Question

Suppose $\sin 2\theta =\cos 3\theta$, here $0<\theta <\frac{\pi }{2}$ then what is the value of $\cos 2\theta ?$

• (A) $\frac{1+\sqrt{5}}{4}$
• (B) $\frac{1-2\sqrt{5}}{8}$
• (C) $\frac{-1+\sqrt{5}}{4}$
• (D) $\frac{-1-\sqrt{5}}{4}$

Answer 1

The Correct Option is A

Explanation:
Given:$\sin 2\theta =\cos 3\theta ,0<\theta <\frac{\pi }{2}$As we know that, $\sin 2\theta =2\sin \theta \cos \theta$ and $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$$\Rightarrow 2\sin \theta \cos \theta =4{\cos }^3\theta -3\cos \theta$$\Rightarrow 2\sin \theta =4{\cos }^2\theta -3$$\Rightarrow 2\sin \theta =4(1-{\sin }^2\theta )-3=4-4{\sin }^2\theta -3$$\Rightarrow 4{\sin }^2\theta +2\sin \theta -1=0$Comparing the above equation with quadratic equation $a{x}^2+bx+c=0,a=4,b=2$ and $c=-1$Now substituting the values in the quadratic formula $\frac{x=(-b\pm {\sqrt{b}}^2-4ac)}{2a}$ we get,$\sin \theta =\frac{-2\pm \sqrt{-2^2-4(4)(-1)}}{2(4)}$$=\frac{-2\pm \sqrt{4+16}}{8}$$=\frac{-2\pm \sqrt{20}}{8}$$=\frac{-1\pm \sqrt{5}}{4}$Thus, $\sin \theta =\frac{-1\pm \sqrt{5}}{4}$Since, $0<\theta <\frac{\pi }{2}\Rightarrow \theta$ lies between $0^{\circ }$ to ${90}^{\circ }\Rightarrow$ all ratios are positive.$\Rightarrow \sin \theta =\frac{-1+\sqrt{5}}{4}$As we know that, $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$$=1-2{\sin }^2\theta$$\Rightarrow \cos 2\theta =1-2{\left(\frac{-1+\sqrt{5}}{4}\right)}^2$$=\frac{1+\sqrt{5}}{4}$$\therefore$ The the value of $\cos 2\theta$ is $\frac{1+\sqrt{5}}{4}$Hence, the correct option is (A).