Question

If $\alpha$ is one of the principal solutions which satisfies the equation $1+{\sin }^2\theta =3\sin \theta \cos \theta ,$ then which of the following is not possible?

• A. $\sin 2\alpha =\frac{4}{5}$
• B. $\cos 2\alpha =0$
• C. $\sin 3\alpha =\frac{1}{\sqrt{2}}$
• D. $\cos 3\alpha =\frac{2}{\sqrt{5}}$

Answer 1

The correct option is D $\cos 3\alpha =\frac{2}{\sqrt{5}}$

$1+{\sin }^2\theta =3\sin \theta \cos \theta \Rightarrow 2{\sin }^2\theta -3\sin \theta \cos \theta +{\cos }^2\theta =0\Rightarrow (\sin \theta -\cos \theta )(2\sin \theta -\cos \theta )=0\Rightarrow \sin \theta =\cos \theta \text{or}2\sin \theta =\cos \theta \Rightarrow \tan \theta =1\text{or}\tan \theta =\frac{1}{2}$
$\to \sin 2\alpha =\frac{2\tan \alpha }{1+{\tan }^2\alpha }=\frac{2\times \frac{1}{2}}{1+\frac{1}{4}}=\frac{4}{5}$

$\to \cos 2\alpha =0$ (if $\tan \alpha =1$)

$\to \sin 3\alpha =\sin {135}^{\circ }=\frac{1}{\sqrt{2}}$ (if $\tan \alpha =1$)

$\to \cos 3\alpha =4{\cos }^3\alpha -3\cos \alpha =4{\left(\frac{2}{\sqrt{5}}\right)}^3-3\left(\frac{2}{\sqrt{5}}\right)$
$=\frac{2}{5\sqrt{5}}$ $(\text{if}\tan \alpha =\frac{1}{2})$