# What is the equation of the line that is normal to f(x)=sin^2x+e^xcotx  at  x=pi/3?

Apr 14, 2017

$y = .776 x - .812$

#### Explanation:

It helps to know some trig derivatives: First, take the derivative of $f \left(x\right)$. I suggest rewriting ${\sin}^{2} x$ as $\sin x \sin x$ so you can use the product rule. You will have to use the product rule on ${e}^{x} \cot x$ too.

$f ' \left(x\right) = \sin x \cos x + \sin x \cos x + {e}^{x} \cot \left(x\right) - {\csc}^{2} \left(x\right) {e}^{x}$

Simplify:

$f ' \left(x\right) = 2 \sin x \cos x + {e}^{x} \cot \left(x\right) - {\csc}^{2} \left(x\right) {e}^{x}$

Plug in $x = \frac{\pi}{3}$:

$f ' \left(\frac{\pi}{3}\right) = 2 \sin \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}\right) + {e}^{\frac{\pi}{3}} \cot \left(\frac{\pi}{3}\right) - {\csc}^{2} \left(\frac{\pi}{3}\right) {e}^{\frac{\pi}{3}} = - 1.288$

That's the slope of the tangent line but we need the slope perpendicular to that which is simply the negative reciprocal.

$- \frac{1}{-} 1.288 = - .776$

We also need the point at $x = \frac{\pi}{3}$ so we plug into the initial equation:

$f \left(\frac{\pi}{3}\right) = {\sin}^{2} \left(\frac{\pi}{3}\right) + {e}^{\frac{\pi}{3}} \left(\cot \left(\frac{\pi}{3}\right)\right) = 2.395$

Now we can use point-slope form to find the equation of the normal line:

$y - \left(\frac{\pi}{3}\right) = .776 \left(x - 2.395\right)$

$y = .776 x - 1.859 + \frac{\pi}{3}$

$y = .776 x - .812$