# What is the slope of the line normal to the tangent line of f(x) = cscx-sin2x  at  x= (pi)/12 ?

Dec 19, 2016

After a long derivation, I got:

${f}_{N} \left(\frac{\pi}{12}\right) = \left[\frac{2 - \sqrt{3}}{\sqrt{3} \left(2 - \sqrt{3}\right) + 2 \sqrt{2 + \sqrt{3}}}\right] x + \frac{12 \sqrt{2 - \sqrt{3}} - 3 \left(2 - \sqrt{3}\right) + \pi \sqrt{2 + \sqrt{3}}}{6 \left(2 - \sqrt{3}\right)} + \frac{\sqrt{3} \pi}{12}$

So the slope is $\frac{2 - \sqrt{3}}{\sqrt{3} \left(2 - \sqrt{3}\right) + 2 \sqrt{2 + \sqrt{3}}}$, or about $0.0619$.

The tangent line shows as:

graph{(cscx - sin(2x) - y)(7.59218 - 16.1516x - y) = 0 [-6.2, 7.844, -2.215, 4.81]}

and you can verify that the slope is indeed the negative reciprocal of $- \frac{\sqrt{3} \left(2 - \sqrt{3}\right) + 2 \sqrt{2 + \sqrt{3}}}{2 - \sqrt{3}}$.

From the following linearization formula:

$\boldsymbol{{f}_{T} \left(a\right) = f \left(a\right) + f \text{'} \left(a\right) \left(x - a\right)}$

we can get the equation of the tangent line. From that, the line normal to it is simply the one with the negative reciprocal slope, making it perpendicular, i.e. normal.

The derivative, $f ' \left(x\right)$, is:

$f ' \left(x\right) = - \csc x \cot x - 2 \cos 2 x$

So, at $x = \frac{\pi}{12}$:

$f ' \left(\frac{\pi}{12}\right) = - \left[\csc \left(\frac{\pi}{12}\right) \cot \left(\frac{\pi}{12}\right) + 2 \cos \left(\frac{\pi}{6}\right)\right]$

and

$f \left(\frac{\pi}{12}\right) = \csc \left(\frac{\pi}{12}\right) - \sin \left(\frac{\pi}{6}\right)$

so that

${f}_{T} \left(\frac{\pi}{12}\right) = f \left(\frac{\pi}{12}\right) + f ' \left(\frac{\pi}{12}\right) \left(x - \frac{\pi}{12}\right)$

$= \csc \left(\frac{\pi}{12}\right) - \sin \left(\frac{\pi}{6}\right) - \left[\csc \left(\frac{\pi}{12}\right) \cot \left(\frac{\pi}{12}\right) + 2 \cos \left(\frac{\pi}{6}\right)\right] \left(x - \frac{\pi}{12}\right)$

Note that $\frac{\pi}{12} = \frac{1}{2} \left(\frac{\pi}{6}\right)$. Let us use a half-angle formula to obtain this from a known trig value. Recall that:

${\sin}^{2} \left(x\right) = \frac{1 - \cos 2 x}{2} \implies {\sin}^{2} \left(\frac{x}{2}\right) = \frac{1 - \cos x}{2}$

$\implies \sin \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos x}{2}}$

Similarly,

$\implies \cos \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 + \cos x}{2}}$.

Since $\sin$ is positive for the first two quadrants, and $\cos x$ is positive for the first quadrant, and $\frac{\pi}{12}$ is in the first quadrant, we take the positive solutions.

This allows us to evaluate $\csc \left(\frac{\pi}{12}\right)$ and $\cot \left(\frac{\pi}{12}\right)$ directly. You should get:

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

$= \frac{1}{\sqrt{\frac{1 - \cos \left(\pi \text{/} 6\right)}{2}}} = \left[\ldots\right] = \frac{2}{\sqrt{2 - \sqrt{3}}}$

Similarly,

$\cot \left(\frac{\pi}{12}\right) = \cos \frac{\frac{\pi}{12}}{\sin} \left(\frac{\pi}{12}\right) = \left[\ldots\right] = \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2 - \sqrt{3}}}$

So, we would then get the following tangent line:

${f}_{T} \left(\frac{\pi}{12}\right) = \csc \left(\frac{\pi}{12}\right) - \sin \left(\frac{\pi}{6}\right) - \left[\csc \left(\frac{\pi}{12}\right) \cot \left(\frac{\pi}{12}\right) \left(x - \frac{\pi}{12}\right) + 2 \cos \left(\frac{\pi}{6}\right) \left(x - \frac{\pi}{12}\right)\right]$

Plug in our previous evaluations to get:

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

$= \frac{4 - \sqrt{2 - \sqrt{3}}}{2 \sqrt{2 - \sqrt{3}}} - \frac{2 \sqrt{2 + \sqrt{3}}}{2 - \sqrt{3}} \left(x - \frac{\pi}{12}\right) - \sqrt{3} \left(x - \frac{\pi}{12}\right)$

$= \frac{4 - \sqrt{2 - \sqrt{3}}}{2 \sqrt{2 - \sqrt{3}}} - \frac{2 \sqrt{2 + \sqrt{3}}}{2 - \sqrt{3}} x + \frac{2 \sqrt{2 + \sqrt{3}}}{2 - \sqrt{3}} \frac{\pi}{12} - \sqrt{3} x + \frac{\sqrt{3} \pi}{12}$

$= \frac{12 - 3 \sqrt{2 - \sqrt{3}}}{6 \sqrt{2 - \sqrt{3}}} + \left[- \sqrt{3} - \frac{2 \sqrt{2 + \sqrt{3}}}{2 - \sqrt{3}}\right] x + \frac{\pi \sqrt{2 + \sqrt{3}}}{6 \left(2 - \sqrt{3}\right)} + \frac{\sqrt{3} \pi}{12}$

$= \left[- \sqrt{3} - \frac{2 \sqrt{2 + \sqrt{3}}}{2 - \sqrt{3}}\right] x + \frac{12 \sqrt{2 - \sqrt{3}} - 3 \left(2 - \sqrt{3}\right) + \pi \sqrt{2 + \sqrt{3}}}{6 \left(2 - \sqrt{3}\right)} + \frac{\sqrt{3} \pi}{12}$

$\implies \textcolor{g r e e n}{{f}_{T} \left(\frac{\pi}{12}\right) = \stackrel{\text{Slope")overbrace(-[(sqrt3(2 - sqrt3) + 2sqrt(2+sqrt3))/(2 - sqrt3)])x + stackrel("y-intercept}}{\overbrace{\frac{12 \sqrt{2 - \sqrt{3}} - 3 \left(2 - \sqrt{3}\right) + \pi \sqrt{2 + \sqrt{3}}}{6 \left(2 - \sqrt{3}\right)} + \frac{\sqrt{3} \pi}{12}}}}$

Finally, when we take the negative reciprocal of the slope, we get the line normal to the tangent line:

$\textcolor{b l u e}{{f}_{N} \left(\frac{\pi}{12}\right) = \left[\frac{2 - \sqrt{3}}{\sqrt{3} \left(2 - \sqrt{3}\right) + 2 \sqrt{2 + \sqrt{3}}}\right] x + \frac{12 \sqrt{2 - \sqrt{3}} - 3 \left(2 - \sqrt{3}\right) + \pi \sqrt{2 + \sqrt{3}}}{6 \left(2 - \sqrt{3}\right)} + \frac{\sqrt{3} \pi}{12}}$

Or, the approximate decimal formula is:

$\textcolor{b l u e}{{f}_{N} \left(\frac{\pi}{12}\right) \approx 0.0619 x + 7.59218}$