The normal line of a point on a function is the line perpendicular to the tangent line at that point. This involves three basic steps: finding the slope of the tangent line, taking the opposite reciprocal of that slope (giving us the slope of the normal line), and finding the y-intercept.
Step 1: Find the Slope of the Tangent Line
All we do here is find the derivative of #sinx# and evaulate it at #x=pi/6# - because the slope of the tangent line at some point is the derivative at that point.
#f(x)=sinx#
#f'(x)=cosx->#the derivative of sine is cosine
Step 2: Opposite Reciprocal
Taking the opposite reciprocal of the tangent line slopes gives us the normal line slope. The reciprocal of #sqrt(3)/2# is #2/sqrt(3)#. We'll have to rationalize this denominator - having a square root down there is a no-no:
#2/sqrt(3)*sqrt(3)/sqrt(3)=(2sqrt(3))/3#
To find the opposite of this result, we simply make it negative:
#(2sqrt(3))/3*-1=-(2sqrt(3))/3#
Step 3: Finding the Equation of the Line
We have the slope, but we need a #y#-intercept. Remember, a normal line has the form #y=mx+b#, where #x# and #y# are points on the line, #m# is the slope, and #b# is the #y#-intercept. We know the normal line passes through the tangent line - they're perpendicular, meaning they intersect at a #90^o# angle. And since we're finding the normal line at #x=pi/6#, we use that for our point:
#f(x)=sinx#
#f(pi/6)=sin(pi/6)=1/2#
We have now identified the point #(pi/6,1/2)#. We will now use this point to find the equation of the line:
#y=mx+b#
#1/2=-(2sqrt(3))/3*pi/6+b#
#1/2=-(pisqrt(3))/9+b#
#1/2+(pisqrt(3))/9=b->b=(9+2pisqrt(3))/18#
Bingo! The #y#-intercept is #(9+2pisqrt(3))/18#. That means the equation of the normal line is #y=-(2sqrt(3))/3x+(9+2pisqrt(3))/18#. We could combine this into one whole fraction, but we don't have to, and I like avoiding unnecessary work :). If you prefer hard numbers, we can approximate this to #y=-1.155x+1.105#.
