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I need help with this questions. I have one part right but can figure the other part.

An equation of the tangent line to the curve $y= \sin⁡(x)$ at $ x = \dfrac{11\pi}{6x} =\dfrac{11\pi}{6} $

$y = -\dfrac{1}{2} + B$

I got the first part which was to find $\frac{d}{dx}$ of $\dfrac{\sin(11\pi)}{6x} = -\dfrac{1}{2}$

But I can't figure out how to get $B$ for the equation.

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    It will be easier to find an equation in point-slope form. – 2017-02-23
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    If you find the slope of the tangent line, then you solve for the intercept using a known point on the line. In this case, solve for $b$ in $y=mx+b$ where $m$ is the slope, and $(x,y)=(11\pi /6,\sin(11\pi /6))$ is a known point on the line. – 2017-02-23

2 Answers 2

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Note general equation of a line:

$y = mx + c$

So we have $y = sinx$

$\Rightarrow {y}' = cosx$

At $x = \frac{11\pi}{6}$, ${y}' = cos(\frac{11\pi}{6}) = \frac{\sqrt{3}}{2}$

Therefore our line of intersection has equation:

$y = \frac{\sqrt{3}}{2}x + c$

To find c, we equate this equation with the original one, since we know they intersect at $x = \frac{11\pi}{6}$.

$\frac{\sqrt{3}}{2}\frac{11\pi}{6} + c = sin\frac{11\pi}{6}$

$ c = -\frac{11\sqrt{3}\pi}{12} -\frac{1}{2}$

Therefore we get the equation of the line as:

$y = \frac{\sqrt{3}}{2}x - \frac{11\sqrt{3}\pi}{12} -\frac{1}{2}$ as required.

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    Take note that your slope wouldn't be $\frac{-1}{2}$ but rather $\frac{\sqrt{3}}{2}$ – 2017-02-23
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    My mistake, I copied that part from original post, will edit it now – 2017-02-23
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    Thank you, I understand it now – 2017-02-23
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    @mrnovice $\cos\frac{11\pi}{6}=\cos 330^\circ=\frac{\sqrt{3}}{2}$? – 2017-02-23
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    @mrnovice Why $y'=-\frac{\sqrt{3}}{2}$? It should be $$y'=\frac{\sqrt{3}}{2}$$ – 2017-02-23
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    In my defence, it's 4am, anyways fixed it lol – 2017-02-23
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    Just edit, and I will up vote you as well.--) – 2017-02-23
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The point slope equation, where a line with slope $m$ and passes through the point $(x_0, y_0)$, has the equation $y-y_0=m(x-x_0)$.

In your case, a line passing through the point $(\frac{11π}{6},-\frac12)$ with slope $\frac{\sqrt{3}}{2}$ can be represented as $y+\frac12=\frac{\sqrt{3}}{2}(x-\frac{11π}{6})$.