Calculus exam review, PLEASE HELP!!
posted by stephanie .
for each function, determine the equation of the tangent line to the graph for the indicated value of the variable.
a) y = 3cosx + 4sinx + 1 , when x = pie
b) y = x  3cosx , when x = (5pie)/3

π is PI not PIE!
a)
y'(x) = 3sinx + 4cosx
y'(π) = 3sin π + 4cos π = 4
y(π) = 3(1) + 0 + 1 = 2
thus, the line through (π,2) having slope 4 is
(y+2) = 4(xpi)
b) y' = 1 + 3sinx
y'(5π/3) = 1 + 3(√3/2) = 1  3√3/2
y(5π/3) = 5π/3  3(1/2) = 5π/3  3/2
(y(5π/3  3/2) = (1  3√3/2)(x5π/3) 
thank you so much steve!
for b) (y(5π/3  3/2) = (1  3√3/2)(x5π/3)
is the square root for the whole 3/2 or is it just root 3 OVER 2. 
I think by now you know the value of cos(pi/3)

(y(5π/3  3/2) = (1  (3√3)/2)(x5π/3)
like that ? :(