**Step 1: Determine the corresponding y coordinate that the function and the tangent pass through**

This can be done by evaluating #x = a# inside the function, where here #a = pi/3#.

#f(pi/3) = sin(2(pi/3) + pi)#

#f(pi/3) = sin((2pi)/3 + pi)#

Evaluate using the sum and difference formulae:

#f(pi/3) = sin((2pi)/3)cospi + sinpicos((2pi)/3)#

#f(pi/3) = sqrt(3)/2 xx -1 + 0 xx -1/2#

#f(pi/3) = -sqrt(3)/2#

Note we could have also combined #(2pi)/3# and #pi# and then evaluated, as well.

So, the function, the tangent and the normal pass through the point #(pi/3, -sqrt(3)/2)#.

**Step 2: Determine the derivative**

We will use the chain rule to find the derivative of this function.

Let #f(x) = sin(u)# and #u= 2x + pi#.

The derivative of #f(x)# is #f'(x) = cos(u)# and the derivative of #u# is #u' = 2# (for #pi# is a constant, therefore its derivative is #0#).

The derivative of the entire function is then #f'(x) = cos(2x + pi) xx 2 =color(turquoise)( 2cos(2x + pi))#.

**Step 3: Determine the slope of the tangent**

The slope of the tangent is given by evaluating #x = a# inside the derivative. You probably find yourself asking at the moment: what does the the tangent have to do with the normal line? The answer: the tangent is perpendicular to the normal line. Or, in other words, the slope of the normal line is the negative reciprocal of that of the tangent.

#f'(pi/3) = 2cos(2(pi/3) + pi)#

This time I will use the combination inside the parentheses method.

#f'(pi/3) = 2cos((2pi)/3 + (3pi)/3)#

#f'(pi/3) = 2cos((5pi)/3)#

#f'(pi/3) = 2 xx 1/2#

#f'(pi/3) = 1#

Hence, the slope of the tangent is #1#. The negative reciprocal of #1# is #-1/1 = -1#.

#:.# The slope of the normal line is of #-1#.

**Step 4: Use point-slope form to determine the equation of the normal line**

We know the point that the function, tangent and normal pass through, and the slope of normal line. We also know that the normal is a line or a linear function. Hence, we can use point-slope form to determine the normal line's equation.

#y - y_1 = m(x - x_1)#

#y - (-sqrt(3)/2) = -1(x - pi/3)#

#y + sqrt(3)/2 = -x + pi/3#

#y = -x + pi/3 - sqrt(3)/2#

#y = -x + (2pi - 3sqrt(3))/6#

If you insist on an approximation, the equation of the line is #y = -x + 0.18#, but normally the exact value answer is better.

Since this is my #900^"th"# answer for Socratic, I have decided to include extra practice exercises. Make good use of them! You'll improve exponentially!

**Practice exercises:**

- Determine the equations of the tangents to the following relations at the given points:

a) #y = sqrt(x^3 + 9)#, at #x = -2#

b) #y = 2^(x - 3)# at #x = 5#

c) #y = x^5 + 3x^4 - 2x^3 - 8x^2 + 2x - 1# at #x = 1#

d) #y = sin(2x)# at #x = pi/4#

#2#. Determine the equations of the normal lines to the following relations at the given points:

a) #y = root(4)(1/3x^3 + 2x + 1)# at the point #x = 3#

b) #y = log_3(2x + 11)# at the point #x = 8#

c) #y = tan(4x + pi/2)# at the point #pi/8#

d) #(x - 3)^2 + (y - 5)^2 = 25# at the point #(6, 1)#

Hopefully this helps, and good luck!