Find an equation of the tangent line to the curve y=sin(3x) +cos(2x)at the point (pi/6, y(pi/6)). Tangent line: 

Expert Answers
sciencesolve eNotes educator| Certified Educator

You need to remember what equation of tangent line to a curve at a point is:

`y - y_0 = f'(x_0)(x - x_0)`

You need to identify what are the `x_0`  and `y_0`  coordinates of the point of tangency, hence, the problem provides that `x_0=pi/6`  and `y_0=sin 3pi/6 + cos 2pi/6 = 1 + 1/2 = 3/2` .

You need to differentiate the function with respect to x such that:

`f'(x) = (sin 3x + cos 2x)'`

You need to use chain rule to differentiate the function such that:

`f'(x) = 3cos 3x - 2 sin 2x`

You need to substitute `pi/6`  for x in `f'(x)`  such that:

`f'(pi/6) = 3 cos 3(pi/6) - 2 sin 2(pi/6)`

`f'(pi/6) = 3 cos (pi/2) - 2 sin (pi/3)`

Since `sin (pi/3) = sqrt3/2`  and `cos(pi/2) =0`  yields:

`f'(pi/6) = - 2 sqrt3/2` 

`f'(pi/6) = -sqrt3`

You need to substitute `pi/6`  for `x_0`  , `3/2`  for `y_0`  and `-sqrt3`  for `f'(x_0)`  in equation of tangent line such that:

`y -3/2 = -sqrt3(x - pi/6)`

You need to use the slope intercept form of equation of tangent line, hence, you need to isolate y to the left side such that:

`y = -sqrt3*x - pi*sqrt3/6 + 3/2`

Hence, evaluating the equation of tangent line to the graph `y=sin(3x) +cos(2x)`  , at (pi/6,3/2) yields`y = -sqrt3*x - pi*sqrt3/6 + 3/2.`