Question #07780

1 Answer
Jun 14, 2017

The equation of the tangent line is: #y = 1/2x#

Explanation:

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

#cos(pi/2-pi/4) + sin(pi/4) = sqrt2#

#cos(pi/4) + sin(pi/4) = sqrt2#

#sqrt2/2 + sqrt2/2 = sqrt2#

#sqrt2=sqrt2#

This checks.

Implicitly differentiate with respect to x:

#(d(cos(x-y)))/dx + (d(sin(y)))/dx = (d(sqrt2))/dx#

#-sin(x-y)(1-dy/dx) + cos(y)dy/dx = 0#

#-sin(x-y) + sin(x-y)dy/dx+cos(y)dy/dx = 0#

#-sin(x-y) + (sin(x-y)+cos(y))dy/dx = 0#

#(sin(x-y)+cos(y))dy/dx = sin(x-y)#

#dy/dx = sin(x-y)/(sin(x-y)+cos(y))#

The slope of the tangent line is the derivative evaluated at the point #(pi/2,pi/4)#:

#m = sin(pi/2-pi/4)/(sin(pi/2-pi/4)+cos(pi/4))#

#m = 1/2#

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

#y = 1/2(x-pi/2)+pi/4#

This simplifies to:

#y = 1/2x#