Question

At what angle does the sine curve y = $\left(1/\sqrt{3}\right)$ sin 3x cut the abscissa axis at the origin?

Answer 1

$y=\frac{1}{\sqrt{3}}\sin 3x$

$⟹\frac{dy}{dx}=\frac{3}{\sqrt{3}}\cos 3x=\sqrt{3}\cos 3x$

At, origin-

$x=0⟹\frac{dy}{dx}=\sqrt{3}$

Hence, the slope of tangent to the curve at origin that is the tangent of the angle $\alpha$ at which the curve cut abscissa at origin is $\sqrt{3}$

$⟹\tan \alpha =\sqrt{3}$

$⟹\alpha ={60}^o$