differentiating trig ratios

[red and white kop!]

find the equation of the tangent to the curve y=cosA + 3sinA at the point where A= pi/2
so i took A=pi/2 to get y= 2sqrt2 and m=sqrt2, with the resulting equation being y=sqrt2(A+2-(pi/2))
apparently that is so wrong, and the answer is y+A=3+(pi/2)
i need help here

 

[daon]

y-y1=m(x-x1)

(x1,y1)=(A,y(A)) = (pi/2, 3)

m = y'(A) = -sinA+3cosA = -1

 

[BigGlenntheHeavy]

\(\displaystyle y \ = \ f(A) \ = \ cos(A)+3sin(A)\)

\(\displaystyle f'(A) \ = \ -sin(A)+3cos(A), \ f'(\pi/2) \ = \ -1 \ = \ m\)

\(\displaystyle f(\pi/2) \ = \ 3\)

\(\displaystyle y-3 \ = \ -1(A-\pi/2), \ y \ = \ -A+\pi/2+3, \ see \ graph\)

[attachment=0:3iiwzk5y]def.jpg[/attachment:3iiwzk5y]