Question

If $\frac{\cos 3x}{\cos x}=\frac{1}{3}$ for some angle $x,0\le x\le \frac{\pi }{2}$, then the value of $\frac{\sin 3x}{\sin x}$ for some x, is

• A. $\frac{7}{3}$
• B. $\frac{5}{3}$
• C. $1$
• D. $\frac{2}{3}$

Answer 1

The correct option is A $\frac{7}{3}$

$\frac{cos3x}{cosx}=\frac{1}{3}\Rightarrow \frac{4{\cos }^{3}x-3\cos x}{cosx}=\frac{1}{3}\Rightarrow 4{\cos }^{2}x-3=\frac{1}{3}\Rightarrow \cos x=\sqrt{\frac{5}{6}}$

we know that, $\sin x=\sqrt{1-{\cos }^{2}x}\Rightarrow \sin x=\sqrt{1-\frac{5}{6}}=\sqrt{\frac{1}{6}}$

Now, $\frac{\sin 3x}{\sin x}=\frac{3\sin x-4{\sin }^{3}x}{\sin x}=3-4{\sin }^{2}x=3-4\ast \frac{1}{6}=\frac{7}{3}$

Therefore, Answer is $\frac{7}{3}$