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#
