If f'(x)=sinx and f(pi)=3, then f(x)=

A. Cosx+4
B. Cosx+3
C. -cosx+2
D. -cosx-2
E. -cosx+4

3 answers

f'(x)=sin(x)
then by the derivative of trig functions, f(x)=-cos(x)+C
Now solve for C in
f(π)=-cos(π)+C=3
-cosx+2
3 + cosx
Similar Questions
  1. Simplify #3:[cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] =
    1. answers icon 1 answer
  2. Trigonometric IdentitiesProve: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx +
    1. answers icon 0 answers
  3. Solve this equation fo rx in the interval 0<=x<=3603sinxtanx=8 I would do it this way: sinxtanx = 8/3 sinx(sinx/cosx)=8/3
    1. answers icon 0 answers
  4. hey, i would really appreciate some help solving for x when:sin2x=cosx Use the identity sin 2A = 2sinAcosA so: sin 2x = cos x
    1. answers icon 0 answers
more similar questions