What is the equation of the tangent line of f(x)=sin x + sin^2 xf(x)=sinx+sin2x at x=0x=0?

1 Answer
May 17, 2016

y=xy=x

Explanation:

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 xx value.
2. Use the given xx value and f(x)f(x) to find CC.
3. Put values into the standard equation of a straight line.

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

1. We'll begin by working out mm, 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:
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