# Question 07780

Jun 14, 2017

The equation of the tangent line is: $y = \frac{1}{2} x$

#### Explanation:

Check that the given point (pi/2,pi/4# lies on the curve:

$\cos \left(\frac{\pi}{2} - \frac{\pi}{4}\right) + \sin \left(\frac{\pi}{4}\right) = \sqrt{2}$

$\cos \left(\frac{\pi}{4}\right) + \sin \left(\frac{\pi}{4}\right) = \sqrt{2}$

$\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$

$\sqrt{2} = \sqrt{2}$

This checks.

Implicitly differentiate with respect to x:

$\frac{d \left(\cos \left(x - y\right)\right)}{\mathrm{dx}} + \frac{d \left(\sin \left(y\right)\right)}{\mathrm{dx}} = \frac{d \left(\sqrt{2}\right)}{\mathrm{dx}}$

$- \sin \left(x - y\right) \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) + \cos \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$- \sin \left(x - y\right) + \sin \left(x - y\right) \frac{\mathrm{dy}}{\mathrm{dx}} + \cos \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$- \sin \left(x - y\right) + \left(\sin \left(x - y\right) + \cos \left(y\right)\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\left(\sin \left(x - y\right) + \cos \left(y\right)\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \sin \left(x - y\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sin \frac{x - y}{\sin \left(x - y\right) + \cos \left(y\right)}$

The slope of the tangent line is the derivative evaluated at the point $\left(\frac{\pi}{2} , \frac{\pi}{4}\right)$:

$m = \sin \frac{\frac{\pi}{2} - \frac{\pi}{4}}{\sin \left(\frac{\pi}{2} - \frac{\pi}{4}\right) + \cos \left(\frac{\pi}{4}\right)}$

$m = \frac{1}{2}$

Use the point-slope form of the equation of a line:

$y = \frac{1}{2} \left(x - \frac{\pi}{2}\right) + \frac{\pi}{4}$

This simplifies to:

$y = \frac{1}{2} x$