Question

Write the equation of a tangent to the graphs of the following curves at the indicated points
$y=\sin 2x$ at the point $x=\frac{\pi }{12}.$

Answer 1

To find : - equation of a tangent of $y=\sin 2x$ at $x=\frac{\pi }{12}$

$y=\sin \left(2x\right)$

at $x=\frac{\pi }{12}$

$y=\sin \left(\frac{2\pi }{12}\right)=\sin \frac{\pi }{6}=\frac{1}{2}$

Now, $y^{\prime }=2\cos 2x$

at $x=\frac{\pi }{12}$

$y^{\prime }=2\cos \left(\frac{\pi }{12}\right)=2\cos \frac{\pi }{6}=2\frac{\sqrt{3}}{2}$

$=\sqrt{3}$

$\therefore slope=m=\sqrt{3}$

Use point - slope form

$(y-y_0)=m(x-x_0)$

where $x_0=\frac{\pi }{12},y_0=\frac{1}{2},m=\sqrt{3}$

$(y-\frac{1}{2})=\sqrt{3}(x-\frac{\pi }{12})$

$y=\sqrt{3}x-\frac{\sqrt{3}\pi }{12}+\frac{1}{2}$

$=\sqrt{3x}+\frac{1}{2}(\frac{-\sqrt{3}\pi }{6}+1)$.