What is the equation of the tangent line of #f(x)=sin x + sin^2 x# at #x=0#?

This problem can be tackled in 3 steps:

1. Figure out the gradient of the line using the derivative of the function and the given #x# value.
2. Use the given #x# value and #f(x)# to find #C#.
3. Put values into the standard equation of a straight line.

The equation of a line #-> y = mx+C#

1. We'll begin by working out #m#, the gradient of the tangent. To do this we simply have to find #f'(x)#, the derivative.

#f(x) = sin(x) + sin^2(x)#
#f'(x) = cos(x) + 2cos(x)sin(x)#

Where the chain rule has been used on the #sin^2(x)# term.

Since we want to find the tangent at #x=0# then the gradient can be given by:
#f'(0)#
#=cos(0)+2cos(0)sin(0) = 1#

Thus #m=1#.
Step 1 is complete.

2. We must now find #C#.

Use #f(0)#
#=sin(0)+sin^2(0)=0#

Thus # {0,0} # Are the co -ordinates which the line intercepts #f(x)# so we know the line passes through #{0,0}#.

So, using #y = 0, x = 0 and m=1# we can substitute these into our equation of a straight line to obtain:

#y = mx + C#
#(0) = (1)(0) +C#
#C=0#

Thus we have our value for #C#. Step 2 is complete.

3. We can now write:

#y = mx+C#
#y = 1(x)+(0)# So the final equation is:
#y = x#

Indeed we see that if we plot both on the same graph we get a line intersection at a tangent exactly where we would expect: