What is the angle between the graph of f(x)=2sin(x)-1 and the x-axis if the graph of f intersects the x-axis at x=pi/6?

I worked out the tangent line to the graph of $f$ at $\left(\frac{\pi}{6} , 0\right)$ and I got $y = 2 \cos \left(x\right) \cdot x - \frac{\pi}{6}$

Aug 9, 2017

The angle is $= \frac{\pi}{3}$

Explanation:

We need

$\left(\sin x\right) ' = \cos x$

$\left(a\right) ' = 0$

Our function is

$f \left(x\right) = 2 \sin x - 1$

The derivative with respect to $x$ is

$f ' \left(x\right) = \left(2 \sin x - 1\right) ' = = \left(2 \sin x\right) ' - \left(1\right) ' = 2 \cos x$

At the point $x = \frac{\pi}{6}$,

$f ' \left(\frac{\pi}{6}\right) = 2 \cos \left(\frac{\pi}{6}\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$

The slope of the tangent at $x = \frac{\pi}{6}$ is

$m = \sqrt{3}$

The equation of the tangent is

$y = \sqrt{3} \left(x - \frac{\pi}{6}\right)$

$m = \tan \alpha = \sqrt{3}$

Therefore,

The angle is $\alpha = \arctan \left(\sqrt{3}\right) = {60}^{\circ} = \frac{\pi}{3}$

graph{((2sinx-1))(y-(sqrt3)(x-pi/6))=0 [-0.915, 1.786, -0.586, 0.765]}