The angle of intersection of the curves y = 2 sin2 x and y = cos 2 x at x = $\pi$/6 is ____________ .

Question

KollegeApply · Accepted Answer

$\pi$/3 \text { Given }: x = $\pi$/6 \text { Now }, y = 2 \sin^2 x \Rightarrow dy/dx = 4\sin x \cos x \Rightarrow dy/dx = 2 \sin 2x \Rightarrow m_1 = \left( dy/dx \right)_{x = $\pi$/6} = 2 $\times$ $\sqrt$(3)/2 = $\sqrt$(3) \text { Also }, y = \cos 2x \Rightarrow dy/dx = - 2 \sin2x \Rightarrow m_2 = \left( dy/dx \right)_{x = $\pi$/6} = - 2 $\times$ $\sqrt$(3)/2 = - $\sqrt$(3) \therefore \tan $\theta$ = \left| m_1 - m_2/1 + m_1 m_2 \right| = \left| $\sqrt$(3) + $\sqrt$(3)/1 - $\sqrt$(3)$\sqrt$(3) \right| = \left| 2$\sqrt$(3)/- 2 \right| = $\sqrt$(3) \Rightarrow $\theta$ = \tan^{- 1} \left( $\sqrt$(3) \right) = $\pi$/3

AChoose Your Answer