 'C' is correct.